Following is the wonderful Euler's partition identity: $$\prod_{i=1}^\infty (1 - x^i) = 1 + \sum_{k=1}^\infty (-1)^k \left (x^{(3k^2-k)/2} + x^{(3k^2+k)/2} \right )$$

I'm wondering if there is similar expansion for infinite product $$\prod_{i=1}^\infty (1 - x^{2i-1})$$

We know that the inverse of that is the generating function for partitions with odd parts.

Edit: After a few computation, the non-zero coefficients seem very dense and quite arbitrary, so an explicit formula might not be plausible. My main question is whether this function is $D$-finite. The notion of $D$-finite function is defined in Stanley's book. From the Euler's formula, we see that the function $$\prod (1-x^i)$$ is not $D$-finite.

distinctodd parts and the number of partitions of $n$ into an odd number ofdistinctodd parts. An even number cannot be the sum of an odd number of odd parts, and an odd number cannot be the sum of an even number of odd parts, so (up to replacing $x$ with $-x$), this is simply the generating function for the number of partitions of $n$ into distinct odd parts. Unhelpful observation: this is well known to be the same as the number of self-conjugate partitions of $n$. $\endgroup$ – Dave Witte Morris Jun 5 '15 at 18:351more comments