Fayers' divisibility conjecture for $(s,s+t,s+2t,\dots)$ -core polynomials

About 1 year old · traced to 2025

Let $s$ and $t$ be positive integers with $\gcd(s,t)=1$ , and let $f_t(s)$ be the polynomial appearing in Fayers' formula for the number of $(s,s+t,s+2t,\dots)$ -core partitions. Fayers' divisibility conjecture. Suppose $t \geq 2$ . Then $f_t(s)$ is divisible by $s+t+(-1)^t$ .

This is the second part of Fayers' conjecture and remains open according to the source.

References

Primary source

William Keith, Rishi Nath and James Sellers, “On simultaneous (s, s+t, s+2t, )-core partitions”, arXiv:2508.00074 (2025).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2025 paper established the polynomial formula, while a reader has posted a complete proof of the divisibility claim that has not been independently checked.

Fayers conjectured that, for coprime positive integers $s,t$ with $t\geq2$ , the polynomial $f_t(s)$ counting these core partitions is divisible by $s+t+(-1)^t$ . Keith, Nath, and Sellers established the polynomial part of the conjecture in 2025, but their paper explicitly records the divisibility statement as open.

Known results

Keith, Nath, and Sellers (2025): the count equals $2^{s-t}f_t(s)/t!$ when $\gcd(s,t)=1$ .
They proved that $f_t(s)$ is monic of degree $t-1$ with nonnegative integer coefficients.
They gave the explicit Lah-number formula $f_t(s)=\sum_{m=1}^{t}L(t,m)(s+1)^{\overline{m-1}}$ , where $L(t,m)=\frac{t!}{m!}\binom{t-1}{m-1}$ .
The same paper states that $s+t+(-1)^t\mid f_t(s)$ remains open.

Posted attempt

A reader claims a complete proof by extracting a coefficient from the generating function and evaluating it at $s=-t-(-1)^t$ , with a further claim about real roots and interlacing. This proof has not been independently verified and is not supported by the primary paper.

Current status (as of August 2026): The polynomial formula and first part are established; the divisibility conjecture has a posted complete-proof claim, but no independent verification, so it remains unresolved.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

Complete proof, with prior generating-function credit. Keith, Nath, and Sellers, Discrete Mathematics 349 (2026), 114958, Theorem 1.8 and Conjecture 1.14, prove the explicit Lah-polynomial formula

f_t(s)=\sum_{m=1}^{t}\frac{t!}{m!}\binom{t-1}{m-1}(s+1)^{\overline{m-1}}\in\mathbb Z[s].

The associated generating function was already recorded by Vladeta Jovovic in 2003 in OEIS A079638. The new step below is its parity-dependent negative-diagonal extraction, which proves the previously open divisibility.

Reordering the coefficientwise finite sums, with $y=z/(1-z)$ , gives

\begin{aligned} \sum_{t\ge1}\frac{f_t(s)}{t!}z^t &=\sum_{m\ge1}\frac{(s+1)^{\overline{m-1}}}{m!} \left(\frac z{1-z}\right)^m\\ &=\frac1s\left[\left(\frac{1-z}{1-2z}\right)^s-1\right]. \end{aligned}

The apparent division by $s$ is removable coefficientwise; the identity follows formally by differentiating in $y$ .

Fix $t\ge2$ , and put

s_0=-t-(-1)^t,\qquad M=-s_0=t+(-1)^t>0.

Since $s_0\ne0$ , it suffices to show

[z^t]\left(\frac{1-2z}{1-z}\right)^M=0.

Make the invertible formal change $w=z/(1-z)$ , so that $z=w/(1+w)$ and $dz=dw/(1+w)^2$ . Formal residue substitution yields

[z^t]\left(\frac{1-2z}{1-z}\right)^M =[w^t](1+w)^{t-1}(1-w)^M.

If $t$ is odd, then $M=t-1$ , and the displayed coefficient is

[w^t](1-w^2)^{t-1}=0.

If $t=2h$ is even, then $M=t+1$ , and the coefficient is

\begin{aligned} [w^{2h}](1-w^2)^{2h-1}(1-w)^2 &=(-1)^h\binom{2h-1}{h} +(-1)^{h-1}\binom{2h-1}{h-1}\\ &=0. \end{aligned}

Consequently $f_t(-t-(-1)^t)=0$ for every $t\ge2$ , and monic polynomial division in $\mathbb Z[s]$ gives

\boxed{\quad s+t+(-1)^t\mid f_t(s)\quad\text{in }\mathbb Z[s],\qquad t\ge2.\quad}

A further strengthening follows from the recurrence already proved in the same source,

f_t(s)=(s+3(t-1))f_{t-1}(s)-2(t-1)(t-2)f_{t-2}(s).

Indeed, $f_t(s)=\det(sI_{t-1}-J_{t-1})$ , where the real symmetric tridiagonal matrix $J$ has diagonal entries $-3j$ and consecutive off-diagonal entries $\sqrt{2j(j+1)}$ . Its nonzero off-diagonal entries imply simple real eigenvalues, while the Lah formula gives strictly positive polynomial coefficients. Thus every root of $f_t$ is simple and strictly negative, the prescribed root is simple, consecutive polynomials strictly interlace, and the quotient $f_t(s)/(s+t+(-1)^t)$ has strictly positive integer coefficients.