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14 votes
0 answers
270 views

A symmetry of lattice paths

The number of $n$-step NSEW lattice paths from $(0,0)$ to $(a,b)$ that intersect the line $y=k$ precisely $t$ times is independent of $k$, for $0\leq k\leq b$, where we assume $b\geq0$ for simplicity. ...
Robin Houston's user avatar
10 votes
0 answers
349 views

A bijective proof for the odd companion to Shapiro's Catalan convolution

Shapiro's Catalan convolution is the following formula (where $C_n$ is the $n$th Catalan number): $$ \sum_{k=0}^{n}{C_{2k}C_{2(n-k)}}=4^nC_n. $$ In other words, letting $C(z)=\sum_{n=0}^{\infty}{C_nz^...
Alexander Burstein's user avatar
6 votes
0 answers
381 views

Reference request: colored Motzkin path interpretation of Catalan numbers

Recall that a Dyck path of length $2n$ is a lattice path in $\mathbb{Z}^2$ from $(0,0)$ to $(2n,0)$ consisting of $n$ up steps $U=(1,1)$ and $n$ down steps $D=(1,-1)$ which never goes below the $x$-...
Sam Hopkins's user avatar
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2 votes
0 answers
413 views

A (really!) cute identity between product of binomials

As an off-shot of my earlier MO question, I have found a "really cute" identity. The connection is revealed in the limit $q\rightarrow 1$. So, I would like to ask: QUESTION. Is there a ...
T. Amdeberhan's user avatar
2 votes
0 answers
495 views

Bijective proof of a combinatorial identity: $\sum\limits_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m$

Identity \begin{equation} \sum_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m \tag{1} \end{equation} was used in an answer here. As shown in that answer, (1) easily reduces to \begin{...
Iosif Pinelis's user avatar