Given the number of partitions of $n$ into distinct parts $q(n)$, with the following generating function
$\displaystyle\prod_{m=1}^\infty (1+x^m) = \sum_{n=0}^\infty q(n)\,x^n\tag{1a}$
Which may be conveniently generalized into
$\displaystyle q_{k}(x)= \sum_{n=0}^\infty q((2k-1)n)\,x^n\tag{1b}$
for positive integers $k$
After some brute-force experimentation with wolfram mathematica, I discovered the following conjecture analogous to the one discussed in this post
The nth coefficient $q((2k-1)n)$ is odd if and only if
the power of the last term included in the partial sum of the generating function $(1b)$ is $n\stackrel{\mathrm{def}}{=}\frac{c}{2}$ where $c$ is the solution to this diophantine equation $d^2-12(2k-1)c=1$ which has two unknowns $d$ and $c$ for every positive integer $k\gt0$
Integer solutions to the diophantine equation above:
Given $e \in \mathbb{Z}$
$k=1: c=3e^2+e, c=3e^2+5e+2$
$k=2: c=9e^2+e, c=9e^2+17e+8$
$k=3: c=15e^2+e, c=15e^2+11e+2, c=15e^2+19e+6, c=15e^2+29e+14$
$k=4:\dots$
$\vdots$
For example let $k=3$, then we have the following generating function $q_3(x)=x^{\color{red}{0}}+3x^{\color{red}{1}}+10x^2+27x^{\color{red}{3}}+64x^4+142x^5+296x^6+585x^{\color{red}{7}}+1113x^{\color{red}{8}}+2048x^9+3658x^{10}+6378x^{11}+\dots$
And as per our conjecture we have the corresponding diophantine equation
$d^2-(12×5)c=1$
Then the coefficient $q(5n)$ from the generating function (1b) is odd iff the powers of x are
$\frac{15e^2+e}{2}: \color{red}{8}, \color{red}{31}, \color{red}{69}, \color{red}{122}, \color{red}{190},\dots$
$\frac{15e^2-e}{2}: \color{red}{7}, \color{red}{29}, \color{red}{66}, \color{red}{118}, \color{red}{185},\dots$
$\frac{15e^2+11e+2}{2}: \color{red}{14}, \color{red}{42}, \color{red}{85}, \color{red}{143}, \color{red}{216},\dots$
$\frac{15e^2-11e+2}{2}: \color{red}{3}, \color{red}{20}, \color{red}{52}, \color{red}{99}, \color{red}{161},\dots$
$\frac{15e^2+19e+6}{2}: \color{red}{20}, \color{red}{52}, \color{red}{99}, \color{red}{161},\dots$
$\frac{15e^2-19e+6}{2}: \color{red}{1}, \color{red}{14}, \color{red}{42}, \color{red}{85}, \color{red}{143},\dots$
$\frac{15e^2+29e+14}{2}: \color{red}{29}, \color{red}{66}, \color{red}{118}, \color{red}{185},\dots$
$\frac{15e^2-29e+14}{2}: \color{red}{0}, \color{red}{8}, \color{red}{31}, \color{red}{69}, \color{red}{122},\dots$
for integers $e\in \mathbb{N_0}$
Thus in general it is clear that the powers at which the coefficient $q((2k-1)n)$ is odd occur when $n\stackrel{\mathrm{def}}{=}\frac{c}{2}$ where $c$ is the solution of the general diophantine equation $d^2-12(2k-1)c=1$
Question: How do we prove the conjecture?
The pentagonal number theorem states
Note that
$$ ((6m-1)^2-1)/24 = m(3m-1)/2 = g_m. \tag1 $$
Equivalently, the Diophantine equation
$$ d^2 - 24n = 1 \tag2 $$
has only solutions $\,d = 6m-1\,$ with $\,n = g_m\,$ for some integer $m$. In other words,
$$ n = g_m \quad\text{iff}\quad m = \frac{1\pm\sqrt{1+24n}}6. \tag3 $$
Define the infinite product power series
$$ \prod_{j=1}^\infty (1 - x^j) = \sum_{n=0}^\infty a(n)x^n. \tag4 $$
The pentagonal number theorem is equivalent to the statement that $\, a(n) = 0\,$ iff $\,n \ne g_m\,$ otherwise, $\,a(n) = (-1)^m.\,$ This implies that $\,a(n) \equiv 1 \pmod 2\,$ iff $\, n = g_m\,$ for some $m$.
Define a related infinite product power series
$$ \prod_{j=1}^\infty (1 + x^j) = \sum_{n=0}^\infty q(n)x^n. \tag5 $$
Note that $\,a(n) \equiv q(n) \pmod 2 \,$ because $\, 1 \equiv -1 \pmod 2.$ This implies that $\,q(n)\,$ is odd iff $\,a(n)\,$ is odd and, as previously noted in equation $(2)$, iff $\,n = g_m\,$ for some $m$.
Thus, the conjecture regarding the parity of $\,q(n)\,$ is true.
For example, if $\,n = 15\,$ in equation $(2),\,$ then $\,d=19.\,$ Now $\,a(n) = -1\,$ is odd and also $\, q(n) = 27.\,$ Now $\,q(n) = q(1\cdot 15) = q(3 \cdot 5) = q(5\cdot 3) = q(15\cdot 1)\,$ which applies to the conjecture