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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.

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    $\begingroup$ Please define what is on the left-hand side of the equality (with the infinite sum)... Thanks! $\endgroup$ – peter a g Aug 13 '15 at 13:19
  • $\begingroup$ @peter a g it's a q-pochhammer symbol $\endgroup$ – Nicco Aug 13 '15 at 13:27
  • $\begingroup$ OK : from en.wikipedia.org/wiki/Q-Pochhammer_symbol: $(a;q)_\infty = \prod_{k=0}^{\infty} (1-aq^k).$ $\endgroup$ – peter a g Aug 13 '15 at 13:46
  • $\begingroup$ By Fermat's little theorem: $(a + b )^p \equiv a^p + b^p \pmod p$. In particular, "mod p", it's not surprising that the power series only has non-zero $q$-coefficients at $q^{pr}$... $\endgroup$ – peter a g Aug 13 '15 at 13:59
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    $\begingroup$ I 'mis-spoke' - it's not Fermat's little theorem (I think the first time I learnt it, the lecturer (jaundiced, no doubt, from having taught calculus too often) called it "the mod p Freshman's dream"). In any event, it's true: the 'middle' binomial coefficients are divisible by p, because the prime $p$ appears in the numerator, but not in the denominator, so the middle binomial coefficients vanish, mod p. Hence $(a + b )^p \equiv a^p + b^p \pmod p$. $\endgroup$ – peter a g Aug 13 '15 at 14:14
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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.

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