All Questions
5 questions with no upvoted or accepted answers
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.
...
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^...
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 $x$-...
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 ...
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{...