This is a follow up from [my earlier MO question](https://mathoverflow.net/questions/353715/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](http://mathworld.wolfram.com/ConjugatePartition.html) $\lambda'$. 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" $\,t\_1t\_2t\_3\cdots=0\prod\_{k=1}^{\infty}01^{2k}02^k$?
> 
> 
> 



Divisibility of polynomials over partitions
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 $k\geqslant 2$.


