Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
5
votes
Accepted
Result attribution for eigenvalues of a matrix of Pascal-type
I don't know a reference. One way to show the eigenvalues starts from the observation (which can be proved using
generating functions)
that $\sum_{i=0}^n {i\choose k} A_{i,j}={2n+1 \choose n-k} {j+k …
8
votes
Accepted
The number of ways to merge a permutation with itself
By @Max Alexeyev's solution above $N_{2k-1}^{\sigma}=tr(M_{k}(P_{\sigma}M_{k}P_{\sigma}^{-1}))$.
The eigenvalues and eigenvectors of $M_k$ are given here: Result attribution for eigenvalues of a matri …
2
votes
Asking for a proof for a sum of products of binomials: an "interesting" identity?
A generating function proof.
As $\arcsin(z)=\sum_{k\geq 0} \frac{1}{2k+1} {2k \choose k} \frac{z^{2k+1}}{4^k}$ and $\frac{1}{\sqrt{1-z^2}}=\sum_{k\geq 0}{2k \choose k}\frac{z^{2k}}{4^k}$
we have that
…