Period-divisor conjecture for tripod Nim orbits

Counterexample at $n=10$, the first value beyond the source's reported checks.

Conjecture 2 in Section 9.3 of Hennessey, Tree and Tripod Nim, arXiv:2401.07943, asserts that every periodic orbit of $D(3,n)$ has period dividing

$$2(4n)(4n+1).$$

Immediately afterward, the paper reports checking $n\le9$. The literal transition rule and its worked Figure 14 yield a counterexample at $n=10$.

A state of $D(3,n)$ is a binary array with three rows and $2n+3$ columns. Encode rows from bottom to top by

$$R_i=\sum_{j=0}^{2n+2}r_{i,j}2^j,$$

so column zero, the leftmost column, is the least significant bit. The source's transition $F$ is:

1. Harvest the leftmost column and shift every row left, appending a zero column on the right.
2. Process rows with harvested bit zero from bottom to top.
3. In each such row, insert a one into its leftmost zero outside the first $n$ columns, without reusing an insertion column in the same transition.

Equivalently, initialize

$$h_i=R_i\bmod2,\qquad Q_i=\left\lfloor\frac{R_i}{2}\right\rfloor, \qquad U=0.$$

For $i=1,2,3$, leave $Q_i$ unchanged when $h_i=1$. When $h_i=0$, take

$$j_i=\min\left\{j\ge n: \operatorname{bit}_j(Q_i)=0,\ \operatorname{bit}_j(U)=0\right\},$$

and replace

$$Q_i\leftarrow Q_i+2^{j_i}, \qquad U\leftarrow U+2^{j_i}.$$

Then $F(R_1,R_2,R_3)=(Q_1,Q_2,Q_3)$. Applying the same rule to the paper's worked $D(8,4)$ example reproduces every one of its 88 displayed before-and-after entries, fixing the shift order, incubator convention, row priority, and insertion restrictions independently.

Now set $n=10$, so the states have $23$ columns. Starting from the all-zero state, iteration reaches

$$S=(1,2047,2042)$$

after exactly $109$ transitions. Exact integer iteration gives

$$\begin{aligned} F^{24}(S)&=(1024,2047,1021)\ne S,\\ F^{88}(S)&=(1,2038,3464)\ne S,\\ F^{132}(S)&=(1,2014,2082)\ne S,\\ F^{264}(S)&=(1,2047,2042)=S. \end{aligned}$$

Since

$$264=2^3\cdot3\cdot11,$$

checking the maximal proper prime-divisor quotients $264/2=132$, $264/3=88$, and $264/11=24$ proves that the minimum period is exactly $264$.

However, the conjectured divisor at $n=10$ is

$$2(4n)(4n+1)=2\cdot40\cdot41=3280,$$

and

$$3280=12\cdot264+112.$$

Therefore $264\nmid3280$, contradicting the conjecture.

There is an unrelated shift-order inconsistency between the source's preceding one-row argument and its explicit Section 9.3 transition. The witness above uses exactly the Section 9.3 rule and reproduces the paper's worked Figure 14. Even if the alternative earlier-proof convention is substituted, the all-zero $D(3,10)$ state instead reaches a cycle of minimum period $220$, and

$$3280=14\cdot220+200,$$

so the proposed divisibility fails under that convention as well.