Generating function. Inverse.

Question

Let $D(x)= (x+1)(x^2 +1 ) (x^3 +1 ).... $ and let F(x) be inverse of $D(x)$

I know, that $ D$ is the number of ways to write n as a sum of positive integers without repeated summands. Sums only differing by the order of the summands are counted only once. But I don't see what is inverse for that. Help me.

$D$ is not "the number of ways...." $D$ is the generating function for the number of ways. Also, do you mean the inverse function, or do you mean $1/D(x)$? — Thomas Andrews, Jan 25 '15 at 22:27

score 2 · Answer 1 · edited Apr 13 '17 at 12:20

2

Using the result of this question, we have:

$$D(x)=\prod_{n=1}^\infty (1+x^n) = \prod_{n=1}^{\infty} \frac{1}{1-x^{2n-1}}$$

Then:

$$\frac{1}{D(x)} = \prod_{n=1}^{\infty} \left(1-x^{2n-1}\right)$$

So this is the generating function for $a_n=(-1)^np_n$ where $p_n$ is the number of ways to partition $n$ into distinct odd numbers.

Aside: Note that $\frac{1}{D(x)} = \frac{D(-x)}{D(x^2)}$.

edited Apr 13 '17 at 12:20

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1

answered Jan 25 '15 at 22:33

Thomas Andrews

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Generating function. Inverse.

1 Answers1