All Questions
43
questions
1
vote
1
answer
54
views
Hint for a combinatorial proof of $n! \binom{n-1}{k-1} = \sum_{j=0}^n \overline{s}_{n,j} \tilde{S}_{j,k}$
To clarify notation, we have $\overline{s}_{n,j}$ the unsigned Stirling numbers of the first kind, meaning the number of permutations on an set with $n$ elements that has exactly $j$ cycles. $\tilde{S}...
4
votes
2
answers
206
views
Find $\sum^{n}_{k=1} \binom{n}{k} \frac{(-1)^k}{k^a}$
I have been trying to find a general representation of the following finite sum.
$$
S(n,a) = -\sum^{n}_{k=1} \binom{n}{k} \frac{(-1)^k}{k^a}
$$
The sum seems to be related to Generalized Harmonic ...
3
votes
1
answer
79
views
Expectation Value of n Bernoulli Variables
Let $X_1,...,X_n$ be i.i.d. Bernoulli Variables with $$E(X_i)=p$$ for $i=1,...,n$ and $$X=\sum_{i=1}^n X_i \, .$$
Furthermore let $k \in \{0,1,...,n\}$ and consider the Expectation Value
$$E_\pi=E\...
4
votes
1
answer
102
views
Gaussian approximation of collision time
In this answer there is a claim that
$$\frac{n!}{(n-k)! n^k} \approx e^{-\frac{k^2}{2n}} \tag{1}$$
which is then used to approximate the sum over $k=1,\ldots, n$ via
$$\sum_{k=1}^n \frac{n!}{(n-k)! n^...
6
votes
4
answers
207
views
Closed form of $ \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k}2^k k^m$?
Is there a closed form of the above equation or something that simplifies it? Here is the same equation copied:
$$\sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k}2^k k^m$$
It looks very similar to the Stirling ...
2
votes
1
answer
53
views
Simple integer sequence with unknown direct generator
I have a sequence (well a sequence of sequences) that I can't figure out how to generate directly, despite the fact I can generate it recursively
Specifically, I can write the elements of a matrix of ...
1
vote
3
answers
237
views
Show the following holds for $c\leq 1$: $e^{ck^2/n} \frac{(n!)^2}{(n+k)!(n-k)!} \leq M, \quad \text{for } k = 1, \dots, n.$ for some constant $M$
Show that the following inequality holds for $c\leq1$:
\begin{equation} \tag{1}
e^{ck^2/n} \frac{(n!)^2}{(n+k)!(n-k)!} \leq M, \quad \text{for $k = 1,
\dots, n$}
\end{equation}
for some constant $M$.
...
7
votes
1
answer
128
views
Finding a closed form for this sequence ((A080416))
I was working on a combinatorics problem that arose from one of my mathematical excursions, and the final formula includes numbers from a sequence that I've never encountered before: https://oeis.org/...
1
vote
1
answer
60
views
A proof for stirling numbers of the second kind... [closed]
So my Prof give me this Statement to proof and i have no idea how i could solve it tho.
$$S(k,n)=S(k−1,n−1) + n \cdot S(k−1,n)$$
My task is to prove it and as hint he said:
Use The binomial ...
3
votes
1
answer
174
views
Order of central moments for Binomial distribution
Suppose $X\sim \operatorname{Bin}(n,p_n)$ (binomial distribution) with $p_n\to 0$ and $np_n\to \infty$. Then
$$\limsup_{n\to\infty} \frac{\mathbb{E}\left[(X-np_n)^{2k}\right]}{(np_n)^k}<\infty, \...
1
vote
1
answer
82
views
definition of stirling numbers
Here is a definition without partition about Stirling numbers of second kind.
"Let $S(n,k)$ be Stirling numbers of second kind.
$1 \leq k \leq n$, and let $k$ and $n$ be positive integers. Let $h(...
1
vote
1
answer
140
views
A formula for $1^m+2^m+3^m+\ldots+n^m$ using binomial coefficients [duplicate]
It is known that
$$
\sum_{k=1}^{n}k^1=\binom{n+1}{2}
$$
and
$$
\sum_{k=1}^{n}k^2=\binom{n+1}{2}+2\binom{n+1}{3}
$$
Is there a formula for
$$
\sum_{k=1}^{n}k^m,
$$
where $m$ is a positive integers, ...
1
vote
2
answers
329
views
Combinatorial arguments for two stirling numbers
I'm trying to use combinatorial arguments to find simple formulas for $\begin{Bmatrix}
n\\
2
\end{Bmatrix}$ and $\begin{Bmatrix}
n\\
n-2
\end{Bmatrix}$
I've used combinatorial arguments to prove ...
0
votes
1
answer
242
views
Expected number of cycles in random permutations
Draw at random a permutation $\pi$ in the set of permutations of $n$ elements, $S_n$, with probability,
$$
P(\pi)= \frac{N^{L(\pi)}}{ \sum_{\pi \in S_n} N^{L(\pi)} },
$$
where $ L(\pi)$ is the number ...
0
votes
0
answers
54
views
Summation ofProduct of r-Stirling numbers of the second kind
I would like to simplify this block
\begin{equation}
\sum_{k=2}^{m-1}\binom{m}{k}S(k+1,3)S_{3}(m-k+3,1+3) \qquad \qquad (1)
\end{equation}
where, $$ n\in \mathbb {N} $$ and
\begin{equation}
S(k+1,3)...