I'd like to know if the infinite product can be expanded as an infinite sum:
$\prod_{n=1}^{\infty}\frac{(1-x^n)(1-x^{adn})}{(1-x^{dn})}$
where $a$ and $d$ are natural numbers, thanks.
Actuelly this infinite product may be solved by some generalisation of Jacobi triple product, which I am looking for, I got this product when I calculate some infinite products of some solutions of Mahler functions. To solve it I tried some basic method used for Jacobi triple product, I considered the function as $f(t)=\prod(1-x^n)(1+tx^{dn}+t^2x^{2dn}...+t^{a-1}x^{(a-1)dn})$ and tried to calculate the coefficients of each $t^n$ but it seems that all the coefficients are power sequences and I can not continue..
Special cases:
$a = 0$ or $d = 0$: the factor of $(1 - x^{adn})$ annihilates the entire product.
$a = 1$: $(1-x^{adn})(1-x^{dn})^{-1}$ cancels leaving $\prod_{n=1}^\infty(1-x^n)$, so we have Euler's pentagonal number sequence: $\sum_{k=-\infty}^\infty (-1)^k x^{k(3k+1)/2}$
$d = 1$: $(1-x^n)(1-x^{dn})^{-1}$ cancels leaving $\prod_{n=1}^\infty (1-(x^a)^n)$, so we have an inflated form of Euler's pentagonal number sequence: $\sum_{k=-\infty}^\infty (-1)^k x^{ak(3k+1)/2}$
$a = d = 2$: $\prod_{n=1}^\infty(1-x^n)(1 + x^{2n}) = (x)_\infty (-x^2; x^2)_\infty$. A search for the first 40 terms in OEIS turns up exactly one hit: A106459, the coefficients of $f(-x, -x^3)$ where $f$ is Ramanujan's general theta function $$f(a,b) = \sum_{k=-\infty}^{\infty} a^{k(k+1)/2} b^{k(k-1)/2} = (-a;ab)_\infty (-b;ab)_\infty (ab;ab)_\infty$$ So $f(-x, -x^3) = (x;x^4)_\infty (x^3;x^4)_\infty (x^4;x^4)_\infty$ and it's fairly easy to regroup the terms $(1-x^{4k+1})(1-x^{4k+3})(1-x^{4k+4}) = (1-x^{4k+1})(1-x^{4k+3})(1-x^{2k+2})(1+x^{2k+2})$ and show that it is indeed a match, so that the sum is $\sum_{k=-\infty}^{\infty} (-1)^{k^2} x^{k(2k-1)}$.
I have not found any hits in OEIS with $a+d > 4$.