* Bill Gosper <billgosper@gmail.com> [Feb 28. 2012 08:14]:
Your perseverance rewards us all. Glitch:
$K(k,\infty)$ for $\lim_{n\to\infty}{K(k,n)}$
would be clearer as
$N(k,\infty)$ for $\lim_{n\to\infty}{N(k,n)}$
--rwg
Thanks! Fixed it as follows Writing $K(k,\infty)$ for $\lim_{n\to\infty}{K(k,n)}$ and $N(\infty,n)$ for $\lim_{k\to\infty}{N(k,n)}$ we obtain \eqref{rel:clausen-lambert} as upper right entries on both sides of \[ \prod_{n\geq{}0}{N(1,n)} \cdot \prod_{k\geq{}1}{K(k,\infty)} = \prod_{k\geq{}1}{K(k,0)} \cdot \prod_{n\geq{}0}{N(\infty,n)} \] And updated
Typset document at http://www.jjj.de/lambert-paper/
Added the following (easily obtained from Osler/Hassen): t * sum(n>=1, x^n*q^n/(1-t*q^n)) == x * sum(n>=1, t^n*q^n/(1-x*q^n)) == sum(n>=1, x^n*t^n*q^(n^2)/(1-t*q^n) + x^(n+1)*t^n*q^(n*(n+1))/(1-x*q^n) ) == sum(n>=1, x^n*t^n*q^(n^2)/(1-x*q^n) + x^n*t^(n+1)*q^(n*(n+1))/(1-t*q^n) ) == sum(n>=1, (1-x*t*q^(2*n)) * x^n*t^n*q^(n^2) / ( (1-x*q^n) * (1-t*q^n) ) ) Also direct derivation of the (less-) generalized Lambert from (Wrench/Knuth)'s beautiful identity: \sum_{n\geq{}1}{ \frac{a_n\,q^n}{ 1 - q^n } } = \sum_{n\geq{}1}{ \left[ a_n + \sum_{k\geq{}1}{ (a_n+a_{n+k})\,q^{k\,n} } \right]\,q^{n^2} } (this is how I originally found it). I'll submit tomorrow unless someone shouts "Stop the presses!"