Conjecture of Pruthviraj Hajari

This generalizes Hahari's first conjecture.

Let k ≥ 3 be an odd integer. The iterates x, t(x), t(t(x)),... of the mapping

t(x) = (k2 - 1)x if x is odd
t(x) = ⌊x/k⌋ if x is even,
where \(\lfloor\quad\rfloor\) denotes the floor function.

\[ t(x)=\left\{\begin{array}{cl} (k^2-1)x & \mbox{ if $x$ is odd,}\\ \left\lfloor x/k\right\rfloor & \mbox{ if $x$ is even,} \end{array} \right. \] are conjectured by Pruthviraj Hajari to eventually reach 0 if x ≥ 0. (See MathOverflow question.)

It seems certain that if x < 0, then tn(x) will eventually reach one of the (k2 - 1)/4 cycles with starting points -(ki +2s), where 1 ≤ i ≤ k - 2, t odd, and 0 ≤ s ≤ (k - 1)/2; each cycle is of length 3:

-(ki +2s) → -(ki +2s)(k2 - 1) → -k2i - 2sk + i → -(ki +2s).

Enter k (3 ≤ k ≤ 99 and odd) :
Enter x :

Last modified 16th December 2020
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