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A generating function $G(x)=-\frac{\frac{1}{x^5}(1+\frac{1}{x})(1-\frac{1}{x^2})}{((1-\frac{1}{x})(1-\frac{1}{x^3}))^2}$ related to partitions of $6n$
Fix a sequence $a_n={n+2\choose 2}$ of triangular numbers with the initial condition $a_0=1$,such that
$1,3,6,10,15,21,\dots$
given by
$F(x)=\frac{1}{(1-x)^3}=\sum_{n=0}^{\infty} a_n x^n\tag1$
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$4\sum_{m,n=1}^{\infty}\frac{q^{n+m}}{(1+q^n)(1+q^m)}(z^{n-m}+z^{m-n})=8\sum_{m,n=1}^{\infty}\frac{q^{n+2m}}{(1+q^n)(1+q^{n+m})}(z^m+z^{-m})$
To prove the identity $$4\sum_{m,n=1}^{\infty}\frac{q^{n+m}}{(1+q^n)(1+q^m)}(z^{n-m}+z^{m-n})=8\sum_{m,n=1}^{\infty}\frac{q^{n+2m}}{(1+q^n)(1+q^{n+m})}(z^m+z^{-m})$$ I replaced $m-n$ by $k$ in LHS ...
