$$G(q)=\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q(1-q^2)^2}{1-q^5+\cfrac{q(1-q^3)^2}{1-q^7+\cfrac{q(1-q^4)^2}{1-q^9}}}}}=\exp(\sum_{n=2}^{\infty} (-1)^n\phi(n)\,q^n)$$

where $|q|\lt1$,and $\phi(n)$ represents the nth coefficients of the series.

Why is $\phi(n)$ integer only when $n$ is prime and non-integer when it is composite?

I made this simple observation using mathematica,a disproof (by providing a counterexample) is very much welcome.

  • $\begingroup$ what is $\phi(n)$? $\endgroup$ Sep 13, 2016 at 16:49
  • $\begingroup$ @user2520938 :$\phi(n)$ is just a symbol I chose to represent the nth coefficients of the series $\endgroup$
    – Nicco
    Sep 13, 2016 at 16:53
  • $\begingroup$ Oke. It's an interesting observation. $\endgroup$ Sep 13, 2016 at 16:58

1 Answer 1


It is a really interesting observation. But sadly I found a counter-example.
As you may know, we can put the following code into Mathematica for the series generation.


where n is the order of the series.

I notice that when $n=143$, $\phi(n)=66650203204753876953026636747858671380118102453197443680828893588$, which is an integer. However, $143=11 \cdot 13$ and thus $143$ is not a prime number.

Therefore, the proposition has been disproved.

I am not able to explain this phenomenon completely, but Ramanujan tau function may help. It is defined by the following identity:

  • $\begingroup$ @ Mc Cheng :thanks for finding the counterexample. But I'm still wondering why does it hold for so many values,there must be a mathematical explanation. $\endgroup$
    – Nicco
    Sep 18, 2016 at 14:21
  • $\begingroup$ @Nicco I will try my best to find an explanation. By the way, I have found some more counter-examples, for instance, 279 and 328. $\endgroup$
    – Mc Cheng
    Sep 18, 2016 at 15:36
  • $\begingroup$ @ Mc Cheng :very interesting $\endgroup$
    – Nicco
    Sep 18, 2016 at 15:48

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