# An identity for polynomials over partitions

Given an integer partition $$\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$$ of $$n$$ where $$\ell(\lambda)$$ is the length of $$\lambda$$, associate its conjugate partition $$\lambda'$$. Denote by $$\lambda''=\lambda',0$$ found by appending one extra zero at the right end of $$\lambda'$$. Further, define the following two numerics $$a(\lambda'')_j=\lambda_j''-\lambda_{j+1}''$$ for $$j=1,2,\dots,\ell(\lambda')$$ and also that $$b(\lambda'')=\#\{j: a(\lambda'')_j>0\}$$.

For example, if $$\lambda=(4,2,1)$$ then $$\lambda'=(3,2,1,1)$$ and $$\lambda''=(3,2,1,1,0)$$ and $$a(\lambda'')=(1,1,0,1)$$ and $$b(\lambda'')=3$$.

QUESTION. If $$n=2^m$$ then are these two polynomials equal? $$\sum_{\lambda\vdash n}(q-1)^{2b(\lambda'')}q^{n-\ell(\lambda)} \prod_{a(\lambda'')_j\geq1}\frac{q^{2a(\lambda'')_j}-1}{q^2-1}=(q-1)(q^{2n-1}-1).\tag1$$

Remark 1. To get some motivation, consider dividing the left-hand side of (1) by $$(q-1)^2$$, for any $$n\in\mathbb{N}$$. Taking the limit $$q\rightarrow1$$ in the resulting expression forces $$b(\lambda'')=1$$ which means the corresponding Young diagram of the partition $$\lambda'$$ (hence $$\lambda$$ itself) must be rectangular. Therefore, the final expression equals the sum of divisors (arithmetic) function $$\sum_{d\,\vert\, n}d.$$

Remark 2. I also observe that if $$q\rightarrow-1$$ in (1), then the left-hand side counts the number of ways of writing $$n\in\mathbb{N}$$ as a sum of two squares, which is this sequence $$r_2(n)$$.

• Some powers of $q-1$ cancel in the l.h.s. to become: $$\sum_{\lambda\vdash n} q^{n-\ell(\lambda)} \prod_{a(\lambda'')_j\geq1}\frac{(q^{2a(\lambda'')_j}-1)(q-1)}{q+1}$$ – Max Alekseyev Feb 28 at 2:12
• @MaxAlekseyev: that's nice. – T. Amdeberhan Feb 28 at 2:56
• Is this true for $n \neq 2^m$ too? – darij grinberg Feb 28 at 23:29
• Equation (1) seems true for $n=2^m$ while Remark 1 and 2 should hold for any $n\in\mathbb{N}$. Does that answer your question? – T. Amdeberhan Feb 29 at 2:17

Yes, your identity $$(1)$$ is true. We can give a proof as follows:
Let's denote the left hand side of your identity $$(1)$$ by $$A_n(q)$$. Starting with the identity $$\prod_{i\geq 1}\left(1+\sum_{r\geq 1}a_r(x_1x_2\cdots x_i)^r\right)=\sum_{\lambda}\left(\prod_{j\geq 1}a_{\lambda_j-\lambda_{j+1}}\right)\left(\prod_{j\geq 1}x_j^{\lambda_j}\right)$$ where $$a_0$$ is taken to be $$1$$, and the $$a_i, x_i$$ are formal variables, we make the substitutions $$a_r=(q-1)^2\frac{q^{2r}-1}{q^2-1}$$ for $$r\geq 1$$, $$x_1=t$$ and $$x_i=qt$$ for $$i\geq 2$$. This turns the right hand side into a generating function for the $$A_n(q)$$ where $$A_0(q)$$ is taken to be $$1$$. More specifically it gives $$\sum_{n\geq 0}A_n(q)t^n=\prod_{i\geq 1}\left(1+(q-1)^2\sum_{r\geq 1}\frac{q^{2r}-1}{q^2-1}(q^{i-1}t^i)^r\right)=\prod_{i\geq 1}\frac{(1-q^it^i)^2}{(1-q^{i-1}t^i)(1-q^{i+1}t^i)}$$ $$=(1-q)\frac{(qt;qt)^2_{\infty}}{(t;qt)_{\infty}(q;qt)_{\infty}}.$$ This final product has a Hecke-Rogers type expansion that was given by Andrews in "Hecke modular forms and the Kac-Peterson identities" (see Lemma 1).Using this expansion we get $$\sum_{n\geq 0}A_n(q)t^n=(1-q)\sum_{N\in \mathbb Z, r\geq |N|}(-1)^{r+N}q^{-N}(qt)^{\frac{(r+N)(r-N+1)}{2}}$$ and if we focus on $$n=2^m$$ we notice that the only way we can have $$(r+N)(r-N+1)=2^{m+1}$$ is if $$(r,N)=(2^m,2^m)$$ or $$(r,N)=(2^m, 1-2^m)$$. This means that $$A_{2^m}(q)=(1-q)(1-q^{2^{m+1}-1})$$ as desired. The observations in the remark can also be deduced from this last summation.