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I have been recently working with generating functions in my discrete mathematics course, and I was interested in one particular generating function. I want to find the generating function for the number of ways one can split $n$ into odd parts. I can see that the first coefficients of the sequence are $$0,1,1,2,2,3,4,5,...$$ And so on. I've been trying to find a recursion, and while it seemed originally that the number of ways looked to be the ceiling of $\frac{n}{2}.$ But it seems like the $6$th coefficient suggests otherwise. Is there another possible recursion going on in this sequence? If so, how would I go about finding the generating function for it?

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    $\begingroup$ The number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts; the generating function is $$\prod_{k\ge 1}\left(1+x^k\right)=\prod_{k\ge 1}\frac1{1-x^{2k-1}}$$ (see for example here). $\endgroup$ – Brian M. Scott Jun 13 '16 at 22:05
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The generating function for this sequence is

$f(x)=\prod\limits_{n=1}^{\infty} \cfrac{1}{1-x^{2n-1}}$

since

$f(x)=(1+x+x^{1+1}+\cdots)(1+x^3+x^{3+3}+\cdots)\cdots (1+x^{2n-1}+x^{(2n-1)+(2n-1)})\cdots = (1+x+x^2+\cdots)(1+x^3+x^{2\cdot3}+\cdots)\cdots (1+x^{2n-1}+x^{2(2n-1)})\cdots=\cfrac{1}{1-x}\cfrac{1}{1-x^3}\cdots\cfrac{1}{1-x^{2n-1}}\dots=\prod\limits_{n=1}^{\infty} \cfrac{1}{1-x^{2n-1}}$

Moreover, you can express this function as follows,

$\prod\limits_{n=1}^{\infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)\cdots=\cfrac{1-x^2}{1-x}\cfrac{1-x^4}{1-x^2}\cfrac{1-x^6}{1-x^3}\cdots=\cfrac{1}{1-x}\cfrac{1}{1-x^3}\cdots=f(x)$

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  • $\begingroup$ The explanation of the last expression of the function $f(x)$ needs more details, but I think it' s fine just like this. $\endgroup$ – richarddedekind Jun 13 '16 at 22:29

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