Summing $n$ numbers so that they equal $0 \mod{n}$

Question

Let $A_n=\{(a_1,a_2,\dots,a_n) :\sum_{i=1}^na_i=0\mod{n}\}$, where $a_i\in[n-1]$. How many elements are in $A_n$? My initial attempt was a stars-and-bars argument.

For example, let $n=4$. Then we need to consider $4k$ stars ($k\in\mathbb{N}$) and $3$ bars.

If $k=1$, then there is only one way, namely $*|*|*|*$, giving $1+1+1+1$.

If $k\geq2$, we need the restriction that there may be no more than $3$ starts between any $2$ bars. I'm not sure how to account for this restriction.

Note that I consider different orderings of the sum as different summations, e.g. $1+2+3+2$ is different than $2+3+2+1$. However, a summation such as $2+2+1+3$, is the same as $2+2+1+3$ (the $2$s being switched).

@Omnomnomnom Yes, I consider that the only representation of $2+2+2+2$. — Bonnaduck, Apr 17 '16 at 19:59
Why did you say then that, for $k=1$, there is only 1+1+1+1? — Bib-lost, Apr 17 '16 at 22:18
@Bib-lost The bars represent addition, and the number of stars between the bars is the number that we are adding. If $k=1$ and $n=4$, we want $4$ natural numbers to add to $4$ -- the only way to do that is $1+1+1+1$. — Bonnaduck, Apr 17 '16 at 22:21
This is not very clear. If you're taking a sum of all $a_1$ through $a_n$ you will get the same thing no matter what the order - it's either $0$ mod $n$ or not $0$ mod $n$. So I don't know what you are asking. — Jair Taylor, Apr 17 '16 at 22:21
@JairTaylor Sorry, that is an error on my part. I will edit the question. — Bonnaduck, Apr 17 '16 at 22:24
Ah, I see. I think you mean the ordered tuple $(a_1, \ldots, a_n)$ not the set ${a_1, \ldots, a_n}$ since the order matters here. — Jair Taylor, Apr 17 '16 at 22:48
I would find the formula for the number of tuples with $\sum a_i = k(n+1)$ and then sum over $k$. — Jair Taylor, Apr 17 '16 at 22:50
If $n=4$ you asked that the sum be a multiple of $n+1=5$, so for $k=1$ the choices are the four orders of $1,1,1,2$ — Ross Millikan, Apr 17 '16 at 22:54
@RossMillikan It should have been $0\mod n$ with $a_i\in[n-1]$ — Bonnaduck, Apr 17 '16 at 22:57

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

You can reduce this to several computations where we count compositions with a bound on sizes of parts. See this previous Question and its related posts for ways to do these computations. Their asymptotics have been explored in this paper.

If $\sum_{i=1}^n a_i \equiv 0 \bmod{n}$, with each $1\le a_i \le n-1$, then the sum is a positive multiple of $n$, but not more than $(n-1)n$. Two selections of $a_i$ are clearly distinct if they give sums with distinct multiples of $n$.

So for each $k=1,\ldots,n-1$, we should count the compositions of $kn$ with parts less than $n$. Adding these together gives the count sought by this Question.

The number of such computations can be cut roughly in half by the observation that $\sum_{i=1}^n a_i = kn$ is an admissible composition of $kn$ if and only if $\sum_{i=1}^n (n-a_i) = (n-k)n$ is an admissible composition of $(n-k)n$. This means we only need to work out the counts for $k=1,\ldots,\lfloor (n-1)/2 \rfloor$, because the remaining counts have equivalence with these by the symmetry just explained.

Summing $n$ numbers so that they equal $0 \mod{n}$

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