Td(n) = n/d if n ≡ 0 (mod d) Td(n) = ((d+1)n - i)/d if n ≡ i (mod d), -d/2 < i ≤ d/2, i ≠ 0.For example, d = 2 gives the 3x-1 mapping:
T2(n) = n/2 if n ≡ 0 (mod 2) T2(n) = (3n - 1)/2 if n ≡ 1 (mod 2).This is a special case of a version of a mapping studied by Herbert Möller and is also an example of a relatively prime mapping, in the language of Matthews and Watts, where m0=1 and mi = d+1 for 1 ≤ i < d and where we have the inequality
m0m1⋯md =(d+1)d-1 < dd.So it seems certain that the sequence of iterates
n, Td(n), Td2(n), ...always eventually enters a cycle and that there are only finitely many such cycles.
Clearly Td(n) = n for -d/2 < n ≤ d/2.
For d = 3, 6 and 10, we appear to get no other cycles.
It would be interesting to determine all d with this property. See the Table, which was constructed using the author's faster CALC program.
We assume that every trajectory will eventually reach a cycle and use Floyd's method of testing for equality of kth and 2kth iterates for find a cycle element. We list the cycle with smallest absolute value as starting point.
Last modified 21st July 2011
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