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" $\,t_1t_2t_3\cdots=0\prod_{k=1}^{\infty}01^{2k}02^k$?

  • $\begingroup$ is not $b(\lambda'')$ simply the number of different (positive) parts in $\lambda$? $\endgroup$ – Fedor Petrov Mar 5 '20 at 18:58
  • $\begingroup$ @FedorPetrov: That is right, Fedor. $\endgroup$ – 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 $k\geqslant 2$.

  • $\begingroup$ Thank you, Fedor! $\endgroup$ – T. Amdeberhan Mar 6 '20 at 14:09
  • $\begingroup$ Following the generating function $A$, we may infer that $-f_n(-1)=(-1)^j$ when $n=j^2$ and $f_n(-1)=0$ otherwise. $\endgroup$ – T. Amdeberhan Mar 13 '20 at 2:16

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