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.



Euler-like identity for partition function
If $F(x)$ is a $D$-finite power series then there are finitely many points $w\_1, \ldots, w\_n \in \mathbb{C}$ such that $F(x)$ extends to a meromorphic function defined on any simply-connected domain in $\mathbb{C}$ not containing any of the $w\_i$. (This is a restatement of (a) on page 185 of Stanley, *[Differentiably finite power series](http://math.mit.edu/~rstan/pubs/pubfiles/45.pdf)*, Europ. J. Combinat 1 (1980), 175-188.) 


Since the singularities of $\prod\_{i=1}^\infty (1-x^{2i-1})^{-1}$ are dense on the unit circle, this function is not $D$-finite. Similarly $\prod\_{i=1}^\infty (1-x^{2i-1})$ is not $D$-finite, since the Taylor expansion of any extension about a point of the unit circle is $0$.


