All Questions
9 questions
8
votes
0
answers
260
views
Efficient listing of ASMs
Famously, the alternating sign matrix theorem gives a product formula for the number $a(n)$ of ASMs of size $n$. There are multiple proofs of this formula, all somewhat involved. My question is ...
- 17k
21
votes
1
answer
766
views
Combinatorial proof of a certain binomial identity
Let $n$, $p$, $q$ be non-negative integers. Then
$$
\sum_{k=0}^n{2k+2p\choose k+p,k,p}{2(n-k)+2q\choose n-k+q,n-k,q}=4^n{2p\choose p}{2q\choose q}{n+p+q\choose n}.\tag{$\heartsuit$}\label{heart}
$$
In ...
- 5,590
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 ...
- 43.2k
14
votes
1
answer
647
views
Bijective proof of recurrence for rooted unlabeled trees
Would've been a better question for Christmas than Thanksgiving, but alas...
Let $t_n$ denote the number of rooted, unlabeled trees on $n$ vertices (OEIS A000081). These are the isomorphism classes of ...
- 5,590
11
votes
5
answers
927
views
The number of ways to merge a permutation with itself
Let $\sigma$ be a permutation of $[k]=\{1,2, \dots , k\}$. Consider all the ordered triples $(\pi, s_{1},s_{2})$, such that $\pi$ is a permutation of length $2k-1$ that is a union of its two ...
- 109
6
votes
1
answer
239
views
Direct bijections for $s,t$-Fibonomial identities
Sagan and Savage gave a combinatorial interpretation of a polynomial generalization of Fibonomial coefficients. Their proof uses the recurrence relation for the Lucas polynomials that generalize the ...
- 116
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.
...
- 2,896
8
votes
1
answer
344
views
Bijective proof of formula for rooted binary forests
For $n\ge 1$, let $f(n)$ be the number of rooted complete (unordered) binary trees with $n$ leaves labeled from $1$ to $n$ ("complete binary" means that every vertex has either $0$ or $2$ children and ...
- 5,822
13
votes
1
answer
929
views
Two to the power of a triangular number: bijections
The numbers $2^{n(n+1)/2}$ come up in various enumerative contexts. In addition to the trivial example (bit-strings of length $n(n+1)/2$) and the old example of domino tilings of Aztec diamonds (...
- 3,587