System of two linear Diophantine equations

Question

Let $n\in\mathbb{N}$ be a positive integer. Denote by $f(n)$ the number of integral solutions of the following system $$ \left\lbrace\begin{array}{l} \sum_{i=1}^nx_i = 3n; \\ \sum_{i=1}^n (2i-1)x_i = n(n+2); \\ x_1\geq x_2\geq \dots \geq x_n\geq 0. \end{array}\right. $$ Is there any result on the asymptotic behavior of $f(n)$?

If we remove from the system se second equations we get the partitions of $3n$ in at most $n$ parts and the question basically boils down to Solutions of a linear diophantine equation which has been fully answered.

Thank you.

Answer 1

Not a complete answer, more like an extended comment.

It is possible to remove the inequalities by introducing new variables $y_i\ge 0$ according to \begin{align} x_n&=y_n\\ x_{n-1}-x_n&=y_{n-1}\\ \ldots\\ x_{1}-x_2&=y_{1}\\ \end{align}

This system is easily solved for $x$ in terms of $y$: $x_i=\sum\limits_{j=i}^ny_j$.

Now the original equations can be restated in terms of $y_i\ge 0$: $$ \sum\limits_{i=1}^niy_i=3n $$ $$ \sum\limits_{i=1}^ni(i+1)y_i=n^2+5n $$ (For the second equation, the simplification given in the comment By Max Alekseev was used). Thus $f(n)$ is given by the coefficient of $z^{3n}q^{n^2+5n}$ in the power series expansion of the function $$ \prod_{k=1}^n\frac{1}{1-z^kq^{k(k+1)}}. $$ From this one sees that the problem is not a standard partition problem. But now one can use this generating function to calculate $f(n)$ and try to make some empirical observations about its behavior for large $n$.