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Business, 30.09.2019 23:20 gafran

Let $$p(x)=\sum_{n=0}^{\infty} p_nx^n=1+x+2x^2+3x^3+5x^4+7x^5+11x^ 6+\cdots$$ be the partition generating function, and let $q(x)=\sum_{n=0}^{\infty} q_nx^n$, where $q_n$ is the number of partitions of $n$ containing no $1$s. then $\frac{q(x)}{p(x)}$ is a polynomial. what polynomial is it?

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Let $$p(x)=\sum_{n=0}^{\infty} p_nx^n=1+x+2x^2+3x^3+5x^4+7x^5+11x^ 6+\cdots$$ be the partition gener...

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