All Questions
9 questions
4
votes
1
answer
379
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Counting permutations with a fixed number of descents and an extra condition
I am computing the volumes of certain polytopes and it turns out that knowing a "closed formula" for the following number would help a lot.
Determine the number of permutations $\sigma\in \...
- 1,889
0
votes
1
answer
403
views
Could you please confirm or deny two identities involving weighted Stirling numbers of the second kind?
In the paper [1] below, among other things, Carlitz introduced weighted Stirling numbers of the second kind $R(n,k,r)$. He also proved that the numbers $R(n,k,r)$ can be generated by
\begin{equation*}%...
- 1,101
2
votes
1
answer
761
views
Prove that $ \sum_{i=0}^{2k}( {n+R-1\choose R+i} + (-1)^{i+1}{ n+R+i\choose R+i } )\sum_{j=0}^i {i\choose j}(-1)^j(i+1-j)^{2k}=0 $
For all $k,R \in \mathbb{N}$ fixed, prove that $ \sum_{i=0}^{2k}( {n+R-1\choose R+i} + (-1)^{i+1}{ n+R+i\choose R+i } )\sum_{j=0}^i {i\choose j}(-1)^j(i+1-j)^{2k}=0 $. I'm quite sure this is true but ...
- 1,109
2
votes
1
answer
345
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Show that $\sum_{i=0}^{2k} [ {n\choose i+1} + (-1)^{i+1}{n+i+1\choose i+1} ] \sum_{j=0}^i {i\choose j}(-1)^j (i+1-j)^{2k} =0.$
Let $u(k,j) = 1$ if $j=0$, $0$ if $j > k$, or else it is $j*u(k-1,j-1) +(j+1)*u(k-1,j) $. Prove that $ \sum_{i=0}^{2k} {n \choose i+1} u(2k,i) +\sum_{i=0}^{2k} {-n-1 \choose i+1} u(2k,i)=0. $ ...
- 1,109
20
votes
4
answers
1k
views
Are surjections $[n]\to [k]$ more common than injections $[k]\to [n]$?
Here's an interesting inequality involving binomial coefficient and Stirling numbers of the second kind that I believe holds for all $n,k$:
$$ k^n {n \choose k} \leq n^k {n \brace k} $$
On the left-...
- 483
2
votes
1
answer
193
views
Bell polynomial with variables 1 and 0
Let $B_{n,k}(x_1,\cdots,x_{n-k+1})$ be the Bell polynomial.
If $x_1=\cdots=x_{n-k+1}=1$, we know that $B_{n,k}(x_1,\cdots,x_{n-k+1})=S(n,k)$, where $S(n,k)$ is the Stirling number of second kind.
...
- 549
4
votes
2
answers
495
views
Showing this formula counts these things
I'm writing an article, and I got stuck trying to prove that some numbers are positive. I have a relatively good intuition for guessing what an expression is counting, but in this case I'm not being ...
- 1,889
4
votes
0
answers
312
views
Positivity of a finite sum involving Stirling numbers of the first kind
Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not that ugly....
- 1,889
0
votes
1
answer
378
views
Simplify a double summation involving binomial coeficient
$$T(N,K)=\sum_{i=2}^{K}\sum_{j=2}^{i}(-1)^{i-j}\binom{i}{j}\frac{j^{N+1}-1}{j-1}$$
Is it possible to evaluate the sum for $K=10^7$ efficiently. If we manage to remove one of the sums, it will be ...
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