Identifying the generating function $G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}.$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:21:58Z http://mathoverflow.net/feeds/question/64817 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64817/identifying-the-generating-function-ga-z-sum-n0-infty-an-zn1n Identifying the generating function $G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}.$ Granger 2011-05-12T17:15:35Z 2011-05-13T12:23:22Z <p>I have computed a generating function for a problem involving a particular series, and would like to know if anyone has any references or a categorisation for it? It's $$G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}.$$ It appears to be related to (mock) theta functions, but seems to be simpler. In particular, I would like to know whether $G(a,z)$ satisfies any identities?</p> <p>Many thanks.</p> http://mathoverflow.net/questions/64817/identifying-the-generating-function-ga-z-sum-n0-infty-an-zn1n/64829#64829 Answer by Simon Rose for Identifying the generating function $G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}.$ Simon Rose 2011-05-12T19:53:07Z 2011-05-12T19:53:07Z <p>It should be noted that using the Jacobi triple product that we have $$H(a,z) = a^{-1}\sum_{n=-\infty}^\infty a^nz^{n(n+1)/2} = a^{-1}\prod_{m=1}^\infty (1 - z^m)(1 - z^ma)(1 + z^{m-1}a)$$ where the main difference is that the indexing shifts and we are doing about twice the sum that you are.</p> <p>If we try relate these, we get $$H(a,z) = \big(a^{-1} + G(a,z)\big) + a^{-2}\big(1 + a^{-1}G(a^{-1},z)\big).$$</p> <p>I'm not sure how much this helps though.</p> http://mathoverflow.net/questions/64817/identifying-the-generating-function-ga-z-sum-n0-infty-an-zn1n/64864#64864 Answer by Brian Hopkins for Identifying the generating function $G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}.$ Brian Hopkins 2011-05-13T04:01:01Z 2011-05-13T04:01:01Z <p>Another possible connection is the following result of Gauss:</p> <p>$$\sum_{n=0}^\infty \ q^{n(n+1)/2} = \prod_{m=1}^\infty \ \frac{1-q^{2m}}{1-q^{2m-1}}$$</p> <p>(Andrews <em>The Theory of Partitions</em> Corollary 2.10), actually a corollary of the Jacobi Triple Identity that Simon used.</p> <p>One close mock theta function is $$\psi_0(q) = \sum_{n=0}^\infty \ q^{(n+1)(n+2)/2}(-q)_n$$</p> <p>(see Andrews chapter 2 examples 12 and 13).</p> http://mathoverflow.net/questions/64817/identifying-the-generating-function-ga-z-sum-n0-infty-an-zn1n/64893#64893 Answer by Johann Cigler for Identifying the generating function $G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}.$ Johann Cigler 2011-05-13T12:23:22Z 2011-05-13T12:23:22Z <p>Your generating function is related to a simple continued fraction expansion due to Touchard: </p> <p>$\sum\limits_{k \ge 0} ( - 1)^k q^{k+1\choose2} v^k$ =$\frac{1}{{1 + v - \frac{{(1 - q)v}}{{1 + v - \frac{{(1 - q^2 )v}}{ \cdots }}}}}.$</p> <p>A simple proof can be found in a paper by H. Prodinger <a href="http://de.arxiv.org/abs/1102.5186" rel="nofollow">http://de.arxiv.org/abs/1102.5186</a></p>