Version C's subtract-three conjecture

Counterexample beyond the source's computationally verified range.

Take seven distinct pile sizes, with multiplicities

$$(a_1,a_2,a_3,a_4,a_5,a_6,a_7) =(2,2,2,2,2,2,3).$$

Every multiplicity is at least two, and this is not the explicitly excluded three-size position $(2,2,3)$. The total number of tokens is

$$\sum_{i=1}^7 i a_i =2(1+2+3+4+5+6)+3\cdot7 =63\equiv0\pmod3.$$

The conjecture therefore predicts that this is a losing position.

In fact it is a winning position. Remove one token from each of the two distinct size-one piles. This legal Version C move leaves

$$(0,2,2,2,2,2,3),$$

which is a losing position by exact backward induction.

For completeness, the entire certificate is reproducible with the following terminating recursion. A state records the multiplicities of piles of sizes $1,2,\ldots$; trailing zero multiplicities are deleted. Every legal move reduces the total token count by one or two.

from functools import cache

def norm(a):
    while a and a[-1] == 0:
        a = a[:-1]
    return a

def followers(a):
    for i, count in enumerate(a):
        if count == 0:
            continue
        b = list(a)
        b[i] -= 1
        if i:
            b[i - 1] += 1
        yield norm(tuple(b))
        for j in range(i, len(a)):
            if a[j] <= (i == j):
                continue
            b = list(a)
            b[i] -= 1
            b[j] -= 1
            if i:
                b[i - 1] += 1
            if j:
                b[j - 1] += 1
            yield norm(tuple(b))

@cache
def losing(a):
    return not any(losing(b) for b in followers(a))

a = (2, 2, 2, 2, 2, 2, 3)
b = (0, 2, 2, 2, 2, 2, 3)
assert b in set(followers(a))
assert losing(b)
assert not losing(a)
assert losing.cache_info().currsize == 35528

The recursion is exact because the empty position has no followers and every other position is losing exactly when every legal follower is winning. Its 35,528 states constitute a finite exhaustive certificate.

The paper reports verification only for at most five distinct pile sizes. The counterexample has seven, satisfies every stated multiplicity condition, and disproves the unrestricted subtract-three characterization.

Source: Danai, Ellis, and Thanatipanonda, Generalizing OOOOOOB, Conjecture 3.1, https://arxiv.org/html/2605.23213 .

Both directions fail. Consider also

$$(a_1,\ldots,a_7)=(2,2,2,2,2,2,4).$$

Again every multiplicity is at least two and the position is not the excluded example. Its token count is

$$2(1+2+3+4+5+6)+4\cdot7=70\equiv1\pmod3.$$

Thus the conjecture predicts a winning position, but the same exact backward-induction recursion above gives

$$\operatorname{losing}(2,2,2,2,2,2,4)=\mathrm{true}.$$

Both examples are verified by the identical recursion over a combined 71,835 states. Therefore divisibility by three is neither sufficient nor necessary for a losing position, with both failures occurring at seven distinct pile sizes beyond the authors' tested range.