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

 A: 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 $k\geqslant 2$.
