Infinite product of $1-q^{n^2}$ Is there anything known about the following product? Is it a known function or related to a known function?
$$\prod_{n\geqslant1}(1-q^{n^2})$$
 A: Write $\theta(q)=\sum_{n \in \mathbb Z} q^{n^2}$. Taking logarithms, we get that the logarithm of your function is
$\sum_n \ln(1-q^{n^2}) = -\sum_n \sum_m \frac 1 m q^{n^2m} = -\sum_m \frac 1 m \frac {\theta(q^m)-1} 2 = \frac 1 2 \sum_{m=1}^\infty \frac {1 - \theta(q^m)} m$.
A: Following up on Sam's comment, A001156 in the OEIS gives the number of partitions of $n$ into squares; that generating function is the reciprocal of your product.  There's an equinumerous type of partitions given: partitions of $n$ where each part $k$ occurs a multiple of $k$ times (which contributes a multiple of $k^2$ to the weight).  You might look through the links and formula entries for the sequence to see if they lead to anything that interests you.
A: Let $m$ be a positive integer. Let $f_{m}(z)=\prod_{k=1}^{\infty}(1-z^{k^{m}})$. Then $f(z)$ is analytic in the disk $\{z:|z|<1\}$ and satisfies the following functional equation:
$$\prod_{j=1}^{\infty}f_{m}(z^{j^{m}})^{\mu(j)}=1.$$
Here $\mu$ denotes the Mobius Mu function which is defined by setting $\mu(M^{2}N)=0$ whenever $M,N$ are positive integers with $M>1$ and $\mu(p_{1}\dots p_{s})=(-1)^{s}$ whenever $p_{1},\dots,p_{s}$ are distinct primes.
