Lemma: sum_{j = 0..n} ( (n choose j) (-1)^j j^m ) = 0 for m < n = (-1)^n n! for m = n Proof: Take the m^th derivative of (1-x)^n, set x = 1, and use induction. Corollary: Let p(j) be a polynomial of degree n with leading coefficient C. Then sum_{j = 0..n} ( (n choose j) (-1)^j p(j) ) = C * (-1)^n * n! Main proof: Let n = 2k+1+D, and rewrite the equation slightly as sum_{j = 0..k} ( (n choose j) (-1)^j (D+k-j)!/(k-j)! (n - 2j)^(2k+1) ) = 2^2k n! Observe that the summand is zero if we take j = k+1, k+2, ..., k+D, and it is unchanged by the substitution j := 2k+1+D-j, so we can extend the sum up to n and the equation becomes sum_{j = 0..n} ( (n choose j) (-1)^j (D+k-j)!/(k-j)! (n - 2j)^(2k+1) ) = 2^(2k+1) n! But (D+k-j)!/(k-j)! (n - 2j)^(2k+1) is a polynomial in j of degree n with leading coefficient (-1)^n 2^(2k+1), so by the corollary the sum is equal to (-1)^n * 2^(2k+1) * (-1)^n * n! = 2^(2k+1) n! as required. J.P. On Wed, Aug 20, 2014 at 11:51 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:

Maple verifies it for k from 1 to 100 with arbitrary D. Perhaps someone who knows WZ theory <http://en.wikipedia.org/wiki/WZ_theory> can prove it.

On Wed, Aug 20, 2014 at 10:55 PM, Warren D Smith <warren.wds@gmail.com> wrote:

...

sum( (D+k-j)! / (k-j)! * (-1)^j * (2*k+1+D-2*j)^(2*k+1) / (j! * (2*k+1+D-j)!), j, 0, k ) = 4^k

claims lots of checking. (only checked when D>0 is integer.)

But... how to prove it?

-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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