# Is there an explanation for the behaviour of this finite continued fraction in connection with prime numbers?

Given

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

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.

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

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.

Series[Log[1/(1-q+(q(1-q)^2)/(1-q^3+(q(1-q^2)^2)/(1-q^5+(q(1-q^3)^2)/(1-q^7+(q(1-q^4)^2)/(1-q^9)))))],{q,0,n}]


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:
$$\sum_{n\ge1}\tau(n)q^n=q\prod_{n\ge1}(1-q^n)^{24}=\eta(z)^{24}=\Delta(z)$$

• @ 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. – Nicco Sep 18 '16 at 14:21
• @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. – Mc Cheng Sep 18 '16 at 15:36
• @ Mc Cheng :very interesting – Nicco Sep 18 '16 at 15:48