conjecture about primes and a certain q-series. Using wolfram Mathematica ,I observed an interesting and surprising property concerning prime numbers and q-series which I could not prove.Yet there is strong evidence supporting it.
I would be happy if anyone can find a counter-example.On the other hand,if it could turn out to be true,it would no doubt have amazing consequences in prime number theory. 
Let $p$ be an odd prime and $$\frac{1}{(q;q^4)_{\infty}^p}=\prod_{n=0}^{\infty}\frac{1}{(1-q^{4n+1})^p}=1+\sum_{n=1}^{\infty} \phi(n)\,q^n$$ 
with the usual convention $$(a;q)_{\infty}=\prod_{k=0}^{\infty} (1-aq^{k})$$ then
$$\phi(n)\equiv 0\pmod{p}$$ is true for all natural numbers $\{1,2,3,\dots\}$ except at multiples of $p$. For example,
$$\frac{1}{(q;q^4)_{\infty}^3}=1 + 3q +6q^2 + \color{brown}{10}q^3 + 15q^4 + 24q^5+ \color{brown}{37}q^6+\dots$$
$$\frac{1}{(q;q^4)_{\infty}^5}=1 + 5q +15q^2 + 35q^3 + 70q^4 + \color{brown}{131}q^5+ 235q^6+405q^7+\dots$$
Note: $\phi(n)$ is just a notation I chose as a matter of personal taste.
Can any one try to prove the conjecture or rather verify it by numerical methods,thanks in advance.

AFAICT my answer in the reasked version shows that the $q^{27}$ term of $\dfrac{1}{(q;q^4)_{\infty}^3}$ and $q^{65}$ term of $\dfrac{1}{(q;q^4)_{\infty}^5}$ are counterexamples to the conjectured non-divisibility.
