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I have been stuck on this question for a while: Let $U(x)=\sum_{n=0}^{\infty} u_nx^n$, where $u_n$ is the number of partitions of $n$ into at most two parts. For example, $u_4=3$ because $4$ can be partitioned into at most two parts as $4$, $3+1$, or $2+2$. Use the convention that $u_0=1$.

Then $\frac 1{U(x)}$ is a polynomial. What polynomial is it? (Enter your answer in expanded form.)

I can't seem to get the polynomial. Could someone help me find the polynomial?

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    The main formulas have been already largely been analysed by Euler. Everything is [there](https://en.wikipedia.org/wiki/Partition_(number_theory)) – 2017-02-14
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    Yes, but I can't figure out how to solve this. – 2017-02-14
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    In the reference I have given it is in "partition function" – 2017-02-14
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    Yes, but I'm not using partitions to solve it, the desired method is generating functions. – 2017-02-14

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The number $u_n$ is simply $n+1$. So $U(x) = \sum_{n=0}^{\infty} (n+1) x^n = 1/(x-1)^2$. Polynomial in expanded form $x^2-2x+1$.

Correction: I had an error for $u_n$, the correct formula is $1+{\rm floor}(n/2)$. Split this in two cases: $n=2k$, then $u_n = 1+k$, and $n=2k+1$, then $u_k = 1+k$. Then $U(x) = U_{\rm even}(x) + U_{\rm odd}(x)$ where $U_{\rm even}(x) = \sum (1+k)x^{2k} = 1/(x^2-1)^2$ and $U_{\rm odd}(x) = x U_{\rm even}$. Then $U(x) = (x+1)/(x^2-1)^2$ and $1/U(x) = (x-1)^2(x+1) = x^3-x^2-x+1$.

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    I already tried that, but it's not right ;-; – 2017-02-17