On the Eight Levels theorem and applications towards Lucas-Lehmer primality test for Mersenne primes, I

Abstract

Lucas-Lehmer test is the current standard algorithm used for testing the primality of Mersenne numbers, but it may have limitations in terms of its efficiency and accuracy. Developing new algorithms or improving upon existing ones could potentially improve the search for Mersenne primes and the understanding of the distribution of Mersenne primes and composites. The development of new versions of the primality test for Mersenne numbers could help to speed up the search for new Mersenne primes by improving the efficiency of the algorithm. This could potentially lead to the discovery of new Mersenne primes that were previously beyond the reach of current computational resources. The current paper proves what the author called the Eight Levels Theorem and then highlights and proves three new different versions for Lucas-Lehmer primality test for Mersenne primes and also gives a new criterion for Mersenne compositeness.

Keywords:

• Mersenne primes
• Mersenne composites
• Lucas-Lehmer primality test
• the Eight Levels theorem
• developing new algorithms
• distribution of primes
• perfect numbers
• combinatorial identities

1. Introduction

Primes of special form have been of perennial interest (Guy, 2004). Among these, the primes of the form $$2^p-1$$ which are called Mersenne prime. It is outstanding in their simplicity.

• Clearly, if $2^p-1$ is prime then p is prime. The number $2^p-1$ is Mersenne composite if p is prime but $2^p-1$ is not prime. For example, the prime number $2^7-1=127$ is called Mersenne prime. However, the number $2^{11}-1=2047=23\times 89$ is called Mersenne composite (see Cambraia, Knapp, Lemos, Moriya, & Rodrigues, 2022; Ferradi & Xagawa, 2020; Jo & Kim, 2022; Kharaghani & Suda, 2020; Skula, 2019; Zheng & Yang, 1985).

• Mathematics is kept alive by the appearance of challenging unsolved problems. The current paper gives new expansions related to the following two major open questions in number theory: Are there infinitely many Mersenne primes? Are there infinitely many Mersenne composite? Many mathematicians believe that there are infinitely many Mersenne primes but a proof of this is still one of the major open problems in number theory (see Deza, 2021; Gallier, 2011; Guy, 2004; Effinger & Mullen, 2021; Kraft & Washington, 2018; Kundu & Mazumder, 2022).

• Mersenne primes have a close connection to perfect numbers, which are numbers that are equal to the sum of their proper divisors. It is known that Euclid and Euler proved that a number N is even perfect number if and only if $N=2^{p-1}(2^p-1)$ for some prime p, and $2^p-1$ is prime. Euclid proved only that this statement was sufficient. Euler, 2000 years later, proved that all even perfect numbers are of the form $2^{p-1}(2^p-1)$ where $2^p-1$ is a Mersenne prime (see Gallier, 2011; Wells, 1987). Thus the theorems of Euclid and Euler characterize all even perfect numbers, reducing their existence to that of Mersenne primes.

• The odd perfect numbers are quite a different story (see Dickson, 1919; Ochem & Rao, 2012; Wells, 1987). An odd perfect number is a hypothetical number that is both an odd integer and a perfect number, meaning that it is equal to the sum of its proper divisors. No odd perfect number has been discovered, and it is not known whether any exist. The search for odd perfect numbers has been ongoing for centuries and has involved many of the greatest mathematicians in history. It is still unknown whether there is any odd perfect number. Recently (Ochem & Rao, 2012), showed that odd perfect numbers, if exist, are greater than 10¹⁵⁰⁰

• There are practical reasons for seeking out bigger and bigger primes. Very big primes are crucial to the most widely used systems for encrypting data such as those that underpin all online banking and shopping. Based on the properties of Mersenne primes, instead of polynomial rings, Aggarwal, Joux, Prakash, and Santha described an elegant public-key encryption (see Aggarwal, Joux, Prakash, & Santha, 2018).

• The discovery of Mersenne primes has been an active area of research for centuries, and the search for larger and more complex prime numbers continues to this day. The discovery of new Mersenne primes is a significant achievement in the field of mathematics, as these numbers have interesting mathematical properties and are very useful in various applications, such as cryptography and computer science. Using the Lucas-Lehmer test, the computation was carried out at the UCLA Computing Facility, it was reported that $2^{4253}-1$ and $2^{4423}-1$ are primes (Hurwitz, 1962). Discovering Mersenne primes is a challenging and time-consuming task, and the discovery of new Mersenne primes often represents a significant achievement in the field of mathematics. The Great Internet Mersenne Prime Search (GIMPS), which is a collaborative project aimed at discovering new Mersenne primes, has discovered the largest known prime number

$$2^{82,589,933}-1,$$ having 24,862,048 digits (see Deza, 2021; GIMPS, 2016). The GIMPS project has been instrumental in discovering the majority of the known Mersenne primes, and it continues to search for new ones. GIMPS is a fascinating example of how the power of distributed computing can be used to advance mathematical knowledge and uncover new discoveries. Each new discovery sets a new record for the largest known prime number. Despite its success in discovering large prime numbers, the Great Internet Mersenne Prime Search (GIMPS) has some limitations. It’s worth noting that Mersenne primes are rare and become increasingly difficult to find as their size increases. Surprisingly, as of March 2023, only 51 Mersenne primes are known. While 51 Mersenne primes are currently known, it is not known whether there are any more beyond those that have been discovered. The question of whether there are infinitely many Mersenne primes is one of the major unsolved problems in mathematics. Whether the list of known Mersenne primes is finite or infinite is an open question in mathematics, and it remains an area of active research (see Deza, 2021; GIMPS, 2016; Kundu & Mazumder, 2022).

1.1. Lucas-Lehmer primality test

Mersenne primes have some special properties that make them useful in certain types of computations, particularly in the field of cryptography and computing. One of these properties is that they are easy to test for primality using the Lucas-Lehmer primality test. This makes it easier to find large prime numbers, which are important for secure cryptographic systems. The Lucas-Lehmer primality test is a primality test specifically designed for testing the primality of Mersenne numbers (i.e. numbers of the form $2^p-1$). It is a fast and efficient test that can quickly determine whether a given Mersenne number is prime or composite. It is well-known the following theorems (see Deza, 2021; Gallier, 2011; Kraft & Washington, 2018):

Theorem 1.

(Lucas-Lehmer primality test)

$2^p-1$ is Mersenne prime if and only if (1) $$2n-1\hspace{1em}|\hspace{1em}{(1+\sqrt{3})}^n+{(1-\sqrt{3})}^n$$(1) where $n:=2^{p-1}.$

1.2. Euclid-Euler-Lucas-Lehmer association

Perfect numbers have fascinated mathematicians for centuries, and many properties and patterns have been discovered about them. One of the most famous results about perfect numbers is that every even perfect number can be written in the form $$2^{p-1}(2^p-1),$$ where $2^p-1$ is a Mersenne prime. Moreover, every Mersenne prime of the form $2^p-1$ gives rise to an even perfect number through this formula. The following theorem tells us that there is a strong association between perfect numbers and Mersenne primes.

Theorem 2.

(Euclid-Euler-Lucas-Lehmer)

A number N is even perfect number if and only if $N=2^{p-1}(2^p-1)$ for some prime p, and (2) $$2n-1\hspace{1em}|\hspace{1em}{(1+\sqrt{3})}^n+{(1-\sqrt{3})}^n$$(2) where $n:=2^{p-1}.$

For example, the first few even perfect numbers are: (3) $$\begin{array}{c}6=2^{2-1}(2^2-1),\hfill \\ 28=2^{3-1}(2^3-1),\hfill \\ 496=2^{5-1}(2^5-1),\hfill \\ 8128=2^{7-1}(2^7-1),\hfill \end{array}\mathrm{ }\mathrm{ }$$(3) and the corresponding Mersenne primes are: (4) $$\begin{array}{c}{2}^2-1=3,\hfill \\ 2^3-1=7,\hfill \\ 2^5-1=31,\hfill \\ 2^7-1=127.\hfill \end{array}$$(4)

1.3. Notations

For a natural number n, we define $\delta (n)=n\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}2).$ For an arbitrary real number x, $\lfloor \frac{x}{2}\rfloor$ is the highest integer less than or equal $\frac{x}{2}.$ We also need the following notation $$\prod _{\lambda =0}^{-1}(n^2-(4\lambda {)}^2):=1.$$

For simplicity, we sometimes write $\prod _{\lambda }{n}^2-{(4\lambda )}^2$ for $\prod _{\lambda }[n^2-{(4\lambda )}^2]$.

2. The purpose of the current paper

The aim of this paper is to study some arithmetical properties of the coefficients of the expansion (5) $$x^n+y^n=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k,$$(5) for $n\equiv 0,\hspace{0.5em}2,\hspace{0.5em}4,\hspace{0.5em}6\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8).$ Then we study the expansion (6) $$\frac{x^n+y^n}{x+y}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k,$$(6) for $n\equiv 1,\hspace{0.5em}3,\hspace{0.5em}5,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8).$

2.1. Reasons for why the expansions of 5 and 6 are interesting

We show in the current paper that the numbers of the form (7) $$\prod _{\lambda }^{}(n^2-(4\lambda {)}^2)$$(7) arise up naturally in the coefficients of the expansions (5, 6), and enjoy some unexpected new interesting arithmetical properties which could be helpful in the study of Mersenne primes and Mersenne composites.

2.2. New results

We state and prove the following new results in order:

1. The Eight Levels theorem

2. The first new version for Lucas-Lehmer primality test

3. The second new version for Lucas-Lehmer primality test

4. The third new version for Lucas-Lehmer primality test

5. New version for Euclid-Euler-Lucas-Lehmer association

6. New criteria for compositeness of Mersenne numbers

7. New combinatorial identities

We end the paper with a discussion about further potential investigations of how this new versions for Lucas-Lehmer primality test may provide a better theoretical understanding of the two major questions about Mersenne numbers; whether there are infinitely many Mersenne primes or Mersenne composites.

3. Summary for the main new results of the paper

This paper proves the following new theorems.

3.1. The first new version for Lucas-Lehmer primality test

The following result proposes a new double-indexed recurrence relation for the Lucas-Lehmer test.

Theorem 3.

Given prime $p\ge 5.$ $2^p-1$ is prime if and only if (8) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(8) where $n:=2^{p-1},{\varphi }_k(n)$ are defined by the double index recurrence relation $${\varphi }_k(m)=4 {\varphi }_{k-1}(m-2)-{\varphi }_k(m-4)$$ and the initial boundary values satisfy (9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9)

3.2. The second new version for Lucas-Lehmer primality test

The following result proposes a new explicit sum of products of difference of squares for the Lucas-Lehmer test.

Theorem 4.

Given prime $p\ge 5,n:=2^{p-1}$. The number $2^p-1$ is prime if and only if (10) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}{(-1)}^{\lfloor \frac{k}{2}\rfloor } \frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{k!}.$$(10)

3.3. The third new version for Lucas-Lehmer primality test

The following result proposes a new nonlinear recurrence relation for the Lucas-Lehmer test.

Theorem 5.

Given prime $p\ge 5$. The number $2^p-1$ is prime if and only if (11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) where $n:=2^{p-1},{\varphi }_k(n)$ are generated by the double index recurrence relation $$\frac{{\varphi }_k(n)}{{\varphi }_{k-2}(n)}=\frac{{(2k-4)}^2-n^2}{k(k-1)},$$ and we can choose either of the following initial values to generate ${\varphi }_k(n)$ from the starting term ${\varphi }_0(n)$ or the last term ${\varphi }_{\lfloor \frac{n}{2}\rfloor }(n):$ (12) $${\varphi }_0(n)=+2\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)=2^n.$$(12)

3.4. New version for Euclid-Euler-Lucas-Lehmer association

Theorem 6.

A number N is even perfect number if and only if $N=2^{p-1}(2^p-1)$ for some prime p, and (13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) where $n:=2^{p-1}.$

3.5. New criteria for compositeness of Mersenne numbers

Theorem 7.

(Criteria for compositeness of Mersenne numbers)

Given prime $p\ge 5$. The number $2n-1=2^p-1$ is Mersenne composite number if (14) $$2\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}1\hspace{1em}\nmid \hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}\left[{(4\lambda )}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{4}^{-1}\right]\hspace{1em}}{k!}.$$(14)

3.6. New combinatorial identities

Theorem 8.

(combinatorial identities)

For any natural number n, the following combinatorial identities are correct $$\begin{array}{ccc}\underset{¯}{n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}4\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\mathrm{\hspace{0.17em}}\left(\lfloor \frac{n}{2}\rfloor \right)!=2\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}\prod _{\lambda =0}^{\lfloor \frac{n-4}{4}\rfloor }\left[n^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda )}^2\right]\hfill \\ \underset{¯}{n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}4\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\mathrm{\hspace{0.17em}}\left(\lfloor \frac{n}{2}\rfloor \right)!=\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}\prod _{\lambda =1}^{\lfloor \frac{n-1}{4}\rfloor }\left[{(n+1)}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda -2)}^2\right]\hfill \\ \underset{¯}{n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}4\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\mathrm{\hspace{0.17em}}\left(\lfloor \frac{n}{2}\rfloor \right)!=2\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\prod _{\lambda =1}^{\lfloor \frac{n-2}{4}\rfloor }\left[n^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda -2)}^2\right]\hfill \\ \underset{¯}{n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}4\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\mathrm{\hspace{0.17em}}\left(\lfloor \frac{n}{2}\rfloor \right)!=\mathrm{\hspace{0.17em}}(n+1)\mathrm{\hspace{0.17em}}\prod _{\lambda =1}^{\lfloor \frac{n-3}{4}\rfloor }\left[{(n+1)}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda )}^2\right]\hfill \end{array}$$

4. Discussion for the proposed method’s theoretical analysis

4.1. Algebraically independent polynomials

Algebraically independent polynomials are important in several areas of mathematics and its applications, including algebraic geometry, commutative algebra, and number theory. Algebraic independent polynomials are important because they provide a way to study and understand the relationships between different algebraic objects, such as numbers, functions, and algebraic structures (Perron, 1932). Two polynomials in two variables, say h(x, y) and g(x, y), are said to be algebraically independent over the field of rational numbers if there exist a polynomial f with rational coefficients such that $$f(h(x,y),g(x,y))=0,$$ then this entails that f = 0. The polynomials xy and $x^2+y^2$ are algebraically independent over the field of rational numbers. Therefore, if we have the following identity, with integer coefficients ${\mu }_k(n),$ (15) $$0=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\mu }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k,$$(15) then this entails that all the coefficients vanish; which means ${\mu }_k(n)=0.$

4.2. Symmetric polynomials

Symmetric polynomials are polynomials that remain unchanged under the permutation of their variables. For example, the polynomial $x^2+y^2+z^2$ is symmetric because it remains the same if we swap the variables x, y, and z. For any natural number n, $x^n+y^n$ is symmetric polynomial. Then we have integer coefficients $f_k(n),$ from the fundamental theorem of symmetric polynomials (Perron, 1932), that satisfy (16) $$x^n+y^n=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{f}_k(n){(x+y)}^{n-2k}{(x^2+y^2)}^k.$$(16)

Dividing by ${(x+y)}^{\delta (n)},$ we get (17) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{f}_k(n){(2xy+x^2+y^2)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k.$$(17)

Hence, we get the integer sequence ${\Psi }_k(n)$ that satisfy (18) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k.$$(18)

5. The Eight Levels theorem

Now we prove what we call the Eight Levels Theorem for the expansion of the polynomial $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}$$ in terms of the symmetric polynomials xy and $x^2+y^2.$ Then we investigate some properties for the coefficients of the expansion (19) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k.$$(19)

Then we study some applications and prove the results of the summary one by one.

5.1. The statement of the Eight Levels theorem

Theorem 9.

(The Eight Levels Theorem)

For any complex numbers x, y, any non negative integers n, k, the coefficients ${\Psi }_k(n)$ of the expansion (20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) are integers and (21) $${\Psi }_0(n)=\{\begin{array}{cc}+2\hfill & n\equiv \hspace{0.5em}0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv \pm 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & n\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv \pm 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & n\equiv \pm 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}$$(21) and, for each $1\le k\le \lfloor \frac{n}{2}\rfloor$, the coefficients satisfy the following statements

• For $n\equiv 0,\hspace{0.5em}2,\hspace{0.5em}4,\hspace{0.5em}6\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$:

  $\underset{¯}{n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\Psi }_k(n)=\{\begin{array}{cc}0\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{odd}\hfill \\ 2\mathrm{\hspace{0.17em}}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}\left[n^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda )}^2\right]}{4^k\mathrm{\hspace{0.17em}}k!}\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{even}\hfill \end{array}$$

$\underset{¯}{n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\Psi }_k(n)=\{\begin{array}{cc}0\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{even}\hfill \\ 2\mathrm{\hspace{0.17em}}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }\left[n^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda -2)}^2\right]}{4^k\mathrm{\hspace{0.17em}}k!}\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{odd}\hfill \end{array}$$

$\underset{¯}{n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\Psi }_k(n)=\{\begin{array}{cc}0\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{odd}\hfill \\ 2\mathrm{\hspace{0.17em}}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}\left[n^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda )}^2\right]}{4^k\mathrm{\hspace{0.17em}}k!}\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{even}\hfill \end{array}$$

$\underset{¯}{n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\Psi }_k(n)=\{\begin{array}{cc}0\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{even}\hfill \\ 2\mathrm{\hspace{0.17em}}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }\left[n^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda -2)}^2\right]}{4^k\mathrm{\hspace{0.17em}}k!}\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{odd}\hfill \end{array}$$

• For $n\equiv 1,\hspace{0.5em}3,\hspace{0.5em}5,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$:

  $\underset{¯}{n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$

$${\Psi }_k(n)={(-1)}^{\lfloor \frac{k}{2}\rfloor }{(n+1-2k)}^{\delta (k)}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }\left[{(n+1)}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda -2)}^2\right]}{4^k\mathrm{\hspace{0.17em}}k!}$$

$\underset{¯}{n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\Psi }_k(n)={(-1)}^{\lfloor \frac{k}{2}\rfloor +\delta (k-1)}(n+1){(n+1-2k)}^{\delta (k-1)}\frac{\prod _{\lambda =1}^{\lfloor \frac{k-1}{2}\rfloor }[{(n+1)}^2-{(4\lambda )}^2]}{4^k\mathrm{\hspace{0.17em}}k!}$$

$\underset{¯}{n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\Psi }_k(n)={(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\mathrm{\hspace{0.17em}}{(n+1-2k)}^{\delta (k)}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }\left[{(n+1)}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda -2)}^2\right]}{4^k\mathrm{\hspace{0.17em}}k!}$$

$\underset{¯}{n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\Psi }_k(n)={(-1)}^{\lfloor \frac{k}{2}\rfloor +\delta (k)}(n+1){(n+1-2k)}^{\delta (k-1)}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k-1}{2}\rfloor }[{(n+1)}^2-{(4\lambda )}^2]}{4^k\mathrm{\hspace{0.17em}}k!}\mathrm{ }\mathrm{ }$$

Proof.

Put $(x,y)=(1,\sqrt{-1})$ in (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) , we immediately get (21)(21) $${\Psi }_0(n)=\{\begin{array}{cc}+2\hfill & n\equiv \hspace{0.5em}0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv \pm 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & n\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv \pm 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & n\equiv \pm 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}$$(21) . To prove the statements of Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we need to prove the following lemmas.

Lemma 10.

For each natural number n, $1\le k\le \lfloor \frac{n}{2}\rfloor$, the coefficients ${\Psi }_k(n)$ of the expansion of (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) are integers, unique and satisfy (22) $${\Psi }_k(n)={\Psi }_{k-1}(n-2)-{\Psi }_k(n-4).$$(22)

Proof.

From the fundamental theorem on symmetric polynomials (Ibrahim, 2004; Perron, 1932), we have a sequence of integers ${\Psi }_k(n)$ satisfy the expansion of (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) . From the algebraic independence of $xy,x^2+y^2,$ the coefficients ${\Psi }_k(n)$ of the expansion of (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) are unique. Now, multiply (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) by $xy(x^2+y^2),$ and noting ${(x+y)}^{\delta (n+2)}={(x+y)}^{\delta (n)}={(x+y)}^{\delta (n-2)}$ and $\lfloor \frac{n+2}{2}\rfloor =\lfloor \frac{n}{2}\rfloor +1$ and $\lfloor \frac{n-2}{2}\rfloor =\lfloor \frac{n}{2}\rfloor -1,$ we get (23) $$\begin{array}{c}\sum _{k=1}^{\lfloor \frac{n}{2}\rfloor +1}{\Psi }_{k-1}(n){(xy)}^{\lfloor \frac{n}{2}\rfloor +2-k}{(x^2+y^2)}^k=\\ \sum _{k=0}^{\lfloor \frac{n}{2}\rfloor +1}{\Psi }_k(n+2){(xy)}^{\lfloor \frac{n}{2}\rfloor +2-k}{(x^2+y^2)}^k+\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor -1}{\Psi }_k(n-2){(xy)}^{\lfloor \frac{n}{2}\rfloor +2-k}{(x^2+y^2)}^k\end{array}$$(23)

Again from the algebraic independence of $xy,x^2+y^2,$ and from (23)(23) $$\begin{array}{c}\sum _{k=1}^{\lfloor \frac{n}{2}\rfloor +1}{\Psi }_{k-1}(n){(xy)}^{\lfloor \frac{n}{2}\rfloor +2-k}{(x^2+y^2)}^k=\\ \sum _{k=0}^{\lfloor \frac{n}{2}\rfloor +1}{\Psi }_k(n+2){(xy)}^{\lfloor \frac{n}{2}\rfloor +2-k}{(x^2+y^2)}^k+\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor -1}{\Psi }_k(n-2){(xy)}^{\lfloor \frac{n}{2}\rfloor +2-k}{(x^2+y^2)}^k\end{array}$$(23) , we get the following identity for any natural number n: (24) $${\Psi }_k(n+2)={\Psi }_{k-1}(n)-{\Psi }_k(n-2).$$(24)

Replace n by n − 2 in (24)(24) $${\Psi }_k(n+2)={\Psi }_{k-1}(n)-{\Psi }_k(n-2).$$(24) , we get (22)(22) $${\Psi }_k(n)={\Psi }_{k-1}(n-2)-{\Psi }_k(n-4).$$(22) . □

Lemma 11.

For every even natural number n, $0\le k\le \lfloor \frac{n}{2}\rfloor$, the following statements are true for each case: (25) $$\begin{array}{ccc}\underset{¯}{n\equiv 0\hspace{0.5em}\text{or}\hspace{0.5em}4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}& \hspace{1em}& {\Psi }_k(n)=0\hspace{1em}\text{for}\hspace{0.5em}k\hspace{0.5em}\text{odd},\\ \underset{¯}{n\equiv 2 \text{or} 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}& \hspace{1em}& {\Psi }_k(n)=0\hspace{1em}\text{for}\hspace{0.5em}k\hspace{0.5em}\text{even}.\end{array}$$(25)

Proof.

Consider n even natural number, and replace $\delta (n)=0$ in (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) to get (26) $$x^n+y^n=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(26)

Then replace x by—x in (26)(26) $$x^n+y^n=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(26) , we get (27) $$x^n+y^n=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(-xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(27)

From the algebraic independence of $xy,x^2+y^2,$ and from (26, 27) we get the proof. □

From (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) , we get the following initial values for ${\Psi }_k(n).$

Lemma 12.

$$\begin{array}{cccc}{\Psi }_0(0)=+2,\hfill & {\Psi }_1(0)=0,\hfill & {\Psi }_2(0)=0,\hfill & {\Psi }_3(0)=0,\hfill \\ {\Psi }_0(2)=0,\hfill & {\Psi }_1(2)=+1,\hfill & {\Psi }_2(2)=0,\hfill & {\Psi }_3(2)=0,\hfill \\ {\Psi }_0(4)=-2,\hfill & {\Psi }_1(4)=0,\hfill & {\Psi }_2(4)=+1,\hfill & {\Psi }_3(4)=0,\hfill \\ {\Psi }_0(6)=0,\hfill & {\Psi }_1(6)=-3,\hfill & {\Psi }_2(6)=0,\hfill & {\Psi }_3(6)=+1.\hfill \end{array}\mathrm{ }\mathrm{ }$$

Lemma 13.

For any odd natural number n, the coefficients ${\Psi }_k(n)$ of the expansion of (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) satisfy the following property (28) $${\Psi }_k(n)=\frac{2(k+1)}{(n+1)}{\Psi }_{k+1}(n+1)+\frac{(\lfloor \frac{n+1}{2}\rfloor -k)}{(n+1)}{\Psi }_k(n+1).$$(28)

Proof.

For n odd, n + 1 is even. Then from the expansion of (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) we get the following (29) $$x^{n+1}+y^{n+1}=\sum _{k=0}^{\lfloor \frac{n+1}{2}\rfloor }{\Psi }_k(n+1){(xy)}^{\lfloor \frac{n+1}{2}\rfloor -k}{(x^2+y^2)}^k.$$(29)

Now acting the differential operator $(\frac{\partial }{\partial x}+\frac{\partial }{\partial y})$ on (29)(29) $$x^{n+1}+y^{n+1}=\sum _{k=0}^{\lfloor \frac{n+1}{2}\rfloor }{\Psi }_k(n+1){(xy)}^{\lfloor \frac{n+1}{2}\rfloor -k}{(x^2+y^2)}^k.$$(29) and noting that $$\begin{array}{c}(\frac{\partial }{\partial x}+\frac{\partial }{\partial y})(x^{n+1}+y^{n+1})=(n+1)(x^n+y^n),\hfill \\ (\frac{\partial }{\partial x}+\frac{\partial }{\partial y})xy=x+y,\hfill \\ (\frac{\partial }{\partial x}+\frac{\partial }{\partial y})(x^2+y^2)=2(x+y),\hfill \end{array}$$ and equating the coefficients, we get the proof. □

5.2. The proof of Theorem (9) for $n\equiv 0,\hspace{0.5em}2,\hspace{0.5em}4,\hspace{0.5em}6(\mathrm{mod\hspace{0.5em}}8)$

Now in this section we prove Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) for n even. The values of ${\Psi }_k(n)$ that comes from the formulas of Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) for $n=0,,2,4,8$ are identical with the correct values that come from Lemma (12)(12) $${\varphi }_0(n)=+2\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)=2^n.$$(12) . So, Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) is correct for the initial vales $n=0,2,4,6.$ Now we assume the validity of Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) for each $0\le m<n,$ with $m\equiv 0,\hspace{0.5em}2,\hspace{0.5em}4,\hspace{0.5em}6\left(\mathrm{mod}8\right)$ and need to prove that Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) for n. Lemma (11)(11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) proves the validity of Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) for $n\equiv 0 \text{or} 4\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ if k odd, and for $n\equiv 2 \text{or} 6\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ if k even. Therefore it remains to prove the validity of the following cases:

• $n\equiv 0\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),k\hspace{0.5em}\text{even}$

• $n\equiv 2\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),k\hspace{0.5em}\text{odd}$

• $n\equiv 4\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),k\hspace{0.5em}\text{even}$

• $n\equiv 6\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),k\hspace{0.5em}\text{odd}$

We prove these cases one by one as following.

Consider $n\equiv 0\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{even}$

Proof.

In this case, $n-2\equiv 6\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and $n-4\equiv 4\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and from Lemma (10)(10) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}{(-1)}^{\lfloor \frac{k}{2}\rfloor } \frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{k!}.$$(10) , we get $$\begin{array}{cc}{\Psi }_k(n)\hfill & ={\Psi }_{k-1}(n-2)-{\Psi }_k(n-4)\hfill \\ \hfill & =2\hspace{0.17em}{(-1)}^{\lfloor \frac{k-1}{2}\rfloor +1}\hspace{0.17em}(n-2)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k-1}{2}\rfloor }{(n-2)}^2-{(4\lambda -2)}^2}{4^{k-1}\hspace{0.17em}(k-1)!}-2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{(n-4)}^2-{(4\lambda )}^2}{4^k\hspace{0.17em}k!}\hfill \\ \hfill & =2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}(n-2)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda )\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda -4)}{4^{k-1}\hspace{0.17em}(k-1)!}+2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda -4)\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda -4)}{4^{k-1}\hspace{0.17em}(k-1)!}\hfill \\ \hfill & =2\hspace{0.17em}\frac{{(-1)}^{\lfloor \frac{k}{2}\rfloor }}{4^k\hspace{0.17em}k!}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda )\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -2}(n+4\lambda )\hspace{0.17em}[4k(n-2)+(n-4\lfloor \frac{k}{2}\rfloor )(n-4)]\hfill \\ \hfill & =2\hspace{0.17em}\frac{{(-1)}^{\lfloor \frac{k}{2}\rfloor }}{4^k\hspace{0.17em}k!}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda )\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -2}(n+4\lambda )\hspace{1em}n\hspace{0.17em}(n+4(\lfloor \frac{k}{2}\rfloor -1))\hfill \\ \hfill & =2\hspace{0.17em}\frac{{(-1)}^{\lfloor \frac{k}{2}\rfloor }}{4^k\hspace{0.17em}k!}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda )\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda )=2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 2\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{odd}$

Proof.

In this case, $n-2\equiv 0\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and $n-4\equiv 6\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and from Lemma (10)(10) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}{(-1)}^{\lfloor \frac{k}{2}\rfloor } \frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{k!}.$$(10) , we get $$\begin{array}{cc}{\Psi }_k(n)\hfill & ={\Psi }_{k-1}(n-2)-{\Psi }_k(n-4)\hfill \\ \hfill & =2\hspace{0.17em}{(-1)}^{\lfloor \frac{k-1}{2}\rfloor }\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k-1}{2}\rfloor -1}{(n-2)}^2-{(4\lambda )}^2}{4^{k-1}\hspace{0.17em}(k-1)!}-2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}(n-4)\frac{\prod _{\lambda =1}^{\lfloor \frac{k-1}{2}\rfloor }{(n-4)}^2-{(4\lambda -2)}^2}{4^k\hspace{0.17em}k!}\hfill \\ \hfill & =2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda -2)\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda -2)}{4^{k-1}\hspace{0.17em}(k-1)!}+2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}(n-4)\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }(n-4\lambda -2)\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }(n+4\lambda -6)}{4^k\hspace{0.17em}k!}\hfill \\ \hfill & =2\hspace{0.17em}\frac{{(-1)}^{\lfloor \frac{k}{2}\rfloor }}{4^k\hspace{0.17em}k!}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda -2)\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda -2)[4k(n-2)+(n-4)(n-4\left(\lfloor \frac{k}{2}\rfloor \right)-2)].\hfill \end{array}\mathrm{ }\mathrm{ }$$

As $$4k(n-2)+(n-4)(n-4(\lfloor \frac{k}{2}\rfloor )-2)=n(n+2k-4),$$ we get the following $$\begin{array}{cc}{\Psi }_k(n)\hfill & =2\hspace{0.17em}\frac{{(-1)}^{\lfloor \frac{k}{2}\rfloor }}{4^k\hspace{0.17em}k!}\hspace{0.17em}n\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }(n-4\lambda +2)\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }(n+4\lambda -2)\hfill \\ \hfill & =2\hspace{0.17em}n\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}\hspace{1em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }{n}^2-{(4\lambda -2)}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 4\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{even}$

Proof.

In this case, $n-2\equiv 2\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and $n-4\equiv 0\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and from Lemma (10)(10) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}{(-1)}^{\lfloor \frac{k}{2}\rfloor } \frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{k!}.$$(10) , we get $$\begin{array}{cc}{\Psi }_k(n)\hfill & ={\Psi }_{k-1}(n-2)-{\Psi }_k(n-4)\hfill \\ \hfill & =2\hspace{0.17em}{(-1)}^{\lfloor \frac{k-1}{2}\rfloor }\hspace{0.17em}(n-2)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k-2}{2}\rfloor }{(n-2)}^2-{(4\lambda -2)}^2}{4^{k-1}\hspace{0.17em}(k-1)!}-2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{(n-4)}^2-{(4\lambda )}^2}{4^k\hspace{0.17em}k!}\hfill \\ \hfill & =2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}(n-2)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda )\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda -4)}{4^{k-1}\hspace{0.17em}(k-1)!}+2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda -4)\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda -4)}{4^k\hspace{0.17em}k!}\hfill \end{array}\mathrm{ }\mathrm{ }$$

Hence $$\begin{array}{cc}{\Psi }_k(n)\hfill & =2\hspace{0.17em}\frac{{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}}{4^k\hspace{0.17em}k!}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda )\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -2}(n+4\lambda )\hspace{1em}n\hspace{0.17em}(n+4(\lfloor \frac{k}{2}\rfloor -1))\hfill \\ \hfill & =2\hspace{0.17em}\frac{{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}}{4^k\hspace{0.17em}k!}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda )\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda )=2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 6\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{odd}$

Proof.

In this case, $n-2\equiv 4\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and $n-4\equiv 2\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and from Lemma (10)(10) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}{(-1)}^{\lfloor \frac{k}{2}\rfloor } \frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{k!}.$$(10) , we get $$\begin{array}{cc}{\Psi }_k(n)\hfill & ={\Psi }_{k-1}(n-2)-{\Psi }_k(n-4)\hfill \\ \hfill & =2\hspace{0.17em}{(-1)}^{\lfloor \frac{k-1}{2}\rfloor +1}\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k-1}{2}\rfloor -1}{(n-2)}^2-{(4\lambda )}^2}{4^{k-1}\hspace{0.17em}(k-1)!}-2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}(n-4)\frac{\prod _{\lambda =1}^{\lfloor \frac{k-1}{2}\rfloor }{(n-4)}^2-{(4\lambda -2)}^2}{4^k\hspace{0.17em}k!}\hfill \\ \hfill & =2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n-4\lambda -2)\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}(n+4\lambda -2)}{4^{k-1}\hspace{0.17em}(k-1)!}+2\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}(n-4)\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }(n-4\lambda -2)\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }(n+4\lambda -6)}{4^k\hspace{0.17em}k!}\hfill \end{array}\mathrm{ }\mathrm{ }$$

Hence $$\begin{array}{cc}{\Psi }_k(n)\hfill & =2\hspace{0.17em}\frac{{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}}{4^k\hspace{0.17em}k!}\hspace{0.17em}n\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }(n-4\lambda +2)\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }(n+4\lambda -2)\hfill \\ \hfill & =2\hspace{0.17em}n\hspace{0.17em}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}\hspace{1em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }{n}^2-{(4\lambda -2)}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

5.3. The proof of Theorem (9) for $n\equiv 1,\hspace{0.5em}3,\hspace{0.5em}5,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$

With the help of Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) and Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) for the even case that we already proved, together with Lemma (11)(11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) , we prove Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) for the odd cases $n\equiv 1,\hspace{0.5em}3,\hspace{0.5em}5,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ one by one, for each parity for k.

Consider $n\equiv 1\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{odd}$

Proof.

In this case, $n+1\equiv 2\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and from Lemma (11)(11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) , we get ${\Psi }_{k+1}(n+1)=0.$ Hence, from Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) , and from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following relation $$\begin{array}{c}{\Psi }_k(n)=\frac{(\lfloor \frac{n+1}{2}\rfloor -k)}{(n+1)}{\Psi }_k(n+1)={(-1)}^{\lfloor \frac{k}{2}\rfloor }(n+1-2k)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }{(n+1)}^2-{(4\lambda -2)}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 1\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{even}$

Proof.

In this case, $n+1\equiv 2\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and from Lemma (11)(11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) , we get ${\Psi }_k(n+1)=0.$ From Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) , and from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following relation $$\begin{array}{c}{\Psi }_k(n)=2\frac{(k+1)}{(n+1)}{\Psi }_{k+1}(n+1)={(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }{(n+1)}^2-{(4\lambda -2)}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 3\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{odd}$

Proof.

In this case, from Lemma (11)(11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) , we get ${\Psi }_k(n+1)=0.$ Hence, from Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) , and from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following relation $$\begin{array}{c}{\Psi }_k(n)=2\frac{(k+1)}{(n+1)}{\Psi }_{k+1}(n+1)={(-1)}^{\lfloor \frac{k}{2}\rfloor }(n+1)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }{(n+1)}^2-{(4\lambda )}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 3\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{even}$

Proof.

In this case, ${\Psi }_{k+1}(n+1)=0.$ From Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) , and from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following relation $$\begin{array}{c}{\Psi }_k(n)=\frac{(\lfloor \frac{n+1}{2}\rfloor -k)}{(n+1)}{\Psi }_k(n+1)={(-1)}^{\lfloor \frac{k}{2}\rfloor +1}(n+1)(n+1-2k)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}{(n+1)}^2-{(4\lambda )}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 5\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{odd}$

Proof.

In this case, from Lemma (11)(11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) , we get ${\Psi }_{k+1}(n+1)=0.$ Hence, from Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) , and from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following relation $$\begin{array}{c}{\Psi }_k(n)=\frac{(\lfloor \frac{n+1}{2}\rfloor -k)}{(n+1)}{\Psi }_k(n+1)={(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}(n+1-2k)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }{(n+1)}^2-{(4\lambda -2)}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 5\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{even}$

Proof.

In this case, from Lemma (11)(11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) , we get ${\Psi }_k(n+1)=0.$ From Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) , and from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following relation $$\begin{array}{c}{\Psi }_k(n)=2\frac{(k+1)}{(n+1)}{\Psi }_{k+1}(n+1)={(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }{(n+1)}^2-{(4\lambda -2)}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{odd}$

Proof.

In this case, from Lemma (11)(11) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.17em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(11) , we get ${\Psi }_k(n+1)=0.$ Hence, from Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) , and from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following relation $$\begin{array}{cc}{\Psi }_k(n)\hfill & =2\frac{(k+1)}{(n+1)}{\Psi }_{k+1}(n+1)\hfill \\ \hfill & ={(-1)}^{\lfloor \frac{k}{2}\rfloor +1}(n+1)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }{(n+1)}^2-{(4\lambda )}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

Consider $n\equiv 7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)\hspace{0.5em}\text{and}\hspace{0.5em}k\hspace{0.5em}\text{even}$

Proof.

In this case, ${\Psi }_{k+1}(n+1)=0.$ From Lemma (13)(13) $$2\mathrm{\hspace{0.17em}}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{(-1)}^{\lfloor \frac{k}{2}\rfloor }\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}$$(13) , and from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following relation $$\begin{array}{cc}{\Psi }_k(n)\hfill & =\frac{(\lfloor \frac{n+1}{2}\rfloor -k)}{(n+1)}{\Psi }_k(n+1)\hfill \\ \hfill & ={(-1)}^{\lfloor \frac{k}{2}\rfloor }(n+1)(n+1-2k)\hspace{0.17em}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -1}{(n+1)}^2-{(4\lambda )}^2}{4^k\hspace{0.17em}k!}.\hfill \end{array}\mathrm{ }\mathrm{ }$$

□

This completes the proof of Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) . □

6. General characteristics for $\Psi -$ sequence

6.1. Examples for $\Psi -$ sequence

From Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we list some examples that show the splendour of the natural factorization of ${\Psi }_k(n)$ for $k=0,1,2,3,4,5,6,7$: (30) $$\begin{array}{c}{\Psi }_0(n)=\{\begin{array}{cc}+2\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{0.5em}0\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{0.5em}0\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\Psi }_1(n)=\{\begin{array}{cc}\hspace{1em}0\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-1)}{4}\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +2\frac{n}{4}\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n+1)}{4}\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-1)}{4}\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\frac{n}{4}\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n+1)}{4}\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\hfill \end{array}\mathrm{ }\mathrm{ }$$(30) (31) $$\begin{array}{c}{\Psi }_2(n)=\{\begin{array}{cc}-2\frac{n^2}{4^{2}2!}\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-1)(n+3)}{4^{2}2!}\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-3)(n+1)}{4^{2}2!}\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +2\frac{n^2}{4^{2}2!}\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-1)(n+3)}{4^{2}2!}\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-3)(n+1)}{4^{2}2!}\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},{\Psi }_3(n)=\{\begin{array}{cc}\hspace{1em}0\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-5)(n-1)(n+3)}{4^{3}3!}\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\frac{(n-2)n(n+2)}{4^{3}3!}\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-3)(n+1)(n+5)}{4^{3}3!}\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-5)(n-1)(n+3)}{4^{3}3!}\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +2\frac{(n-2)n(n+2)}{4^{3}3!}\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-3)(n+1)(n+5)}{4^{3}3!}\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\hfill \end{array}\mathrm{ }\mathrm{ }$$(31) (32) $${\Psi }_4(n)=\{\begin{array}{cc}+2\frac{(n-4)n^2(n+4)}{4^{4}4!}\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-5)(n-1)(n+3)(n+7)}{4^{4}4!}\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-7)(n-3)(n+1)(n+5)}{4^{4}4!}\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\frac{(n-4)n^2(n+4)}{4^{4}4!}\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-5)(n-1)(n+3)(n+7)}{4^{4}4!}\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-7)(n-3)(n+1)(n+5)}{4^{4}4!}\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},$$(32) and (33) $${\Psi }_5(n)=\{\begin{array}{cc}\hspace{1em}0\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-9)(n-5)(n-1)(n+3)(n+7)}{4^{5}5!}\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +2\frac{(n-6)(n-2)n(n+2)(n+6)}{4^{5}5!}\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-7)(n-3)(n+1)(n+5)(n+9)}{4^{5}5!}\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-9)(n-5)(n-1)(n+3)(n+7)}{4^{5}5!}\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\frac{(n-6)(n-2)n(n+2)(n+6)}{4^{5}5!}\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-7)(n-3)(n+1)(n+5)(n+9)}{4^{5}5!}\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},$$(33) and (34) $${\Psi }_6(n)=\{\begin{array}{cc}-2\frac{(n-8)(n-4)n^2(n+4)(n+8)}{4^{6}6!}\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-9)(n-5)(n-1)(n+3)(n+7)(n+11)}{4^{6}6!}\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-11)(n-7)(n-3)(n+1)(n+5)(n+9)}{4^{6}6!}\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +2\frac{(n-8)(n-4)n^2(n+4)(n+8)}{4^{6}6!}\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-9)(n-5)(n-1)(n+3)(n+7)(n+11)}{4^{6}6!}\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-11)(n-7)(n-3)(n+1)(n+5)(n+9)}{4^{6}6!}\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right),\hfill \end{array},$$(34) and the following example should give us a better vision about that sequence: (35) $$\begin{array}{c}{\Psi }_7(n)=\{\begin{array}{cc}\hspace{1em}0\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-13)(n-9)(n-5)(n-1)(n+3)(n+7)(n+11)}{4^{7}7!}\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\frac{(n-10)(n-6)(n-2)n(n+2)(n+6)(n+10)}{4^{7}7!}\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-11)(n-7)(n-3)(n+1)(n+5)(n+9)(n+13)}{4^{7}7!}\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-13)(n-9)(n-5)(n-1)(n+3)(n+7)(n+11)}{4^{7}7!}\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +2\frac{(n-10)(n-6)(n-2)n(n+2)(n+6)(n+10)}{4^{7}7!}\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-11)(n-7)(n-3)(n+1)(n+5)(n+9)(n+13)}{4^{7}7!}\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(35)

As we see, sometimes the numerator of the sequence ${\Psi }_k(n)$ has a centre factor that all the other factors in the numerators get around it. For $n\equiv 0,2,4,6\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ it is symmetric in this sense; meaning if $(n-a)$ is a factor of the numerator then $(n+a)$ would be a factor of that numerator and vice versa. However, it gives a different story for $n\equiv 1,3,5,7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ and a natural phenomenon for these numbers arises up here and needs a closer attention.

6.2. The right tendency concept for ${\Psi }_k(n)$

For $n\equiv \pm 1\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ from the data above, and from the formulas of ${\Psi }_k(n),$ we can observe that the factors of the numerators for $k=1,2,3,4,5,6,\dots$ always fill the right part first then go around the centre factor to fill the left part and so on. For example, for k = 1 the centre factor is $(n-1).$ Then for k = 2, the next factor $(n+3)$ is located on the right of $(n-1).$ We call this natural behaviour by the right tendency which is explained in table (36)(36) $$\text{Right}\mathrm{ }\text{Tendency}\mathrm{ }\text{For}\mathrm{ }{\Psi }_k(n)\hspace{1em}\begin{array}{cccccccc}k& & & & \text{The}\mathrm{ }\text{Center}& & & \\ 1& & & & (n-1)& & & \\ 2& & & & (n-1)& \mathrm{ }\left(\mathit{n\; +\; 3}\right)& & \\ 3& & & (n-5)& (n-1)& (n+3)& & \\ 4& & & (n-5)& (n-1)& (n+3)& (n+7)& \\ 5& & (n-9)& (n-5)& (n-1)& (n+3)& (n+7)& \\ 6& & (n-9)& (n-5)& (n-1)& (n+3)& (n+7)& (n+11)\end{array}$$(36) . (36) $$\text{Right}\mathrm{ }\text{Tendency}\mathrm{ }\text{For}\mathrm{ }{\Psi }_k(n)\hspace{1em}\begin{array}{cccccccc}k& & & & \text{The}\mathrm{ }\text{Center}& & & \\ 1& & & & (n-1)& & & \\ 2& & & & (n-1)& \mathrm{ }\left(\mathit{n\; +\; 3}\right)& & \\ 3& & & (n-5)& (n-1)& (n+3)& & \\ 4& & & (n-5)& (n-1)& (n+3)& (n+7)& \\ 5& & (n-9)& (n-5)& (n-1)& (n+3)& (n+7)& \\ 6& & (n-9)& (n-5)& (n-1)& (n+3)& (n+7)& (n+11)\end{array}$$(36)

6.3. The left tendency concept for ${\Psi }_k(n)$

However, for $n\equiv \pm 3\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ and from the data above, and the formulas of ${\Psi }_k(n),$ we can also observe that the factors of the numerators for $k=1,2,3,4,5,6,\dots$ always fill the left part first then go around the centre factor to fill the right part and so on. For example, for k = 1 the centre factor is $(n+1).$ Then for k = 2, the next factor $(n-3)$ is located on the left of $(n+1).$ We call this natural behaviour by the left tendency which is explained in table (37)(37) $$\text{Left}\mathrm{ }\text{Tendency}\mathrm{ }\text{For}\mathrm{ }{\Psi }_k(n)\hspace{1em}\begin{array}{cccccccc}k& & & & \text{The}\mathrm{ }\text{Center}& & & \\ 1& & & & (n+1)& & & \\ 2& & & \mathrm{ }\left(\mathit{n-3}\right)& (n+1)& & & \\ 3& & & (n-3)& (n+1)& (n+5)& & \\ 4& & (n-7)& (n-3)& (n+1)& (n+5)& & \\ 5& & (n-7)& (n-3)& (n+1)& (n+5)& (n+9)& \\ 6& (n-11)& (n-7)& (n-3)& (n+1)& (n+5)& (n+9)& \end{array}$$(37) . (37) $$\text{Left}\mathrm{ }\text{Tendency}\mathrm{ }\text{For}\mathrm{ }{\Psi }_k(n)\hspace{1em}\begin{array}{cccccccc}k& & & & \text{The}\mathrm{ }\text{Center}& & & \\ 1& & & & (n+1)& & & \\ 2& & & \mathrm{ }\left(\mathit{n-3}\right)& (n+1)& & & \\ 3& & & (n-3)& (n+1)& (n+5)& & \\ 4& & (n-7)& (n-3)& (n+1)& (n+5)& & \\ 5& & (n-7)& (n-3)& (n+1)& (n+5)& (n+9)& \\ 6& (n-11)& (n-7)& (n-3)& (n+1)& (n+5)& (n+9)& \end{array}$$(37)

6.4. The signs phenomena of $\Psi -$ integers

For $k\equiv 0,\hspace{0.5em}1,\hspace{0.5em}2,\hspace{0.5em}3,\hspace{0.5em}4,\hspace{0.5em}5,\hspace{0.5em}6,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ we get the following data (38) $$\begin{array}{ccccc}& & & & \\ k\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)& {(-1)}^{\lfloor \frac{k}{2}\rfloor }& {(-1)}^{\lfloor \frac{k}{2}\rfloor +1}& {(-1)}^{\lfloor \frac{k}{2}\rfloor +\delta (k)}& {(-1)}^{\lfloor \frac{k}{2}\rfloor +\delta (k-1)}\\ 0& +1& -1& +1& -1\\ 1& +1& -1& -1& +1\\ 2& -1& +1& -1& +1\\ 3& -1& +1& +1& -1\\ 4& +1& -1& +1& -1\\ 5& +1& -1& -1& +1\\ 6& -1& +1& -1& +1\\ 7& -1& +1& +1& -1\end{array}$$(38)

Therefore, the signs of the $\Psi -$ sequence get back periodically every 8 steps. Table (39)(39) $$\text{Signs}\mathrm{ }\text{of}\mathrm{ }{\Psi }_k(n)\begin{array}{ccccccccc}n\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)& {\Psi }_0(n)& {\Psi }_1(n)& {\Psi }_2(n)& {\Psi }_3(n)& {\Psi }_4(n)& {\Psi }_5(n)& {\Psi }_6(n)& {\Psi }_7(n)\\ 0& +& 0& -& 0& +& 0& -& 0\\ 1& +& +& -& -& +& +& -& -\\ 2& 0& +& 0& -& 0& +& 0& -\\ 3& -& +& +& -& -& +& +& -\\ 4& -& 0& +& 0& -& 0& +& 0\\ 5& -& -& +& +& -& -& +& +\\ 6& 0& -& 0& +& 0& -& 0& +\\ 7& +& -& -& +& +& -& -& +\end{array}$$(39) shows that three plus are followed by zero then it followed by three minus then followed by zero and so on. For $n\equiv 0,\hspace{0.5em}1,\hspace{0.5em}2,\hspace{0.5em}3,\hspace{0.5em}4,\hspace{0.5em}5,\hspace{0.5em}6,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8)$ and for $k\equiv 0,\hspace{0.5em}1,\hspace{0.5em}2,\hspace{0.5em}3,\hspace{0.5em}4,\hspace{0.5em}5,\hspace{0.5em}6,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ the following data is useful (39) $$\text{Signs}\mathrm{ }\text{of}\mathrm{ }{\Psi }_k(n)\begin{array}{ccccccccc}n\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)& {\Psi }_0(n)& {\Psi }_1(n)& {\Psi }_2(n)& {\Psi }_3(n)& {\Psi }_4(n)& {\Psi }_5(n)& {\Psi }_6(n)& {\Psi }_7(n)\\ 0& +& 0& -& 0& +& 0& -& 0\\ 1& +& +& -& -& +& +& -& -\\ 2& 0& +& 0& -& 0& +& 0& -\\ 3& -& +& +& -& -& +& +& -\\ 4& -& 0& +& 0& -& 0& +& 0\\ 5& -& -& +& +& -& -& +& +\\ 6& 0& -& 0& +& 0& -& 0& +\\ 7& +& -& -& +& +& -& -& +\end{array}$$(39)

6.5. General divisibility relationship for the $\Psi -$ integers

Whether $n\equiv 0,\hspace{0.5em}1,\hspace{0.5em}2,\hspace{0.5em}3,\hspace{0.5em}4,\hspace{0.5em}5,\hspace{0.5em}6,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ it is rather surprising that the quantity $${\Psi }_k(n)÷{\Psi }_{k-2}(n)$$ always gives $$-\frac{n^2-{(2k-4)}^2}{16k(k-1)}\hspace{1em}\text{or}\hspace{1em}-\frac{{(n\pm 1)}^2-{(2k-2)}^2}{16k(k-1)}\hspace{1em}\text{or}\hspace{1em}0÷0.$$

Therefore, computing these ratios, at the eight levels $n\equiv 0,\hspace{0.5em}1,\hspace{0.5em}2,\hspace{0.5em}3,\hspace{0.5em}4,\hspace{0.5em}5,\hspace{0.5em}6,\hspace{0.5em}7\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ immediately proves the following desirable theorem that gives a fraction with a difference of two squares divided by a multiplication of the factors 16, k, and k − 1.

Theorem 14.

(Generating The $\Psi -$integers From The Previous Term)

If ${\Psi }_k(n)$ not identically zero, then $$\frac{{\Psi }_k(n)}{{\Psi }_{k-2}(n)}=-\frac{{[n+{(-1)}^{\lfloor \frac{n}{2}\rfloor +k}\delta (n)]}^2-{[2k-2-2\delta (n-1)]}^2}{16k(k-1)},$$ and we can choose either of the following initial values to generate ${\Psi }_k(n)$ from the starting term ${\Psi }_0(n)$ or the last term ${\Psi }_{\lfloor \frac{n}{2}\rfloor }(n)$: $$\begin{array}{c}{\Psi }_0(n)=\{\begin{array}{cc}+2\hfill & n\equiv \hspace{0.5em}0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv \pm 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{0.5em}0\hfill & n\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv \pm 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & n\equiv \pm 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\Psi }_{\lfloor \frac{n}{2}\rfloor }(n)=1.\hfill \end{array}\mathrm{ }\mathrm{ }$$

7. The emergence of $\varphi -$ sequence

Another natural sequence that emerges naturally from ${\Psi }_k(n)$ is the integer sequence ${\varphi }_k(n)$ which is defined as follows

Definition 15.

$${\varphi }_k(n):=4^k\hspace{0.17em}{\Psi }_k(n).$$

7.1. Recurrence relation of order 4 to generate $\varphi -$ sequence

From (10)(10) $$2n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}{(-1)}^{\lfloor \frac{k}{2}\rfloor } \frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{k!}.$$(10) , we get the following recurrence relation

Lemma 16.

For each natural number n, $0\le k\le \lfloor \frac{n}{2}\rfloor$, the integers ${\varphi }_k(n)$ satisfy the following property (40) $${\varphi }_k(n)=4\hspace{0.17em}{\varphi }_{k-1}(n-2)-{\varphi }_k(n-4).$$(40)

From (30)(30) $$\begin{array}{c}{\Psi }_0(n)=\{\begin{array}{cc}+2\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{0.5em}0\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{0.5em}0\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\Psi }_1(n)=\{\begin{array}{cc}\hspace{1em}0\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n-1)}{4}\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +2\frac{n}{4}\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +\frac{(n+1)}{4}\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n-1)}{4}\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\frac{n}{4}\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -\frac{(n+1)}{4}\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\hfill \end{array}\mathrm{ }\mathrm{ }$$(30) , we easily get the following initial values

Lemma 17.

For each natural number n (41) $$\begin{array}{c}{\varphi }_0(n)=\{\begin{array}{cc}+2\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv \pm 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{0.5em}0\hfill & n\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv \pm 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & n\equiv \pm 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(n)=\{\begin{array}{cc}\hspace{1em}0\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +(n-1)\hfill & n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +2\hspace{0.17em}n\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +(n+1)\hfill & n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -(n-1)\hfill & n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hspace{0.17em}n\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -(n+1)\hfill & n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(41)

7.2. Explicit formulas for $\varphi -$ sequence

Now, from Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following explicit formulas for the integer sequence ${\varphi }_k(n).$

Lemma 18.

For any non negative integers n, k, the sequences ${\varphi }_k(n)$ satisfy the following statements

$\underset{¯}{n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\varphi }_k(n)=\{\begin{array}{cc}0\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{odd}\hfill \\ 2\mathrm{\hspace{0.17em}}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{even}\hfill \end{array}$$ $\underset{¯}{n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\varphi }_k(n)={(-1)}^{\lfloor \frac{k}{2}\rfloor }{(n+1-2k)}^{\delta (k)}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }[{(n+1)}^2-{(4\lambda -2)}^2]}{k!}\mathrm{ }\mathrm{ }$$ $\underset{¯}{n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\varphi }_k(n)=\{\begin{array}{cc}0\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{even}\hfill \\ 2\mathrm{\hspace{0.17em}}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }[n^2-{(4\lambda -2)}^2]}{k!}\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{odd}\hfill \end{array}$$ $\underset{¯}{n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\varphi }_k(n)={(-1)}^{\lfloor \frac{k}{2}\rfloor +\delta (k-1)}(n+1){(n+1-2k)}^{\delta (k-1)}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -\delta (k-1)}[{(n+1)}^2-{(4\lambda )}^2]}{k!}\mathrm{ }\mathrm{ }$$ $\underset{¯}{n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\varphi }_k(n)=\{\begin{array}{cc}0\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{odd}\hfill \\ 2\mathrm{\hspace{0.17em}}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}[n^2-{(4\lambda )}^2]}{k!}\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{even}\hfill \end{array}$$ $\underset{¯}{n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\varphi }_k(n)={(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\mathrm{\hspace{0.17em}}{(n+1-2k)}^{\delta (k)}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }[{(n+1)}^2-{(4\lambda -2)}^2]}{k!}\mathrm{ }\mathrm{ }$$ $\underset{¯}{n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\varphi }_k(n)=\{\begin{array}{cc}0\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{even}\hfill \\ 2\mathrm{\hspace{0.17em}}{(-1)}^{\lfloor \frac{k}{2}\rfloor +1}\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor }[n^2-{(4\lambda -2)}^2]}{k!}\hfill & \text{for}\mathrm{ }k\mathrm{ }\text{odd}\hfill \end{array}$$ $\underset{¯}{n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}$ $${\varphi }_k(n)={(-1)}^{\lfloor \frac{k}{2}\rfloor +\delta (k)}(n+1){(n+1-2k)}^{\delta (k-1)}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =1}^{\lfloor \frac{k}{2}\rfloor -\delta (k-1)}[{(n+1)}^2-{(4\lambda )}^2]}{k!}$$

7.3. Nonlinear recurrence relation to generate $\varphi -$ sequence

To study the arithmetic properties of $\varphi -$ Sequence, we need to generate ${\varphi }_k(n)$ from the previous one, ${\varphi }_{k-2}(n),$ or generate ${\varphi }_k(n)$ from the next one, ${\varphi }_{k+2}(n).$ As $$\frac{{\Psi }_k(n)}{{\Psi }_{k-2}(n)}=\frac{{\varphi }_k(n)}{4^2\hspace{0.17em}{\varphi }_{k-2}(n)},$$ and noting that $\delta (n)=0,\delta (n-1)=1,$ for n even, we immediately get, from Theorem (14)(14) $$2\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}1\hspace{1em}\nmid \hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}\left[{(4\lambda )}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{4}^{-1}\right]\hspace{1em}}{k!}.$$(14) , the following desirable theorem.

Theorem 19.

(Generating The $\varphi -$integers From The Previous Term)

If ${\varphi }_k(n)$ not identically zero, n even, then $$\frac{{\varphi }_k(n)}{{\varphi }_{k-2}(n)}=-\frac{n^2-{(2k-4)}^2}{k\hspace{0.17em}(k-1)},$$ and depending on the parity of k we can choose either of the following initial values to generate ${\varphi }_k(n)$ from the starting terms ${\varphi }_0(n)$ or ${\varphi }_1(n)$: (42) $$\begin{array}{c}{\varphi }_0(n)=\{\begin{array}{cc}+2\hfill & n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(n)=\{\begin{array}{cc}+2 n\hfill & n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 n\hfill & n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }$$(42)

8. Three new versions for Lucas-Lehmer primality tests for Mersenne numbers

From Lemmas (16–18), we are ready now to prove the following new results.

8.1. The proof of the first new version

Theorem 20.

Given prime $p\ge 5.$ $2^p-1$ is prime if and only if (43) $$2^p-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(43) where $n:=2^{p-1},{\varphi }_k(n)$ are defined by the double index recurrence relation $${\varphi }_k(m)=4 {\varphi }_{k-1}(m-2)-{\varphi }_k(m-4)$$ and the initial boundary values satisfy (44) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{0.5em}0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2\hspace{0.17em}m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0,\hspace{0.5em}4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hspace{0.17em}m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(44)

Proof.

Given prime $p\ge 5,$ let $n:=2^{p-1}.$ From Lucas-Lehmer-Test, see Gallier(2011) and Deza (2021), we have $$\begin{array}{c}{2}^p-1\hspace{1em}\text{is}\mathrm{ }\text{prime}\hspace{1em}\iff 2^p-1\hspace{1em}|\hspace{1em}{(1+\sqrt{3})}^n+{(1-\sqrt{3})}^n.\hfill \end{array}$$

Hence, as $n\equiv 0\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8),$ replace $x=1+\sqrt{3},\hspace{1em}y=1-\sqrt{3}$ in Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , we get the following equivalent statements: $$\begin{array}{cc}{2}^p-1\hspace{1em}\text{is}\mathrm{ }\text{prime}\hspace{1em}\hfill & \iff 2^p-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n) {(-2)}^{\lfloor \frac{n}{2}\rfloor -k}{(8)}^k\hfill \\ \hfill & \iff 2^p-1\hspace{1em}|\hspace{1em}{2}^{\lfloor \frac{n}{2}\rfloor }\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n)\hspace{1em}{4}^k\hfill \\ \hfill & \iff 2^p-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n).\hfill \end{array}$$

Lemmas (16, 17) already proved the rest of Theorem (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) . □

8.2. The proof of the second new version

From Lemmas (16, 17), we should observe that the recurrence relation of ${\varphi }_k(n)$ is always even integer for $n\equiv 0\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}8).$ Hence from Lemma (18)(18) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k.$$(18) , we get the following theorem.

Theorem 21.

Given prime $p\ge 5,n:=2^{p-1}$. The number $2^p-1$ is prime if and only if (45) $$2 n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}{(-1)}^{\lfloor \frac{k}{2}\rfloor }\hspace{0.17em}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}{n}^2-{(4\lambda )}^2}{k!}.$$(45)

8.3. The proof of the third new version

When we generate the $\varphi -$ integers needed for Mersenne numbers, we should notice that for $n:=2^p-1,p\ge 5,$ we have $$n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right).$$

From the proof of Theorem (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) , we know that $$\begin{array}{cc}{2}^p-1\hspace{1em}\text{is}\mathrm{ }\text{prime}\hspace{1em}\hfill & \iff 2^p-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n).\hfill \end{array}$$

It is plain that, from Eq. (20)(20) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k$$(20) , ${\Psi }_{\lfloor \frac{n}{2}\rfloor }(n)=1.$ Consequently, ${\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)=4^{\lfloor \frac{n}{2}\rfloor }=2^n.$ Therefore, from Theorem (19)(19) $$\frac{x^n+y^n}{{(x+y)}^{\delta (n)}}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\Psi }_k(n){(xy)}^{\lfloor \frac{n}{2}\rfloor -k}{(x^2+y^2)}^k.$$(19) , we get the following result.

Theorem 22.

Given prime $p\ge 5, n:=2^{p-1}$. The number $2^p-1$ is prime if and only if (46) $$2\hspace{0.17em}n-1\hspace{1em}|\hspace{1em}\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(46) where ${\varphi }_k(n)$ are defined and generated by the double index recurrence relation $$\frac{{\varphi }_k(n)}{{\varphi }_{k-2}(n)}=\frac{{(2k-4)}^2-n^2}{k(k-1)},$$ and we can choose either of the following initial values to generate ${\varphi }_k(n)$ from the starting term ${\varphi }_0(n)$ or the last term ${\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)$: $$\begin{array}{c}{\varphi }_0(n)=+2\mathrm{\hspace{1em}}\mathrm{\hspace{1em}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)=2^n.\hfill \end{array}\mathrm{ }$$

9. Criteria for compositeness of Mersenne numbers

The following theorem is an immediate consequence of Theorem (21)(21) $${\Psi }_0(n)=\{\begin{array}{cc}+2\hfill & n\equiv \hspace{0.5em}0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ +1\hfill & n\equiv \pm 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & n\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -1\hfill & n\equiv \pm 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & n\equiv \pm 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}$$(21) .

Theorem 23.

(Criteria for compositeness of Mersenne numbers)

Given prime $p\ge 5$. The number $2n-1=2^p-1$ is Mersenne composite number if (47) $$2\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}1\hspace{1em}\nmid \hspace{1em}\sum _{\stackrel{k=0,}{k\mathrm{\hspace{0.17em}}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }\hspace{1em}\mathrm{\hspace{0.17em}}\frac{\prod _{\lambda =0}^{\lfloor \frac{k}{2}\rfloor -1}\left[{(4\lambda )}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{4}^{-1}\right]\hspace{1em}}{k!}.$$(47)

10. The proof of the combinatorial identities

Choosing $k=\lfloor \frac{n}{2}\rfloor$ in the Eight Levels Theorem (9)(9) $$\begin{array}{c}{\varphi }_0(m)=\{\begin{array}{cc}+2\hfill & m\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ 0\hfill & m\equiv \pm 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2\hfill & m\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{\varphi }_1(m)=\{\begin{array}{cc}+2 m\hfill & m\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ \hspace{1em}0\hfill & m\equiv 0, 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \\ -2 m\hfill & m\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)\hfill \end{array}.\hfill \end{array}\mathrm{ }\mathrm{ }$$(9) , and noting ${\Psi }_{\lfloor \frac{n}{2}\rfloor }(n)=1,$ we surprisingly get the following eight combinatorial identities which reflect some unexpected facts about the nature of numbers. $$\begin{array}{ccc}\underset{¯}{n\equiv 0\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\hspace{0.17em}\left(\lfloor \frac{n}{2}\rfloor \right)!=2\hspace{0.17em}\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{n}{4}\rfloor -1}{n}^2\hspace{0.17em}-\hspace{0.17em}{(4\lambda )}^2\hfill \\ \underset{¯}{n\equiv 1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\hspace{0.17em}\left(\lfloor \frac{n}{2}\rfloor \right)!=\hspace{0.17em}\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{n-1}{4}\rfloor }{(n+1)}^2\hspace{0.17em}-\hspace{0.17em}{(4\lambda -2)}^2\hfill \\ \underset{¯}{n\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\hspace{0.17em}\left(\lfloor \frac{n}{2}\rfloor \right)!=2\hspace{0.17em}\hspace{0.17em}n\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{n-2}{4}\rfloor }{n}^2\hspace{0.17em}-\hspace{0.17em}{(4\lambda -2)}^2\hfill \\ \underset{¯}{n\equiv 3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\hspace{0.17em}\left(\lfloor \frac{n}{2}\rfloor \right)!=\hspace{0.17em}(n+1)\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{n-3}{4}\rfloor }{(n+1)}^2\hspace{0.17em}-\hspace{0.17em}{(4\lambda )}^2\hfill \\ \underset{¯}{n\equiv 4\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\hspace{0.17em}\left(\lfloor \frac{n}{2}\rfloor \right)!=\hspace{0.17em}2\hspace{0.17em}\prod _{\lambda =0}^{\lfloor \frac{n-4}{4}\rfloor }{n}^2\hspace{0.17em}-\hspace{0.17em}{(4\lambda )}^2\hfill \\ \underset{¯}{n\equiv 5\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\hspace{0.17em}\left(\lfloor \frac{n}{2}\rfloor \right)!=\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{n-1}{4}\rfloor }{(n+1)}^2\hspace{0.17em}-\hspace{0.17em}{(4\lambda -2)}^2\hfill \\ \underset{¯}{n\equiv 6\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\hspace{0.17em}\left(\lfloor \frac{n}{2}\rfloor \right)!=2\hspace{0.17em}\hspace{0.17em}n\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{n-2}{4}\rfloor }{n}^2\hspace{0.17em}-\hspace{0.17em}{(4\lambda -2)}^2\hfill \\ \underset{¯}{n\equiv 7\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}8\right)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\hspace{0.17em}\left(\lfloor \frac{n}{2}\rfloor \right)!=\hspace{0.17em}(n+1)\hspace{0.17em}\prod _{\lambda =1}^{\lfloor \frac{n-3}{4}\rfloor }{(n+1)}^2\hspace{0.17em}-\hspace{0.17em}{(4\lambda )}^2\hfill \end{array}$$

Writing only the different identities, which are 4, we get the following theorem which gives formulas to factor any factorial in terms of a product of difference of squares.

Theorem 24.

(combinatorial identities) For any natural number n, the following combinatorial identities are correct $$\begin{array}{ccc}\underset{¯}{n\equiv 0\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}4)}\hfill & \hspace{1em}\hfill & 4^{\frac{n}{2}}\mathrm{\hspace{0.17em}}(\lfloor \frac{n}{2}\rfloor )!=2\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}\prod _{\lambda =0}^{\lfloor \frac{n-4}{4}\rfloor }[n^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda )}^2]\hfill \\ \underset{¯}{n\equiv 1\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}4)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\mathrm{\hspace{0.17em}}(\lfloor \frac{n}{2}\rfloor )!=\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}\prod _{\lambda =1}^{\lfloor \frac{n-1}{4}\rfloor }[{(n+1)}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda -2)}^2]\hfill \\ \underset{¯}{n\equiv 2\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}4)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\mathrm{\hspace{0.17em}}(\lfloor \frac{n}{2}\rfloor )!=2\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}n\mathrm{\hspace{0.17em}}\prod _{\lambda =1}^{\lfloor \frac{n-2}{4}\rfloor }[n^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda -2)}^2]\hfill \\ \underset{¯}{n\equiv 3\hspace{0.5em}(\mathrm{mod\hspace{0.5em}}4)}\hfill & \hspace{1em}\hfill & 4^{\lfloor \frac{n}{2}\rfloor }\mathrm{\hspace{0.17em}}(\lfloor \frac{n}{2}\rfloor )!=\mathrm{\hspace{0.17em}}(n+1)\mathrm{\hspace{0.17em}}\prod _{\lambda =1}^{\lfloor \frac{n-3}{4}\rfloor }[{(n+1)}^2\mathrm{\hspace{0.17em}}\text{-}\mathrm{\hspace{0.17em}}{(4\lambda )}^2]\hfill \end{array}$$

11. Formulas

Now we compute the summation (48) $$\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(48) for the first few terms from the starting and from the end. From Theorem (22)(22) $${\Psi }_k(n)={\Psi }_{k-1}(n-2)-{\Psi }_k(n-4).$$(22) , we get the following explicit formulas: $${\varphi }_0(n)=+2.$$

Now, compute ${\varphi }_2(n)$ as following $${\varphi }_2(n)=-\frac{n^2-{(4-4)}^2}{2(2-1)}{\varphi }_0(n)=-n^2.$$

Proceeding this way, we get $${\varphi }_4(n)=-\frac{n^2-{(8-4)}^2}{4(4-1)}{\varphi }_2(n)=+\frac{n^2(n^2-4^2)}{12}.$$

Similarly $${\varphi }_6(n)=-\frac{n^2(n^2-4^2)(n^2-8^2)}{360},$$ $${\varphi }_8(n)=+\frac{n^2(n^2-4^2)(n^2-8^2)(n^2-{12}^2)}{20160},\hspace{1em}\mathit{etc}$$

Now, compute ${\varphi }_k(n)$ from the end, and from Theorem (22)(22) $${\Psi }_k(n)={\Psi }_{k-1}(n-2)-{\Psi }_k(n-4).$$(22) , we get $${\varphi }_{k-2}(n)=-\frac{k(k-1)}{n^2-{(2k-4)}^2}{\varphi }_k(n).$$

Initially $${\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)=2^n.$$

Hence $${\varphi }_{\lfloor \frac{n}{2}\rfloor -2}(n)=-n{2}^{n-5},$$ and $${\varphi }_{\lfloor \frac{n}{2}\rfloor -4}(n)=+n(n-6)2^{n-11}.$$

Similarly $${\varphi }_{\lfloor \frac{n}{2}\rfloor -6}(n)=-\frac{n(n-8)(n-10)}{3}{2}^{n-16},$$ $${\varphi }_{\lfloor \frac{n}{2}\rfloor -8}(n)=+\frac{n(n-10)(n-12)(n-14)}{3}{2}^{n-23},\hspace{1em}\mathit{etc}.$$

The author feels that we need a clever way to evaluate the sum (48)(48) $$\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(48) . We may like to add the terms in a way reflect some elegant arithmetic. Remember that we do not need to compute the sum (48)(48) $$\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)$$(48) exactly; but we just need to find the sum modulo $2n-1.$ According to the following theorem, and working modulo $2n-1,$ the last term always gives the value of the first term.

Theorem 25.

For $p\ge 5$ prime, and $n:=2^{p-1},$ (49) $${\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}2n-1\right).$$(49)

Proof.

We should observe that if $p\ge 5$ prime, then $n=2^{p-1}=1+\zeta \hspace{0.17em}p,$ for some positive integer $\zeta .$ Then $${\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)=2^n=2^{1+\zeta p}=2^1{(2^p)}^{\zeta }\equiv 2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}2n-1\right).$$

□

Hence, this encourages one to compute the partial sums of (50) $$\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}2n-1\right),$$(50) in the following order $${\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)+{\varphi }_0(n)+{\varphi }_2(n)+{\varphi }_4(n)+\cdots +{\varphi }_{\lfloor \frac{n}{2}\rfloor -1}(n).$$

12. The 5 scenario

For example, take p = 5, then $n=2^4.$ Hence (51) $$\begin{array}{cc}{\varphi }_8(16)\hfill & \equiv \hspace{0.5em}+2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}31\right),\hfill \\ {\varphi }_0(16)\hfill & \equiv \hspace{0.5em}+2\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}31\right),\hfill \\ {\varphi }_2(16)\hfill & \equiv -2^3\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}31\right),\hfill \\ {\varphi }_4(16)\hfill & \equiv +2^2+1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}31\right),\hfill \\ {\varphi }_6(16)\hfill & \equiv -1\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}31\right).\hfill \end{array}$$(51)

Hence, we get the partial sums (52) $$\begin{array}{cc}{\varphi }_8(16)\hfill & \equiv +2,\hfill \\ {\varphi }_8(16)+{\varphi }_0(16)\hfill & \equiv +2^2,\hfill \\ {\varphi }_8(16)+{\varphi }_0(16)+{\varphi }_2(16)\hfill & \equiv -2^2,\hfill \\ {\varphi }_8(16)+{\varphi }_0(16)+{\varphi }_2(16)+{\varphi }_4(16)\hfill & \equiv +1,\hfill \\ {\varphi }_8(16)+{\varphi }_0(16)+{\varphi }_2(16)+{\varphi }_4(16)+{\varphi }_6(16)\hfill & \equiv \hspace{0.5em}0.\hfill \end{array}$$(52)

As we ended up with zero, this shows that $2^5-1=31$ is Mersenne prime.

13. Further research investigations

Now, consider p prime greater than 3, and $n:=2^{p-1}.$ The previous sections give various explicit formulas and techniques to compute and generate all of the terms ${\varphi }_k(n)$ needed for checking the primality of the Mersenne number $2^p-1.$

13.1. Searching for a hypothetical pattern

This previous particular example, for $p=5=2^2+1,$ is illuminating and should motivate us for further theoretical investigations for other similar scenarios. We need to find a general pattern for n similar to this special case, for p = 5, such that, working modulo $2n-1,$ the partial sums give: (53) $$\begin{array}{c}{\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)\equiv +2,\hfill \\ {\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)+{\varphi }_0(n)\equiv +2^2,\hfill \\ {\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)+{\varphi }_0(n)+{\varphi }_2(n)\equiv -2^2+{ϵ}_1,\hfill \\ {\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)+{\varphi }_0(n)+{\varphi }_2(n)+{\varphi }_4(n)\equiv +2^3+{ϵ}_2,\hfill \\ {\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)+{\varphi }_0(n)+{\varphi }_2(n)+{\varphi }_4(n)+{\varphi }_6(n)\equiv -2^3+{ϵ}_3,\hfill \\ {\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)+{\varphi }_0(n)+{\varphi }_2(n)+{\varphi }_4(n)+{\varphi }_6(n)+{\varphi }_8(n)\equiv +2^4+{ϵ}_4,\hfill \\ {\varphi }_{\lfloor \frac{n}{2}\rfloor }(n)+{\varphi }_0(n)+{\varphi }_2(n)+{\varphi }_4(n)+{\varphi }_6(n)+{\varphi }_8(n)+{\varphi }_{10}(n)\equiv -2^4+{ϵ}_5,\hfill \\ \cdots \hfill & \hfill \end{array}$$(53) where $${ϵ}_k\in \{0,-1,+1\}$$ for each k. We need to investigate certain types of prime p. We should try to investigate the primes p of the form $2^a\pm 1$ or $2^a\pm 2^b\pm 1$ to make the sum (54) $$\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)\hspace{1em}\left(\mathrm{mod\hspace{0.5em}}2n-1\right),$$(54) easily to understand and help us identify a general pattern for p for which this sum gives zero modulo $2n-1$ infinitely many times (in this case we could prove that Mersenne primes are infinite), or gives nonzero modulo $2n-1$ infinitely many times (in this case we could prove that Mersenne composites are infinite).

13.2. New identities to help understand the sums of (48) and (54)

To motivate the readers about this future vital research and investigations, and in the spirit of the previous results, I feel compelled to mention, even though succinctly, the following new identities that I found recently for ${\varphi }_k(n):$

For any natural number p greater than 3, and $n:=2^{p-1},$ we get the identities (55) $$\begin{array}{cc}\mathrm{ }\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n) 2^{-k}\hfill & =2,\hfill \\ \mathrm{ }\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n) 2^{-2k}\hfill & =-1,\hfill \\ \mathrm{ }\sum _{\stackrel{k=0,}{k\hspace{0.5em}\mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n) 2^{-2k} 3^k\hfill & =L(n),\hfill \end{array}$$(55) where L(n) is Lucas sequence defined by $L(m+1)=L(m)+L(m-1),L(1)=1,L(0)=2.$

13.3. The nature of the prime factors of the sum

We use SAGE (SAGE Mathematical Software, Version 2.6), with double-checking provided by Mathematica, to carry out all the numerical computations of the current paper. For $p=5, n=16,$ the following sum (56) $$\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n)= 2\times 31\times 607.$$(56)

This shows that the number 31 is Mersenne prime because it is one of the factors of this sum. Surprisingly, although the prime number 607 is not a Mersenne prime, the number 607 is the exponent of the Mersenne prime $2^{607}-1.$ Moreover, the prime 607 is an irregular prime since it divides the numerator of the Bernoulli number B₅₉₂. So, for a given prime $p, n:=2^{p-1},$ we should also investigate the arithmetical nature of all the other prime factors for the sum (57) $$\sum _{\stackrel{k=0,}{k \mathit{even}}}^{\lfloor \frac{n}{2}\rfloor }{\varphi }_k(n).$$(57)

14. Conclusions

The discovery of new Mersenne primes is significant in the field of mathematics because they are relatively rare and difficult to find. Moreover, they have important applications in areas such as cryptography, number theory, and computer science. As of April 2023, there are currently only 51 known Mersenne primes. The largest known Mersenne prime as of this date is $2^{82,589,933}-1,$ which has 24,862,048 digits. The discovery of Mersenne primes is a very challenging, computationally intensive process, and time-consuming task, and the search for new Mersenne primes via GIMPS is an ongoing effort by many mathematicians and computer scientists around the world. While the Great Internet Mersenne Prime Search (GIMPS) has been successful in discovering 51 Mersenne primes to date, there are several limitations to the search. The search for Mersenne primes is a probabilistic process, and it is not guaranteed that a new Mersenne prime will be found. Even if a new Mersenne prime exists, there is a chance that it may not be discovered in a given search due to the limitations of the algorithm or computational resources. The search becomes increasingly difficult as the size of the potential primes increases, and it may become impractical or impossible to search for larger Mersenne primes without significant advances in theoretical understanding of the nature of Mersenne primes. Therefore, the current paper developed three new versions of primality tests for Mersenne numbers which could potentially help in the search for new Mersenne primes or provide insights into the open questions surrounding the infinitude of Mersenne primes or Mersenne composites.

Disclosure statement

The author declares that he has no conflict of interest.