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9 votes
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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
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8 votes
0 answers
253 views

q-analog of $(d/dx) \binom{x}{k}$?

It is not hard to find easy formulas for the derivative of the function $\binom{x}{k}$, for instance it is not too hard to see (for $k$ an integer) that $\frac{d}{dx} \binom{x}{k} = \sum_{i=1}^k \...
Ratio Bound's user avatar
6 votes
0 answers
214 views

Looking for a combinatorial proof for an identity involving $q$-Catalan triangles

Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Following my earlier post on MO, one fine colleague asked me if there is a $q$-analogue of the identity formed by the so-called Shapiro's ...
T. Amdeberhan's user avatar
6 votes
0 answers
132 views

Q-analogue of an inequality

Pick integers $b\geq a \geq 0$ and $k\geq j\geq 0$. It is not super-difficult to prove the inequality $$ \binom{kb}{ka}^j \geq \binom{jb}{ja}^k. $$ This is actually quite a nice inequality that was ...
Per Alexandersson's user avatar
3 votes
0 answers
136 views

A recursion involving binomial coefficients: looking for a q-analog

Let $a_n := \frac{1}{2n+1}\binom{3n}{n}$. Then it is known that (one can find references in the OEIS for this.) $$ a_n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n-1} } a_i a_j a_k. $$ Is there a natural ...
Per Alexandersson's user avatar
0 votes
0 answers
174 views

3D generalization of Gaussian q-binomial coefficient

It is known that the coefficient of $q^t$ in Gaussian binomial coefficient $\binom{m+n}m_q$ equals the number of permutations of the multiset $\{0^m, 1^n\}$ with $t$ inversions. Is there a closed ...
Max Alekseyev's user avatar