I posed a conjecture about odd primes and a certain q-serieshere.I thought it would be more appropriate ,if I could ask the converse of the aforementioned problem .
Is $p$ an odd prime iff $$\prod_{n=0}^{\infty}\frac{1}{(1-q^{4n+1})^p}=1+\sum_{n=1}^{\infty}\phi(n)\,q^n$$
such that $$\phi(n)\equiv 0\pmod{p}$$ is true for all natural numbers $n\in\mathbb{N}$ except at multiples of $p$.
I've experimentally verified the conjecture for many values of p,and it seems to hold.Independent verification would be greatly appreciated.This conjecture seems to point to a deep connection between primes and modular forms,which I was not able to prove.Outlining a proof or connection between the particular q-series above and odd primes will be appreciated.
In the ring of formal power series $\Bbb{Z}[[q]]$ the left hand side is (by Freshman's Dream) $$ \sum_{k=0}^\infty \phi_p(k)q^k=\prod_{n=0}^{\infty}\frac{1}{(1-q^{4n+1})^p}\equiv\prod_{n=0}^{\infty}\frac{1}{(1-q^{p(4n+1)})}\pmod p. $$ This is clearly a power series in $q^p$, so $\phi_p(k)$ is divisible by $p$ whenever $k$ is not divisible by $p$.
Let $p=3$. We have $$ \psi(Q):=\prod_{n=0}^\infty\frac1{1-Q^{4n+1}}=\frac1{(1-Q)(1-Q^5)(1-Q^9)}+\ \text{terms with $Q^{13}$ or higher powers.} $$ Expanding this gives $$ \begin{aligned} \frac1{(1-Q)(1-Q^5)(1-Q^9)}&=(1+Q+Q^2+\cdots)(1+Q^5+Q^{10}+\cdots)(1+Q^9+\cdots)\\ &=1+Q+Q^2+Q^3+Q^4+2Q^5+2Q^6+2Q^7+2Q^8+3Q^9\\ &+\ \text{terms with $Q^{10}$ or higher}. \end{aligned} $$ Again the Freshman's dream implies that $$ \sum_{k=0}^\infty \phi_3(k)q^k=\prod_{n=0}^{\infty}\frac{1}{(1-q^{4n+1})^3}\equiv\psi(q^3)\pmod 3. $$ Here $$ \psi(q^3)=1+q^3+q^6+q^9+q^{12}+2q^{15}+2q^{18}+2q^{21}+2q^{24}+3q^{27} $$ plus higher degree terms. Therefore $\phi_3(27)$ is divisible by three even though $27$ is also divisible by three. So we cannot say that $\phi_3(k)$ is divisible by three if and only if $3\nmid k$.