Fix an integer d≥ 1. For an unordered partition (d_1,…,d_r)vdash d into strictly positive parts, let #operatornameAut(d_1,…,d_r) be the order of its automorphism group. Multiple-cover partition identity. One has ∑_(d_1,…,d_r)vdash dfrac2^r-1d^r-2#operatornameAut(d_1,…,d_r)∏_i=1^r frac(-1)^d_i-1d_i3d_i choose d_i=frac1d^24d-1 choose d. The identity is equivalent to the preceding multiple-cover formula and was verified computationally through d=50 in the source; no general proof is stated, so it remains open.

The partition identity for multiple-cover contributions

Fix an integer $d\geq 1$ . For an unordered partition $(d_1,\ldots,d_r)\vdash d$ into strictly positive parts, let $\#\operatorname{Aut}(d_1,\ldots,d_r)$ be the order of its automorphism group. Multiple-cover partition identity. One has

\sum_{(d_1,\ldots,d_r)\vdash d}\frac{2^{r-1}d^{r-2}}{\#\operatorname{Aut}(d_1,\ldots,d_r)}\prod_{i=1}^r\frac{(-1)^{d_i-1}}{d_i}{3d_i\choose d_i}=\frac{1}{d^2}{4d-1\choose d}.

The identity is equivalent to the preceding multiple-cover formula and was verified computationally through $d=50$ in the source; no general proof is stated, so it remains open.

Progress summary

Solved

A complete proof has been posted and the source is said to contain a proof, but neither claim has independent verification in the retrieved record.

The identity was originally checked computationally through $d=50$ without a general proof. The cited paper by Michel van Garrel, Navid Nabijou, and Yannik Schuler is the relevant source, but its retrieved abstract does not state the identity or its proof.

Posted attempt

A reader claims that the current paper proves the identity as Theorem $3.8$ and gives a complete formal-power-series proof: encode the partition sum by $A(x)=\sum_{m\geq1}(-1)^{m-1}\binom{3m}{m}x^m/m$ , identify $A(x)$ by Lagrange inversion, and apply inversion again to obtain the binomial coefficient on the right. The argument has not been independently verified.

Current status (as of August 2026): A complete proof is claimed, but the retrieved evidence does not independently verify it, so the identity is not certified resolved.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Michel van Garrel, Navid Nabijou and Yannik Schuler, “Gromov-Witten theory of bicyclic pairs”, arXiv:2310.06058 (2025).

1 week ago

Proof

The MathDB status reflects the original 2023 version of the source. The current version proves this identity as Theorem 3.8. Here is also a direct formal-power-series proof.

Put

a_m=\frac{(-1)^{m-1}}m{3m\choose m}, \qquad A(x)=\sum_{m\geq1}a_mx^m.

For partitions of length $r$ , the multinomial theorem gives

\frac1{r!}[x^d]A(x)^r = \sum_{\substack{(d_1,\ldots,d_r)\vdash d\\ \text{length }r}} \frac1{\#\operatorname{Aut}(d_1,\ldots,d_r)} \prod_{i=1}^r a_{d_i}.

Therefore the left-hand side of the conjectured identity equals

\frac1{2d^2}[x^d]\exp(2dA(x)).

(The omitted constant term does not contribute because $d\geq1$ .)

Let $T=T(z)$ be the formal series determined by

T=z(1+T)^3.

Lagrange inversion gives, for $m\geq1$ ,

[z^m]\,3\log(1+T) = \frac3m[u^{m-1}](1+u)^{3m-1} = \frac1m{3m\choose m}.

Now let $V=V(x)$ satisfy

V=x(1-V)^3.

Since $T(-x)=-V(x)$ , the preceding coefficient identity yields

A(x)=-3\log(1-V(x)).

Hence

\exp(2dA(x))=(1-V(x))^{-6d}.

Applying Lagrange inversion once more,

\begin{aligned} [x^d](1-V)^{-6d} &=\frac1d[u^{d-1}] 6d(1-u)^{-6d-1}(1-u)^{3d}\\ &=6[u^{d-1}](1-u)^{-3d-1}\\ &=6{4d-1\choose d-1}\\ &=2{4d-1\choose d}. \end{aligned}

Substitution therefore gives

\frac1{2d^2}[x^d]\exp(2dA(x)) = \frac1{d^2}{4d-1\choose d},

which is exactly the required identity.

The current source independently proves the conjecture geometrically as Theorem 3.8:

https://arxiv.org/abs/2310.06058

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Samuel Schlesinger · 5 days ago