Alternating sums of alternate Stirling numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:11:26Z http://mathoverflow.net/feeds/question/103092 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103092/alternating-sums-of-alternate-stirling-numbers Alternating sums of alternate Stirling numbers Adam 2012-07-25T12:09:49Z 2012-07-25T13:08:45Z <p>Does anybody know of any identities or combinatorial interpretations for alternating sums of alternate Stirling numbers?</p> <p>I am particularly interested in expressions of the form:</p> <p>$$\pm\sum_{k}(-1)^k|s(n,2k)|=\mp|s(n,2)|\pm|s(n,4)|\mp|s(n,6)|\pm\ldots$$</p> <p>or:</p> <p>$$\pm\sum_{k}(-1)^k|s(n,2k+1)|=\pm|s(n,1)|\mp|s(n,3)|\pm|s(n,5)|\mp\ldots,$$</p> <p>where $|s(n,k)|$ is an unsigned Stirling number of the first kind. </p> <p>However, I would be happy with any related identities or information (including an argument as to why sums like this might not have nice formulae or count anything interesting). As far as I can tell these do not appear on the OEIS, or in any of the literature.</p> http://mathoverflow.net/questions/103092/alternating-sums-of-alternate-stirling-numbers/103102#103102 Answer by Gjergji Zaimi for Alternating sums of alternate Stirling numbers Gjergji Zaimi 2012-07-25T13:08:45Z 2012-07-25T13:08:45Z <p>The generating function here is $$\sum_{n\geq 0}s(n,k) x^n y^k=\sum_{n\geq 0} \frac{x^n}{n!}y(y-1)\cdots (y-n+1)=e^{y\log(1+x)}.$$ If we put $y=i$ the coefficient of $x^n$ becomes $A_n+iB_n$ where $A_n$ and $B_n$ are your sequences. It is pretty clear from here that the exponential generating function for $A_n$ is $$\mathfrak {Re}\left(e^{i\log(1+x)}\right)=\cos(\log(1+x)),$$ and for $B_n$ it is $$\mathfrak{Im}\left(e^{i\log(1+x)}\right)=\sin(\log(1+x)).$$ Also these are <a href="http://oeis.org/A009454" rel="nofollow">A009454</a> and <a href="http://oeis.org/A003703" rel="nofollow">A003703</a>.</p>