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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 ...

**4**

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**1**answer

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### Reference request for simple $q-$ identities

I stumbled upon the following simple $q-$identities:
$$\frac{1}{(-q;q)_\infty}\sum \limits_{j =0}^{\infty}\frac{q^{(2r+1)j}}{(q^2;q^2)_j}=(q;q^2)_r$$
and
$$\frac{1}{(q;q^2)_\infty}\sum \limits_{j =0}^...

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votes

**1**answer

378 views

### Quest for a human proof of a $q-$binomial identity

Let $$f(n,k) = \sum\limits_{j = - k}^k {{{( - 1)}^{k - j}}}
\binom{n-j}{k-j}\binom{n+j}{k+j}.$$
Then $f(n,k)=\binom{n}{k}$
because it satisfies $f(n,k)=f(n-1,k)+f(n-1,k-1)$ and the obvious ...

**4**

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### Identities for ${~}_3\phi_1$?

I am looking for some source of summation formulas for the $q$-hypergeometric function ${~}_3\phi_1$ in the sense of Gasper-Rahman book. For some reason, the book focuses heavily on ${~}_{r+1}\phi_r$ ...

**2**

votes

**1**answer

93 views

### Expressions involving $q$-binomial coefficients?

I have bumped into the following expressions involving $q$-binomial coefficients.
$$
\sum_{s=0}^a (-1)^s q^{s^2-s} \left(\begin{array}{c}2b+1-2s\\2a-2s\end{array}\right)_q
\left(\begin{array}{c}b\\s\...

**4**

votes

**2**answers

1k views

### series expansion of the q-Pochhammer symbol

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 ...

**4**

votes

**1**answer

605 views

### Are the following q-Genocchi numbers known?

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{...

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220 views

### 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 ...