# Divisibility of polynomials over partitions

This is a follow up from my earlier MO question.

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 numeric $$b(\lambda'')=\#\{j: \lambda_j''-\lambda_{j+1}''>0\}$$.

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

Consider the polynomials $$f_n(q):=\sum_{\lambda\vdash n}(q-1)^{b(\lambda'')-1}\,q^{\ell(\lambda)-b(\lambda'')}.\tag1$$

Denote by $$t_n$$ the largest $$t$$ such that $$q^t$$ divides $$f_n(q)$$.

QUESTION 1. Is it true that $$t_n\in\{0,1,2\}$$?

QUESTION 2. (stronger) Is it true that the "infinite word" $$\,t_1t_2t_3\cdots=0\prod_{k=1}^{\infty}01^{2k}02^k$$?

• is not $b(\lambda'')$ simply the number of different (positive) parts in $\lambda$? – Fedor Petrov Mar 5 '20 at 18:58
• @FedorPetrov: That is right, Fedor. – T. Amdeberhan Mar 5 '20 at 19:00

Both are true and these follow routinely from Euler's Pentagonal Theorem (PT). We have \begin{align} A:&=1+\sum_{n=1}^\infty (q-1)f_n(q)x^n\\ &=\prod_{j=1}^\infty(1+(q-1)x^j+q(q-1)x^{2j}+q^2(q-1)x^{3j}+\ldots) \\ &= \prod_{j=1}^\infty\frac{1-x^j}{1-qx^j}. \end{align} Consider it modulo small powers of $$q$$. Modulo $$q$$ we get $$\prod(1-x^j)=1-x-x^2+x^5+x^7-\dots$$, so the zeroes of the sequence $$(t_n)$$ are as you predict in Question 2 due to PT.
Modulo $$q^2$$ we get \begin{align} A&\equiv\prod (1-x^j)\cdot(1-q(x+x^2+\ldots)) \\ &=(1-x-x^2+x^5+\ldots)-q(x-x^3-x^4-\ldots)),\end{align} and see where are $$t_i$$'s equal to 1 (that is, which coefficients of $$x^i$$ are divisible by $$q$$ but not by $$q^2$$). It is again as predicted in Question 2.
Finally to confirm the conjecture in Question 2, we should prove that the coefficients $$[x^m]A$$ which are divisible by $$q^2$$ (this happens when $$m=k(3k-1)/2+\ell$$, $$1\leqslant \ell\leqslant k-1$$ for certain integer $$k\geqslant 2$$) are not divisible by $$q^3$$. We have modulo $$q^3$$: $$\prod (1-qx^j)^{-1}\equiv \prod(1+qx^j+q^2x^{2j})\equiv 1+q\sum_{j=1}^\infty x^j+q^2\sum_{s=0}^\infty\lfloor s/2\rfloor x^s.$$ Multiplying this by Euler's product $$\prod(1-x^j)$$ we get modulo $$q^3$$: $$[x^m]A\equiv q^2\left(\sum_{i=0}^k (-1)^i\left\lfloor\frac{m-i(3i-1)/2}2\right\rfloor+ \sum_{i=1}^{k-1} (-1)^i\left\lfloor\frac{m-i(3i+1)/2}2\right\rfloor\right).$$ We substitute the formula $$\lfloor x/2\rfloor=x/2-1/4+(-1)^x/4$$ in above sums. Since $$\sum_{i=0}^k (-1)^i+\sum_{i=1}^{k-1}(-1)^i=0$$, and $$\sum_{i=0}^k (-1)^{i+1} \frac{i(3i-1)}4+ \sum_{i=1}^k (-1)^{i+1} \frac{i(3i+1)}4=(-1)^{k+1}k/2,$$ we get the expression $$(-1)^{k+1}k/2+\frac14\sum_{i=0}^k (-1)^{m+i(3i+1)/2}+ \frac14\sum_{i=1}^{k-1} (-1)^{m+i(3i-1)/2}$$ which is obviously non-zero for $$k\geqslant 2$$.
• Following the generating function $A$, we may infer that $-f_n(-1)=(-1)^j$ when $n=j^2$ and $f_n(-1)=0$ otherwise. – T. Amdeberhan Mar 13 '20 at 2:16