Formula:6171

From MaRDI portal

Digital Library of Mathematical Functions ID 17.14.E5

n = 0 q n 2 + 2 n ( q 2 ; q 2 ) n ( q ; q 2 ) n + 1 =  coeff. of  z 0  in  ( - z q ; q 2 ) ( - z - 1 q ; q 2 ) ( q 2 ; q 2 ) ( - q 2 z - 1 ; q 2 ) ( q ; q 2 ) ( z - 1 q 2 ; q 2 ) = 1 ( q ; q 2 )  coeff. of  z 0  in  ( - z q ; q 2 ) ( - z - 1 q ; q 2 ) ( q 2 ; q 2 ) ( q 4 z - 2 ; q 4 ) = H ( q 4 ) ( q ; q 2 ) . superscript subscript 𝑛 0 superscript 𝑞 superscript 𝑛 2 2 𝑛 q-Pochhammer-symbol superscript 𝑞 2 superscript 𝑞 2 𝑛 q-Pochhammer-symbol 𝑞 superscript 𝑞 2 𝑛 1  coeff. of  superscript 𝑧 0  in  q-Pochhammer-symbol 𝑧 𝑞 superscript 𝑞 2 q-Pochhammer-symbol superscript 𝑧 1 𝑞 superscript 𝑞 2 q-Pochhammer-symbol superscript 𝑞 2 superscript 𝑞 2 q-Pochhammer-symbol superscript 𝑞 2 superscript 𝑧 1 superscript 𝑞 2 q-Pochhammer-symbol 𝑞 superscript 𝑞 2 q-Pochhammer-symbol superscript 𝑧 1 superscript 𝑞 2 superscript 𝑞 2 1 q-Pochhammer-symbol 𝑞 superscript 𝑞 2  coeff. of  superscript 𝑧 0  in  q-Pochhammer-symbol 𝑧 𝑞 superscript 𝑞 2 q-Pochhammer-symbol superscript 𝑧 1 𝑞 superscript 𝑞 2 q-Pochhammer-symbol superscript 𝑞 2 superscript 𝑞 2 q-Pochhammer-symbol superscript 𝑞 4 superscript 𝑧 2 superscript 𝑞 4 𝐻 superscript 𝑞 4 q-Pochhammer-symbol 𝑞 superscript 𝑞 2 {\displaystyle{\displaystyle\sum_{n=0}^{\infty}\frac{q^{n^{2}+2n}}{\left(q^{2}% ;q^{2}\right)_{n}\left(q;q^{2}\right)_{n+1}}=\mbox{ coeff. of }z^{0}\mbox{ in % }\frac{\left(-zq;q^{2}\right)_{\infty}\left(-z^{-1}q;q^{2}\right)_{\infty}% \left(q^{2};q^{2}\right)_{\infty}}{\left(-q^{2}z^{-1};q^{2}\right)_{\infty}% \left(q;q^{2}\right)_{\infty}\left(z^{-1}q^{2};q^{2}\right)_{\infty}}=\frac{1}% {\left(q;q^{2}\right)_{\infty}}\mbox{ coeff. of }z^{0}\mbox{ in }\frac{\left(-% zq;q^{2}\right)_{\infty}\left(-z^{-1}q;q^{2}\right)_{\infty}\left(q^{2};q^{2}% \right)_{\infty}}{\left(q^{4}z^{-2};q^{4}\right)_{\infty}}=\frac{H(q^{4})}{% \left(q;q^{2}\right)_{\infty}}.}}


Constraint(s)

Symbols List

Resources that cite this formula

No records found.