TITLE: Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$
QUESTION [26 upvotes]: I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. This inequality has been verified by computer for $k\le40$.
Some clues that might work (kindly provided by Doron Zeilberger) are as follows:

Let $Ef(x):=f(x-1)$, let $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}\binom{2k+1}{j}E^j$;
These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2P_{k-1}(E) =$ some binomial in $E$;
The original sum can be expressed as $P_k(E)x^{(2k-2)} |_{x=k} $;
Try to derive a recurrence for $P_k(E)x^{(2k-2)}$ before plugging in
$x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.

Unfortunately I do not know how to find a recurrence such as suggested by clue 4.

REPLY [20 votes]: This is a clarification of Pietro Majer's beautiful and insightful, yet a bit cryptic answer.
The Eulerian numbers are expressible as 
$$\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+1\choose i}(m+1-i)^n.$$
View them as functions of $m$ and let $\Delta$ be the backward difference operator, 
$$\Delta f(m)=f(m)-f(m-1).$$
Claim  The $r$th iterated backward difference of the Eulerian number is given by the formula
$$\Delta^r\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+r+1\choose i}(m+1-i)^n.$$
Proof This is proved by induction in $r$ using the binomial identity $${n+r\choose i}+{n+r\choose i-1}={n+r+1\choose i}. \quad\square$$
Setting $m=k-1$ and comparing with the definition of the sequence, we see that 
$$A(k)=\sum_{j=0}^{k-1}(-1)^{j+1}{2k+1 \choose j}(k-j)^{2k-2}=-\Delta^2\left\langle {n\atop k-1}\right\rangle\ \text{evaluated at }\ n=2k-2.$$
Thus 
$$A(k) = -\Delta\left\langle {n\atop k-1}\right\rangle + \Delta\left\langle {n\atop k-2}\right\rangle = \Delta\left\langle {n\atop k-2}\right\rangle\ \text{evaluated at }\ n=2k-2$$
and the first summand vanishes due to the symmetry of the Eulerian numbers, $\left\langle {n\atop m}\right\rangle=\left\langle {n\atop n-1-m}\right\rangle$, which implies that $\left\langle {2k-2\atop k-1}\right\rangle=\left\langle {2k-2\atop k-2}\right\rangle.$ 
Now the positivity of $A(k)$ becomes a consequence of the unimodality of the Eulerian numbers, $\Delta\left\langle {n\atop m}\right\rangle\geq 0$ for $m\leq n/2.$ Explicitly, 
$$A(k)=\left\langle {2k-2\atop k-2}\right\rangle-\left\langle {2k-2\atop k-3}\right\rangle > 0\ \text{for}\ k\geq 2.$$