Timeline for Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$

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Jun 25, 2010 at 3:51	comment	added	Alexandra Seceleanu		In my case, these numbers appeared as follows: Consider the algebra $A_{r,t}=k[x_1,\ldots x_n]/(l_1^t,\ldots l_{r+1}^t)$ whose Hilbert function is given by $HF(A_{r,t},i)=\sum\limits_{j=0}^m(-1)^j\binom{r-1+i-tj}{r-1} \cdot \binom{r+1}{j} $ where $m=\mbox{min}\{\lfloor \frac{i}{t} \rfloor,r \}$. Now set $r=2k$. I am interested in the asymptotics of $HF(A_{r,t},k(t-1)-1)$. Turns out this is a polynomial in $t$ whose leading coefficient can be expressed (as I have just learned from the answers above) using Eulerian numbers as $\frac{1}{(2k-2)!}A(2k-2,k-2)$.
Jun 22, 2010 at 21:39	comment	added	David Carchedi		P.P.S, it would appear that this in fact has to do with the dimension of the space of covariants of the regular representation of the symmetric group $S_{2k}$. If this rings a bell, we should talk (if not, maybe we should talk anyway).
Jun 22, 2010 at 21:32	comment	added	David Carchedi		P.S. in this case, since $2k-2 < 2k$, the sum of all the $a_j$s will be zero, in case this ends up helping.
Jun 22, 2010 at 21:29	history	answered	David Carchedi	CC BY-SA 2.5