# Hook lengths, contents and recurrence

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 $$(n+2)f_{n+2}(t)=tf_{n+1}(t)+(t^2+n)f_n(t).$$

• It seems that $$\sum_{n=0}^\infty f_n(t) x^n = \frac{1}{(1-x)^t (1-x^2)^{\binom{k}{2}}}.$$ Mar 15 at 20:27
• 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).
– user35313
Mar 15 at 20:49
• 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. Mar 15 at 20:56
• The generating function via Littlewood's identity was discussed here mathoverflow.net/questions/263643 Mar 15 at 21:15

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.