On the post Is there an explanation for the behaviour of this finite continued fraction in connection with prime numbers? I asked this question in a non-generalized form focusing only on the $4th$ partial convergent but now generalizing the problem and clarifying it ,we have

Given the continued fraction which satisfies the property proposed in one of my old posts


and $kth$ partial convergent of the continued fraction

$\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q(1-q^2)^2}{1-q^5+\cfrac{q(1-q^3)^2}{\ddots+\cfrac{q(1-q^k)^2}{1-q^{2k+1}}}}}}=\exp\Big(\sum_{n=2}^{\infty} (-1)^n\phi_{k}(n)\,q^n\Big)$

where $|q|\lt\frac{1}{4}$,and $\phi_{k}(n)$ is our symbol of choice that represents coefficients of the series(please note that it doesn't represent any standard function) depending on the $kth$ partial convergent $k\gt2$.

For $k\gt2$,every partial convergent of the continued fraction seems to have the property that:

For all values of $n$ but a few(that are exceptions to the rule) ,$\phi_{k}(n)$ is integer when $n$ is prime and non-integer when $n$ is composite.

For example on the $7th$ partial convergent of the continued fraction,there's only one exception $n=15$ in $1\lt n\lt200$

Formally we may call $\phi_{k}(n)$ an arithmetic function which returns an integer when $n$ is a prime number and non-integer when $n$ is a composite number for all natural numbers $n$ but a few for $k\gt2$.

So the question is

Why is $\phi_{k}(n)$ integer when $n$ is prime and non-integer when it is composite for all values of $n$ but a few for $k\gt2$?

We may be led to conjecture that whenever $n=prime$,the arithmetic function $\phi_{k}(n)$ is always integer.

  • $\begingroup$ Perhaps a case of the "law of small numbers". Extend the search in order to see whether this is the case. $\endgroup$
    – Peter
    Aug 9, 2017 at 14:19
  • $\begingroup$ It would help to write it as a sequence of $q$-series and see what are their coefficients $\endgroup$
    – reuns
    Aug 12, 2017 at 8:04
  • $\begingroup$ @reuns:I have tried it already ,but unfortunately the OEIS doesn't seem to recognize it.Moreover the radius of convergence for q-series is the unit circle $|q|\lt1$ ,while for this particular series is $|q|\lt\frac{1}{4}$ $\endgroup$
    – Nicco
    Aug 12, 2017 at 8:13
  • $\begingroup$ Did you find an induction rule for the (coefficients of) the sequence of $q$-series ? $\endgroup$
    – reuns
    Aug 12, 2017 at 8:17

1 Answer 1


Take any power series on the form $F(q)=1+a_2q^2+a_3q^3+\cdots$ with integer coefficients: ie, the constant term is $1$ and the linear term is zero, otherwise it is a general power series. Your $G(q)$ takes this form, as do the $k$th partial convergents.

Next, rewrite $F(q)$ on the form $$ F(q)=\prod_{k=2}^{\infty}(1+\alpha_k q^k)=(1+\alpha_2 q^2)(1+\alpha_3 q^3)\cdots $$ which can always be done and gives $\alpha_i\in\mathbb{Z}$.

Now, for $F(q)=\exp f(q)$, you ask why $f(q)=f_1q+f_2q^2+\cdots$ gives integer coefficients for $f_n$ whenever $n$ is a prime, but not when $n$ is composite.

To see why, write $f(q)=\ln F(q)$ and take the power expansion $$ f(q) = \ln F(q) = \sum_{k=2}^\infty \ln(1+\alpha_k q^k) = \sum_{k=2}^\infty \sum_{i=1}^\infty (-1)^{i-1}\frac{\alpha_k^i q^{ki}}{i}. $$ Contributions to the coefficient of $q^n$ come from pairs $(k,i)$ where $ki=n$. If $n$ is prime, then $k=n$ and $i=1$: the opposite, $k=1$ and $i=n$, does not contribute as $k\ge2$ (or, equivalently, the coefficient $\alpha_1=0$). When $k=n$ and $i=1$ is the only contributing term to the coefficient of $q^n$, the coefficient $f_n=\alpha_n$.

When $n$ is not prime, there may be terms contributing to the coefficient of $q^n$ where $i>1$ and which may therefore be non-integral.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.