Consider subtraction Nim with subtraction set s_1,s_2,s_3 and one pass. Let G(x,0) denote the Grundy number without a pass and G(x,1) the Grundy number with a pass. The source defines the relevant reverse-mex recurrences by equations (1)–(3). The three-family prediction. (a) For subtraction set s_1,s_2,s_3=a,2an,2an+a with a,n∈N, equation (1) holds, and for x≥6an+(4a+1) equations (2) and (3) hold; the Grundy sequence enters a loop at that threshold. (b) For subtraction set a,(2n+1)a,(2n+3)a with a,n∈N, equation (1) holds, and for x≥ a(2n+3)+1 equations (2) and (3) hold; moreover G(x,1)<4 for every x, and the sequence enters a loop at that threshold. (iii) For subtraction set a,(2n+1)a,(2n+5)a with a,n∈N, equation (1) holds, G(x,1)<4 for every x, and for x≥ a(2n+5)+1 equations (2) and (3) hold; the sequence enters a loop at that threshold. These are examples and predictions based on computer calculations; the source does not establish them as theorems or report a resolution.

Predicted reverse-mex behavior for three families of subtraction Nim with a pass

Less than 1 year old · traced to 2026

Consider subtraction Nim with subtraction set $\{s_1,s_2,s_3\}$ and one pass. Let $\mathcal{G}(x,0)$ denote the Grundy number without a pass and $\mathcal{G}(x,1)$ the Grundy number with a pass. The source defines the relevant reverse- $mex$ recurrences by equations $(1)$ – $(3)$ . The three-family prediction. (a) For subtraction set $\{s_1,s_2,s_3\}=\{a,2an,2an+a\}$ with $a,n\in\mathbb{N}$ , equation $(1)$ holds, and for $x\geq6an+(4a+1)$ equations $(2)$ and $(3)$ hold; the Grundy sequence enters a loop at that threshold. (b) For subtraction set $\{a,(2n+1)a,(2n+3)a\}$ with $a,n\in\mathbb{N}$ , equation $(1)$ holds, and for $x\geq a(2n+3)+1$ equations $(2)$ and $(3)$ hold; moreover $\mathcal{G}(x,1)<4$ for every $x$ , and the sequence enters a loop at that threshold. (iii) For subtraction set $\{a,(2n+1)a,(2n+5)a\}$ with $a,n\in\mathbb{N}$ , equation $(1)$ holds, $\mathcal{G}(x,1)<4$ for every $x$ , and for $x\geq a(2n+5)+1$ equations $(2)$ and $(3)$ hold; the sequence enters a loop at that threshold. These are examples and predictions based on computer calculations; the source does not establish them as theorems or report a resolution.

References

Primary source

Urban Larsson, Hikaru Manabe and Ryohei Miyadera, “A Subtraction Nim with a Pass”, arXiv:2605.14321 (2026).

combinatorial-games subtraction-nim

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

An unverified posted calculation claims all three predictions fail, while the only published work proves a different special case.

Larsson, Manabe, and Miyadera (2026) state the three families as Conjecture 3, based on examples and computer calculations rather than proofs. The conjecture predicts reverse- $\mathrm{mex}$ recurrences, bounds, and eventual looping for the specified subtraction sets.

Known results

Larsson, Manabe, and Miyadera (2026) prove the separate case $\{2,4n,4n+2\}$ for $n\geq3$ , including the one-pass game and eventual periodic behavior from $x\geq12n+9$ ; this does not establish Conjecture 3.

Posted attempt

An unverified calculation claims explicit counterexamples within the stated ranges: $(a,n)=(1,2)$ for family (a), $(1,1)$ for family (b), and $(1,2)$ for family (iii), with scaling extending them to every positive $a$ . If correct, this disproves all three universal predictions; it has not been independently verified.

Current status (as of August 2026): The three-family assertions remain unproved, and an unverified calculation claims counterexamples to all three; no verified resolution is recorded.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

All three stated families have infinitely many counterexamples.

For a subtraction set $S$ , let $g_S(x)$ and $h_S(x)$ be the Grundy values without and with one available pass. The source's recurrences are

g_S(0)=h_S(0)=0,

g_S(x)=\operatorname{mex}\{g_S(x-s):s\in S,\ s\le x\},

and, for $x>0$ ,

h_S(x) =\operatorname{mex} \left( \{h_S(x-s):s\in S,\ s\le x\} \cup\{g_S(x)\} \right). \tag{1}

All three conjectured families claim that after their respective thresholds the pass option can be omitted:

h_S(x) =\operatorname{mex}\{h_S(x-s):s\in S\}. \tag{2}

Each assertion fails inside its explicitly claimed range.

Family (a). Choose $n=2$ , $a=1$ , so $S=\{1,4,5\}$ and the threshold is $6an+4a+1=17$ . Exact mex recursion gives

h_S(18)=4, \qquad \operatorname{mex}\{h_S(17),h_S(14),h_S(13)\} =\operatorname{mex}\{2,3,1\}=0.

Family (b). Choose $n=1$ , $a=1$ , so $S=\{1,3,5\}$ and the threshold is $a(2n+3)+1=6$ . Then

h_S(6)=1, \qquad \operatorname{mex}\{h_S(5),h_S(3),h_S(1)\} =\operatorname{mex}\{2,2,2\}=0.

Family (iii). Choose $n=2$ , $a=1$ , so $S=\{1,5,9\}$ and the threshold is $a(2n+5)+1=10$ . Then

h_S(10)=1, \qquad \operatorname{mex}\{h_S(9),h_S(5),h_S(1)\} =\operatorname{mex}\{2,2,2\}=0.

Moreover, direct induction in (1) proves, for every positive integer $a$ ,

g_{aS}(aj)=g_S(j), \qquad h_{aS}(aj)=h_S(j).

Therefore these counterexamples extend to all positive $a$ :

\begin{array}{c|c|c|c|c} \text{family}&n&S&x&\text{threshold}\\ \hline (a)&2&\{a,4a,5a\}&18a&16a+1\\ (b)&1&\{a,3a,5a\}&6a&5a+1\\ (iii)&2&\{a,5a,9a\}&10a&9a+1 . \end{array}

In every row $x$ meets its claimed threshold and (2) fails. Conjecture 3 explicitly quantifies $a,n\in\mathbb N$ ; the earlier restriction $n\ge3$ belongs to a different proved special case. These counterexamples concern the explicitly included small- $n$ cases; they do not assert failure after adding the extra restriction $n\ge3$ .

Source: Larsson, Manabe, and Miyadera, A Subtraction Nim with a Pass, equations (35)–(37) and Conjecture 3, https://arxiv.org/html/2605.14321 .