I came across the following alternating sum $$ \sum_{k=0}^n (-1)^k \binom{2n}{k} (n-k)^r,\quad 1\leq r < n. $$ It seems that when $r$ is an even integer the sum is $0$ and when $r$ is an odd integer the sum is not zero (regardless of the parity of $n$).

[Edited] The case when $r$ is even is easy by symmetry as Darij Grinberg pointed out below. So the question left is how to show that the sum is nonzero when $r$ is odd.

The main difficulty I have proving this is that the sum is only from $0$ to $n$ instead of to $2n$, and I don't see how to apply the classical methods such as finite differences, integral representation, series, etc.

  • 1
    $\begingroup$ Oh. I'm pretty sure you can just extend this sum to $k=0$ to $2n$ by doubling it (for symmetry reasons). And then it should follow from finite differences. $\endgroup$ Oct 14 '18 at 23:36
  • $\begingroup$ The sum from $k=0$ to $n$ and the corresponding sum from $k=n$ to $2n$ are equal, because $\left(n-k\right)^r = \left(n-\left(2n-k\right)\right)^r$ (since $r$ is even) and $\dbinom{2n}{k} = \dbinom{n}{2n-k}$ and $\left(-1\right)^k = \left(-1\right)^{2n-k}$. $\endgroup$ Oct 14 '18 at 23:41
  • $\begingroup$ @darijgrinberg I see, thanks! How do you show that the sum is not zero if $r$ is not even? $\endgroup$
    – user58955
    Oct 14 '18 at 23:52
  • $\begingroup$ Oh, that isn't a consequence of anything I did, and I don't know how to show it. I thought you were just mentioning it as a limitation of your claim. $\endgroup$ Oct 14 '18 at 23:53
  • $\begingroup$ My proof that the sum is $0$ when $r$ is even can now be found in full detail in the solutions to UMN Fall 2018 Math 4707 Homework #4 (where the claim appears as Exercise 6 (b)). $\endgroup$ Nov 4 '18 at 2:11

In a post on Math Stack Exchange, MSE 2827591, I prove the following:

$$\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = \binom{2n}{n} \sin(\pi s/2) \int_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}. $$

The OP's formula can be put in the form on the LHS of this equation. By inspection the questions of concern can be answered; namely, the value of zero for $s$ an even integer, and non-zero otherwise.

  • $\begingroup$ Thanks! The equation [2] in that post of yours is already enough. $\endgroup$
    – user58955
    Oct 15 '18 at 5:21
  • $\begingroup$ what is $(n+ix)!$? $\endgroup$ Oct 15 '18 at 7:06
  • $\begingroup$ @FedorPetrov I think it means $\Gamma(n+ix+1)$ $\endgroup$
    – user58955
    Oct 15 '18 at 8:24
  • $\begingroup$ @user58955 I thought about it, but that text contains also Gamma-function notation, why use both? It looks confusing. $\endgroup$ Oct 15 '18 at 9:51

Just for the curious mind.

Let's suppose $r\rightarrow 2r+1$. Then, the constant term in Brendan McKay's experiment can be given as follows: $$\frac{2(-1)^{r-1}(2r+1)!B(2r)}{r!\,2^r}$$ where $B(n)$ are the Bernoulli numbers.


Here's a very clunky approach that might not be close to a general solution. However, it is a proof for $r\le 99$. Define $p_r(n)$ by $$ \sum_{k=0}^n (-1)^k \binom {2n}k (n-k)^r = (-1)^n\binom{2n}{n}\frac{n^2\,p_r(n)}{2\prod_{t=0}^{(r-1)/2} (2n-2t+1)}. $$

For $r=3,5,\ldots,99$, $p_r(n)$ is a polynomial of degree $\frac{r-3}2$ with integer coefficients.

For example $p_3(n) = 1$, $p_5(n)=-4n+1$, $p_7(n)=34n^2-24n+5$.

For odd $r\le 99$, $p_r(n)$ is irreducible, but that is likely to be very hard to prove (or false) for all $r$. A simpler observation is that, for $r\le 99$, all the coefficients of $p_r(n)$ are even, except that the constant term is odd. Therefore, $p_r(n)$ is an odd integer when $n$ is an integer.

Without caring about its coefficients, maybe there is a direct way to show that $p_r(n)$ is odd whenever $r$ is odd.

  • $\begingroup$ How did you find that the non-constant terms in $p_r(n)$ have even coefficients? With computer aid? $\endgroup$
    – user58955
    Oct 15 '18 at 2:55
  • $\begingroup$ I used Maple. It took about 10 mins to get up to $r=99$. $\endgroup$ Oct 15 '18 at 4:38

There is already a fine answer. Here are a few comments and generalizations that might lead to another. Together they are slightly redundant but I don't know which (if any) would work best.

Along with the given sum $$f(n,r)=\sum_{k=0}^{n} (-1)^k \binom{2n}{k} (n-k)^r$$ consider also $$g(n,r)=\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}(n-k)^r.$$

The observation (ignoring the part about odd $r$) is that, seemingly,

$f(n,r)=0$ for even $1 \le r \lt n.$

Here is a claim which looks nicer although it is really no stronger and no weaker:

$g(n,r)=0$ for $0 \le r \lt n.$

The reason the two are the same (aside from the well known case of $r=0$) is that for odd $r,$ $g(n,r)=f(n,r)-f(n,r)$ for odd $r$ and for even $r$ $g(n,r)=f(n,r)+f(n,r)+(-1)^n\binom{2n}{n}0^r.$

However there might be a nice proof for all $r$ which ignores the easier cancellation for odd $r.$

Also, we will note below that "$(n-k)^r$ for $r \lt N$" can be replaced by "$p(k)$ for any polynomial $p(t)$ of degree less than $N.$"

The bounds on $r$ are more restrictive than needed. Computations leave one highly confident that

$f(n,r)=0$ for even $1 \le r \lt 2n.$


$g(n,r)=0$ for $0 \le r \lt 2n.$

of course for the rather trivial reason $g(n,r)=0$ for any odd $r.$

Since the functions $t^0,t^1,\cdots,t^{2n-1}$ are a basis for the space $P_{2n-1}(t)$ of polynomials of degree less than $2n,$ the last claim that

$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}(n-k)^r=0$ for $0 \le r \lt 2n$

is equivalent to

$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}p(n-k)=0$ for all $p \in P_{2n-1}(t)$

and hence also to

$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}q(k)=0$ for all $q \in P_{2n-1}(t)$

This fits in nicely with the remark about finite differences.

Using a different basis

$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}\binom{k}{r}=0$ for $0 \le r \lt 2n.$

Now consider for a polynomial $p=p(t)$ the sum

$$G(N,p)=\sum_{k=0}^{N} (-1)^k \binom{N}{k}p(k).$$

The previous claim

$g(n,r)=G(2n,(n-t)^r)=0$ for $0 \le r \lt 2n$

which , as explained, is equivalent to

$G(2n,p(k))=0$ for any $p\in P_{2n-1}$

Does not depend on the bound being even:

$G(N,p(k))=0$ for any $p \in P_{N-1}$

Finally, here is another perspective which follows from (and imples) some of the above. Consider the polynomials

$$h_{n,r}(t)=\sum_{k=0}^{n} t^k \binom{2n}{k}(n-k)^r.$$

The starting claim was

For even $1 \lt r \lt 2n,$ $h_{n,r}(-1)=0$ i.e. $(t+1)$ divides $h_{n,r}.$

In fact

$(t+1)^{2n-r}$ divides $h_{n,r}.$

Again, the moves of changing to $\sum_{k=0}^{2n} t^k \binom{2n}{k}(n-k)^r$ and then $\sum_{k=0}^{N} t^k \binom{N}{k}p(k)$ are possible with the last thing being divisible by $(t+1)^{N-r}$ where $r \leq N$ is the degree of $p(t).$


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