alternating sum of binomial coefficients - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T08:26:41Z http://mathoverflow.net/feeds/question/88573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88573/alternating-sum-of-binomial-coefficients alternating sum of binomial coefficients JM Landsberg 2012-02-15T22:52:37Z 2012-03-21T02:00:11Z <p>I would like to know a closed formula for $\sum_{j=0}^{p-n } (-1)^j\binom{n^2}{p-n-j} \binom{n+j-1}j\binom{2n+j}{n+j+1}$, especially in the case $p$ is near $n^2/2$. Similarly, I would like a closed formula for: setting $q=2\cdot\lceil\frac{n(n+1)}{4}\rceil -1$, and setting $p=\lceil\frac{q}{2}\rceil-1$, what is the sum $\sum_{j=0}^{p-n } (-1)^j\binom{q}{p-n-j} \binom{n+j-1}j\binom{2n+j}{n+j+1}$? </p> <p>In either case I would be happy for an estimate of the growth of the sum (divided by $\binom {n^2-1}p$ in the first case, and divided by $\binom{q-1}p$ in the second).</p> http://mathoverflow.net/questions/88573/alternating-sum-of-binomial-coefficients/91507#91507 Answer by sqz for alternating sum of binomial coefficients sqz 2012-03-18T03:27:56Z 2012-03-18T03:27:56Z <p><code>$\sum_{j=0}^{p-n} (-1)^j a_{n,p}(j) = \sum_{j=0}^{\frac{p-n}{2}} a_{n,p}(2j)-a_{n,p}(2j+1)$</code></p> <p>so examine this difference <code>$a_{n,p}(2j)-a_{n,p}(2j+1)$</code>, we can factor out <code>$c_{n,p}(j):=\frac{(n^2)!(2n+2j)!(n+2j)!}{(n-1)!(n-1)!(2j+1)!(n+2j+2)!(p-n-2j)!(n^2-p+n+2j)!}$</code></p> <p>giving <code>$c_{n,p}(j) \left[ \frac{(n+2j+2)(2j+1)}{(n+2j)} - \frac{(2n+2j+1)(p-n-2j)}{(n^2-p+n+2j+1)}\right]$</code></p> <p><code>$=c_{n,p}(j) \left[ (2j+1)\left(1+\frac{2}{n+2j}- \frac{p-n-2j}{n^2-p+n+2j+1}\right)-\frac{2n(p-n-2j)}{n^2-p+n+2j+1}\right]\approx -2nc_{n,p}(j)$</code></p> <p>this is assuming $p\approx \frac{n^2}{2}$ is large. not sure if this helps</p> http://mathoverflow.net/questions/88573/alternating-sum-of-binomial-coefficients/91526#91526 Answer by Jacques Carette for alternating sum of binomial coefficients Jacques Carette 2012-03-18T12:57:02Z 2012-03-21T02:00:11Z <p>I played around with your sum in Maple and got</p> <p>$$\frac{2n}{n+1}{2n-1 \choose n-1}{n^2 \choose p-n} 3F_{2}([n,n-p,2n+1],[n+2,n^2+n+1-p],1)$$</p> <p>I make no guarantees that this is correct (especially as the original answer contained a $\binom{n^2}{-1}$ in it).</p>