A generalization of the 3x-1 mapping due to Lu Pei

Consider the mapping Td: ℤ → ℤ Let d ≥ 2. Then

                       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.

Experiment with the special case of d = 3.

We search all trajectories x, Td(x), Td2(x),... where |x| ≤ R ⁄ 2 = 60000 ⁄ 2 for cycling by Floyd's method of testing for equality of kth and 2kth iterates for k ≤ U = 1000000.
Here we list cycles found other than the above, for all d satisfying m ≤ d ≤ n, where m and n satisfy 2 ≤ m ≤ n ≤ 100.
We choose the cycle element with smallest absolute value as starting point.
Finally, we limit the number of cycles for each d to 500.

Enter m (2 ≤ m ≤ 100 ):
Enter n (m ≤ n ≤ 100 ):
print a complete cycle and the cycle-length
print only the start of a cycle and the cycle-length

Last modified 17th January 2011
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