upper bound for the number of integer solutions of a special equation

Question

we know that the number of integer solutions for the following equation with integers $m, r$:

$\begin{equation} z_1 + \cdots + z_m = 2m, \quad 1 \leq z_i \leq r \end{equation}$

is $N = \binom{2m-1}{m-1} - \binom{m}{1} \binom{2m-r-1}{m-1} + \binom{m}{2} \binom{2m-2r-1}{m-1} - \binom{m}{3} \binom{2m-3r-1}{m-1}+ \binom{m}{4} \binom{2m-4r-1}{m-1}- \cdots$

I am interesting in the asymptotic behavior of $N$ when $m$ goes to infinity. My first try is to derive a upper bound for $N$, but I do not know how to do that. Can anyone give me some hints?

Answer 1

Hint: it's always worth checking the Online Encyclopedia of Integer Sequences.

For $r=3$ the values of $N$ are the sequence A002426. There's a wealth of literature references, a number of comments which you could try generalising, and the asymptotic $N \sim \sqrt{\frac{3}{8\pi}} 3^m m^{-1/2}$.

For $r=4$ the sequence is A005725. There's a recurrence for the g.f., and the asymptotic $N \sim k \alpha^m m^{-1/2}$ where $k = \sqrt{\frac{39 (117+2\sqrt{78})^{1/3} +7\times 39^{2/3}+39^{1/3}(117+2\sqrt{78})^{2/3}}{156\pi(117+2\sqrt{78})^{1/3}}}$, $\alpha = \frac{(6371+624\sqrt{78})^{2/3}+11(6371+624\sqrt{78})^{1/3}+217}{12(6371+624\sqrt{78})^{1/3}}$.

For $r=5$ the sequence is A187925. There's a statement that the g.f. is $1 + x\frac{A'(x)}{A(x)}$ where $A(x) = \frac{1 - x^5 A(x)^5}{1 - xA(x)}$ which looks like a very interesting avenue of investigation, and the asymptotic $N \sim k \alpha^m m^{-1/2}$ where $\alpha = 3.834437249\ldots$ is a root of the equation $27\alpha^4 - 94\alpha^3 - 15\alpha^2 - 50\alpha - 125= 0$, $k = 0.340444098\ldots$

Some higher values of $r$ are also present, but their entries are rather spartan.