Number of solutions of $x_1 + 2x_2 + 3x_3 + \cdots + n x_n = p$

Question

Given a non-negative integer $p$. What is the number of solutions of $x_1+2x_2+3x_3 + \cdots + nx_n = p$, where the $x_i$'s are non-negative integers.

Can we answer this by using number of solutions of $x_1+x_2+x_3 + \cdots + x_m = q$ for any $m,q$?

Answer 1

The number of solutions of $x_1+2x_2+ \cdots +nx_n=p$ is given by the coefficient of $x^{p}$ in \begin{eqnarray*} \frac{1}{(1-x)(1-x^2) \cdots (1-x^n) } \end{eqnarray*}

The number of solutions of $x_1+x_2+ \cdots +x_m=q$ is given by the coefficient of $x^{q}$ in \begin{eqnarray*} \frac{1}{(1-x)^m } \end{eqnarray*} So the answer to your question is no.

Answer 2

The number of nonnegative integer solutions of $x_1 + 2 x_2 + \cdots + n x_n = p$, where $p \geq 0$, is the coefficient of $t^p$ in the following generating function [JDL]

$$\prod_{k=1}^n \frac{1}{1-t^k} = \frac{1}{(1-t) (1-t^2) \cdots (1-t^n)}$$

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[JDL] Jesús A. De Loera, The Many Aspects of Counting Lattice Points in Polytopes.