# Questions tagged [q-identities]

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### For $q$-analogues of a known curious identity

In 2002 I published the folllowing curious combinatorial identity: $$(x+m+1)\sum_{i=0}^m(-1)^i\binom{x+y+i}{m-i}\binom{y+2i}i-\sum_{i=0}^m\binom{x+i}{m-i}(-4)^i=(x-m)\binom xm.$$ My original proof is ...
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### On Ramanujan's beautiful cubic identity

Let $a_i, b_i, c_i$ be defined by the following$\colon$ $\frac{1 + 53X + 9X^2}{1 - 82X - 82X^2 + X^3} = a_0 + a_1X + \ldots$. $\frac{2 - 26X - 12X^2}{1 - 82X - 82X^2 + X^3} = b_0 + b_1X + \ldots$. ...
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### notation for $(a-b)(a-qb)\dots (a-q^{n-1}b)$

I wonder whether there is a notation for such thing, which I denote $[a;b]_q^n$ for a moment: $$[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,$$ this last equation uses $q$-Pochhammer symbol ...
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### references for q-series identities

I need references for $\sum_{n=0}^N\frac{q^n}{(q^2;q^2)_n(q^2;q^2)_{N-n}}=\frac{(-q,q)_N}{(q^2;q^2)_N}$ and $\sum_{n=0}^N\frac{(-1)^nq^{n^2}}{(q^2;q^2)_n(q;q)_{N-n}}=\frac1{(q^2;q^2)_N}$ A ...
The following identity arose while I was working on a recent MO question: $-\sum_{n=1}^{\infty}\frac{1}{n}\frac{(-x)^n}{1-x^n}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^n}{1-x^{2n}}.$ I have no doubt ...
The sequence of Genocchi numbers ${({G_{2n}})_{n \ge 0}}=$ $(0,1,1,3,17,155,2073,...)$ can be defined by the generating function \$z\frac{{1 - {e^z}}}{{1 + {e^z}}} = \sum {{{( - 1)}^n}{G_{2n}}\frac{...