The main question is: Are there exist infinitely many odd numbers and even numbers in $p(an+b)$? Where $an+b\ (n\geq1)$ is an arbitrary arithmetic sequence with $a\in\mathbb{Z}_{>0}$, $b\in\{0,\cdots,a-1\}$, and $p(n)$ denotes the partition function, the numbers of unrestricted representations of $n$ as a sum of positive integers.
The idea came from some known results. Srinivasa Ramanujan proved that
$$p(5n+4)\equiv0\ (\mathrm{mod}\ 5),$$
$$p(7n+5)\equiv0\ (\mathrm{mod}\ 7),$$
$$p(11n+6)\equiv0\ (\mathrm{mod}\ 11).$$
In 2000, Kenneth Ono proved that for all primes $p\geq5$, there exists congruence of this form. For example,
$$p(11864749n+56062)\equiv0\ (\mathrm{mod}\ 13),$$
$$p(48037937n+1122838)\equiv0\ (\mathrm{mod}\ 17).$$
Later, Scott Algren proved for all integers $m$ with $\mathrm{gcd}(m,6)=1$, there exists congruence of this form.
So it's natural to ask what happen when $\mathrm{gcd}(m,6)\neq1$. Is it no any Ramanujan-type Congruence exists?
We take $m=2$ as an example. It's equivalent to ask that whether any arithmetic sequences include both odd numbers and even numbers. We can consider a stronger version: Are there exist infinitely many odd numbers and even numbers in $p(an+b)$? It's exactly the question and it seems to be true from calculation. For $(a,b)=(1,0)$, we already know there exist infinitely many odd numbers and even numbers in $p(n)$ via works of Kohlberg.
We can ask a similar question for $m=3$ and etc. But the case $m=2$ is quite troublesome enough. Any comments, answers or links is appreciated. :)