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For a cell $\square$ in the Young diagram of a partition $\lambda$, let $h_{\square}$ and $c_{\square}$ denote the hook length and content of $\square$, respectively.

Define the functions $$f_n(t):=\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\frac{t+c_{\square}}{h_{\square}}.$$

QUESTION. Does this recurrence hold true? $f_0(t)=1, f_1(t)=t$ and $$(n+2)f_{n+2}(t)=tf_{n+1}(t)+(t^2+n)f_n(t).$$

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    $\begingroup$ It seems that $$\sum_{n=0}^\infty f_n(t) x^n = \frac{1}{(1-x)^t (1-x^2)^{\binom{k}{2}}}.$$ $\endgroup$
    – Ira Gessel
    Commented Mar 15, 2021 at 20:27
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    $\begingroup$ The generating function mentioned by Ira Gessel is the specialization of an identity of Littlewood: on one side is the sum of all Schur polynomials in a finite fixed set of variables and on the other side $\prod_{1\leq i\leq n}(1-x_i)^{-1}\prod_{1\leq i<j\leq n}(1-x_ix_j)^{-1}$. So $f_n(t)$ is counting semistandard tableau according to their evaluations (the other meaning of content). $\endgroup$
    – user35313
    Commented Mar 15, 2021 at 20:49
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    $\begingroup$ So together with @user61318's observation, Ira's formula follows from Stanley's Hook Content Formula. By the way for discussion of Littlewood's identity, in fact "bounded" versions of it, see arxiv.org/abs/1506.02755. $\endgroup$
    – Sam Hopkins
    Commented Mar 15, 2021 at 20:56
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    $\begingroup$ The generating function via Littlewood's identity was discussed here mathoverflow.net/questions/263643 $\endgroup$
    – Gjergji Zaimi
    Commented Mar 15, 2021 at 21:15

1 Answer 1

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Just to complete the picture, here is a wrap up.

The comments by user61318, Sam Hopkins and Gjergji Zaimi justify the generating function by Ira Gessel $$\sum_{n\geq0}f_n(2t)x^n=\frac1{(1-x)^{2t}(1-x^2)^{t(2t-1)}},$$ where I replace $t\rightarrow 2t$ for convenience. Expanding the RHS into a Taylor series, we get $$f_n(2t)=\sum_{k=0}^{\lfloor\frac{n}2\rfloor}\binom{2t^2-t-1+k}k\binom{2t-1+n-2k}{n-2k}.\tag1$$ So, we derive the recurrence $(n+2)f_{n+2}(2t)-2tf_{n+1}(2t)-(4t^2+n)f_n(2t)=0$ for the RHS of (1). Denote the summand in (1) by $F(n,k)$ (suppressing $t$) and introduce the function $$G(n,k)=-2k\binom{2t^2-t-1+k}k\binom{2t+1+n-2k}{n-2k+2}.$$ It's routine to check (preferably using a symbolic software) that \begin{align*} &(n+2)F(n+2,k)-2tF(n+1,k)-(4t^2+n)F(n,k) \\ =&G(n,k+1)-F(n,k). \tag2 \end{align*} Summing both sides of (2) over all integers and noting that the RHS telescopes to $0$ reveals the desired recursive relation.

The above proof-technique is called the Wilf-Zeiberger methodology.

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