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
$G(q)=\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q(1-q^2)^2}{1-q^5+\cfrac{q(1-q^3)^2}{1-q^7+\cfrac{q(1-q^4)^2}{1-q^9+\dots}}}}}$
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.