Abstract
Let T be a tree with n vertices. Let be continuous and suppose that the n vertices form a periodic orbit under f. We show:
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If n is not a divisor of 2 k then f has a periodic point with period 2 k .
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If
, where
is odd and
, then f has a periodic point with period 2 p r for any
.
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The map f also has periodic orbits of any period m where m can be obtained from n by removing ones from the right of the binary expansion of n and changing them to zeroes.
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Conversely, given any n, there is a tree with n vertices and a map f such that the vertices form a periodic orbit and f has no other periods apart from the ones given above.
1991 Mathematics Subject Classification::