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Let P(x) be the partition generating function, and let \(R(x)=\sum_{n=0}^{\infty} r_nx^n\), where \(r_n\) is the number of partitions of n containing no 1s or 2s.

Then \(\displaystyle \frac{R(x)}{P(x)}\) is a polynomial. What polynomial(expanded form) is it?

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I find that the generating function for partitions is \(P(x)=\sum_{n=0}^{\infty} p_nx^n=1+x+2x^2+3x^3+5x^4+7x^5+11x^6+\cdots\)
I also know that \(P(x)=\frac{1}{\prod_{n=1}^{\infty}\left(1-x^n\right)},\) and \( \frac{1}{1-x}\cdot \frac{1}{1-x^2} \cdot \frac{1}{1-x^3}\cdot\ldots =1+x^1+\left(x^{1+1}+x^2\right)+\left(x^{1+1+1}+x^{2+1}+x^3\right)+\\+\left(x^{1+1+1+1}+x^{2+1+1}+x^{2+2}+x^{3+1}+x^4\right)+\ldots\)

How do I continue?

 Nov 12, 2018
 #1
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Since you exclude 1's and 2's, by the theory of partitions, R(x)/P(x) = 1 + x + x^2.

 Dec 8, 2020

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