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.

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    $\begingroup$ Would it be of any help that $f(x)f(-x) = f(x^2)$? $\endgroup$ – Ilya Jun 5 '15 at 15:08
  • $\begingroup$ It should be, but I don't see how it would yield explicit expansion. $\endgroup$ – Thanh Vu Jun 5 '15 at 15:25
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    $\begingroup$ By Gauss's identity $$ \frac{(1-x^2)(1-x^4)\ldots }{(1-x)(1-x^3) \ldots} = \sum_{k=0}^\infty x^{k(k+1)/2} $$ we have $$ \prod_{i=1}^\infty (1-x^{2i-1}) = \frac{1+\sum_{k=1}^\infty (-1)^k (x^{3k^2+k} + x^{3k^2-k})}{\sum_{k=0}^\infty x^{k(k+1)/2}}. $$ Not posting as an answer because while the power series on the right-hand side are sparse, they appear in a quotient. $\endgroup$ – Mark Wildon Jun 5 '15 at 16:32
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    $\begingroup$ Relevant: mathcs.emory.edu/~ono/publications-cv/pdfs/017.pdf $\endgroup$ – Lucia Jun 5 '15 at 17:24
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    $\begingroup$ Correction: $f_n$ is the difference between the number of partitions of $n$ into an even number of distinct odd parts and the number of partitions of $n$ into an odd number of distinct odd 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 18:35

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, 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 $0$.


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