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12 votes
3 answers
1k views

A "quantum" identity: in search of a proof -Part II

As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$ As a follow up on this ...
T. Amdeberhan's user avatar
12 votes
0 answers
629 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
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
325 views

Looking for a $q$-analogue of a binomial identity

The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae): $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \...
T. Amdeberhan's user avatar
3 votes
1 answer
186 views

Is there a $q$-analogue to Shapiro's convolution identity?

Let $C_n=\frac1{n+1}\binom{2n}n$ denote the Catalan numbers. This question is motivated by the (unanswered) MO post by Alexander Burstein and my own (answered by Fedor Petrov) MO post. Specifically, ...
T. Amdeberhan's user avatar