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12 votes
0 answers
631 views

$q$-analogue of the multinomial theorem?

The $q$-binomial theorem states that $$ \prod_{k=0}^{n-1}(1+q^kt) = \sum_{k=0}^n q^{\binom k2}{n\brack k}_q t^k. $$ This identity is a $q$-analogue of the binomial theorem $$ (1+t)^n = \sum_{k=0}^n \...
Amritanshu Prasad's user avatar
9 votes
1 answer
420 views

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 ...
Fedor Petrov's user avatar
9 votes
0 answers
192 views

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 ...
Zhi-Wei Sun's user avatar
  • 15.6k
7 votes
1 answer
320 views

A curious $q$-series identity on a truncated Euler function

Recall that a $q$-Pochhammer symbol is defined as $$ (x)_n = (x;q)_n := \prod_{l=0}^{n-1}(1-q^l x). $$ I found the following curious $q$-series identity that seems to hold for any $n\geq 0$: $$ (-1)^{...
Henry's user avatar
  • 1,430
4 votes
1 answer
672 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{...
Johann Cigler's user avatar