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

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

$$$$z_1 + \cdots + z_m = 2m, \quad 1 \leq z_i \leq r$$$$

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?

• To get bounds and asymptotics on the sequence $N(m)$ it is much easier to work with its generating function. Check e.g. Richard Stanley's Enumerative Combinatorics Commented May 13, 2021 at 5:14
• Thanks, I tried this method, it remains to bound the coefficient of $x^m$ on $\left(1+x+\cdots+x^{r-1}\right)^m$
– fs l
Commented May 13, 2021 at 5:56
• @fsI if $r$ is fixed, the asymptotics is obtained by the saddle point method Commented May 13, 2021 at 17:53

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.

• Thanks for sharing the idea, this is very helpful!
– fs l
Commented May 13, 2021 at 12:46
• @fsl, talking of literature references, A005725 points to an exercise in Comtet's Advanced Combinatorics which gives an integral expression for the coefficient of $x^k$ in $(1 + x + \cdots + x^{r-1})^m$. Commented May 13, 2021 at 19:46
• Thanks! I will have a look
– fs l
Commented May 14, 2021 at 1:57