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

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.