A 3-branched generalized-Collatz Conjecture

Consider the function T1:

T1(x) = 2x if 3 divides x,
T1(x) = (7x+2)/3 if 3 divides x-1,
T1(x) = (x-2)/3 if 3 divides x+1.

Clearly T1(x) ≥ 0 if x ≥ 1 and T1(x) < 0 if x < 0.
It is conjectured by Keith Matthews that
Similarly, for the function

T2(x) = 2x if 3 divides x,
T2(x) = (5x-2)/3 if 3 divides x-1,
T2(x) = (x-2)/3 if 3 divides x+1,

we conjecture that (See article on generalized 3x+1 mappings.)

A $100 Australian prize for a resolution of one of these conjectures is offered.

The algorithms are performed with starting values y= 3M+1 and y=3M-1.

The iterates y, Ti(y), Ti(Ti(y)),... of each function are printed and the number of steps taken to reach either a multiple of 3, or else one of the cycles, is recorded.

Enter M:

Mapping T1
Mapping T2

Last modified 4th December 2013
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