The conjecture on P-positions for Slow SetNim with at least $k$ playable stacks

Counterexample for every $k\ge3$: infinitely many full-support reduced losing positions omitted by the proposed classification.

Consider Conjecture 1 of Dufour and Heubach, INTEGERS 26 (2026), #G3, doi:10.5281/zenodo.20931511. In the game with

$$A=\{k,k+1,\ldots,n\},$$

a move selects any allowed number of positive heaps and removes exactly one token from each selected heap. The conjecture concerns reduced positions and uses

$$s=\sum_i p_i\pmod{2k},\qquad o=\#\{i:p_i\text{ is odd}\}.$$

At the central residue $s=k-1$, it asserts that the losing positions are exactly those with

$$o\equiv n\pmod2,\qquad 1\le o\le k-1. \tag{1}$$

The other two proposed classes require $s<k-1$ or $s>k-1$.

For every $k\ge3$, choose any integer $n$ satisfying

$$\left\lceil\frac{3k}{2}\right\rceil\le n\le2k-1, \qquad n\equiv k\pmod2.$$

Define

$$a=2n-3k+1,\qquad b=3k-1-n, \qquad p=(\underbrace{1,\ldots,1}_{a}, \underbrace{2,\ldots,2}_{b}).$$

Then

$$a\ge1,\qquad b\ge k,\qquad a+b=n,\qquad \sum_i p_i=a+2b=3k-1.$$

Every heap is positive, and the exact reduction criterion in Theorem 4 of the source holds:

$$k\max_i p_i=2k\le3k-1=\sum_i p_i.$$

Thus $p$ belongs to the source's reduced playable game, with $k>n/2$.

We now prove its outcome without any computational assumption. After every possible first move, each of the $b\ge k$ heaps initially equal to $2$ remains positive: a selected such heap becomes $1$, and an unselected one remains $2$. Therefore at least $k$ positive heaps remain, and a second move is always possible.

Conversely, each move removes at least $k$ tokens, so after every possible pair of moves at most

$$(3k-1)-2k=k-1$$

tokens remain. There are then fewer than $k$ positive heaps, so no third move is possible. Hence every legal play lasts exactly two moves, irrespective of the selected move sizes. Therefore $p$ is a losing position. Theorem 5 of the source transfers this outcome to its equivalent reduced playable game.

However,

$$s=(3k-1)\bmod(2k)=k-1,\qquad o=a=2n-3k+1.$$

Although

$$1\le o\le k-1,$$

its parity satisfies

$$o\equiv k-1\not\equiv n\pmod2.$$

Therefore the conjectured central class (1) excludes $p$; the other two classes also exclude it because their residues differ. The conjecture thus incorrectly labels this genuine losing position as winning.

The parameter range is nonempty for every $k\ge3$. Explicit one-parameter families are

$$\begin{aligned} k\ge3\text{ odd}:&\qquad n=2k-1,\quad p=(1^{k-1},2^k),\\ k\ge4\text{ even}:&\qquad n=2k-2,\quad p=(1^{k-3},2^{k+1}). \end{aligned}$$

The smallest instances are

$$(n,k,p)=(5,3,(1,1,2,2,2))$$

and

$$(n,k,p)=(6,4,(1,2,2,2,2,2)).$$

All counterexamples have full support, a positive number of odd heaps, satisfy the exact reduced-position criterion, and lie within the intended nontrivial regime $k>n/2$. The failure is therefore a structural error in the middle-slice parity condition, not a missing zero-odd-heaps endpoint.