Questions tagged [bell-numbers]
For questions related to the Bell numbers, a sequence of natural numbers that occur in partitioning a finite set.
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Partitions and Bell numbers
Let $F(n)$ be the number of all partitions of $[n]$ with no singleton blocks.
Find the recursive formula for the numbers $F(n)$ in terms of the numbers $F(i)$, with $i ≤ n − 1$
Find a formula for $F(...
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Calculating large Bell number modulo a composite number
I have been trying to solve http://www.javaist.com/rosecode/problem-511-Bell-Numbers-Modulo-Factorial-askyear-2018
It is not an ongoing contest problem.
We can calculate $n$th Bell number modulo a ...
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About the Binomial Transform of Incomplete Bell Polynomials $\displaystyle \sum_{j=1}^k(-1)^{j-1}\binom{k}{j} B_{m,j}$?
The binomial transform is the shift operator for the Bell numbers. That is,
$$ \sum_{j=0}^k {k\choose j} B_j =B_{k+1} $$
where the $B_n$ are the Bell numbers.
Is there a somewhat similar ...
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Exponential Generating Function Stirling Numbers
In class we found the exponential generating function for the Bell numbers $B_n$ which are defined by the recurrence $B(0) = 1$, $B(1) = 1$ and $B(n+1) =\sum_{i=1}^n\dbinom{n}{i}B(n-i)$ for all$ n\geq ...
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Bell number modulo prime power
I'd like to ask how to fastly calculate the Bell number $B_n$ modulo a prime power, where $n$ is around one million.
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Compute $S_n=\sum\limits_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}$
It is tagged as an open problem in the book Fractional parts,series and integrals. If this proof is valid , I don't have any idea how to get it published so I posted it here .
$\displaystyle \sum_{...
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Extending Bell Numbers to Fractional Values
An identity of the Bell numbers is given by
$$B_n=\frac{1}{e}\sum_{x=1}^\infty \frac{x^n}{x!}$$
and I was wondering if it would be valid to define fractional Bell numbers in the same way, to preserve ...
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Sum of Bell Polynomials of the Second Kind
A problem of interest that has come up for me recently is solving the following.
$$\frac{d^{n}}{dt^{n}}e^{g(t)}$$
There is a formula for a general $n$-th order derivative of a composition as shown ...
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Combinatorial proof for Touchard's congruence
Bell number denoted $B_n$ is the number of ways to partition a set with cardinality $n$ into $k$ indistinguishable sets , where $0\le k\le n$
It's known that Bell numbers obey Touchard's Congruence ...
user715522
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How many different equivalence relations with exactly two different equivalence classes are there on a set with $n$ elements
I came across with this topic. It looks straight forward for $5$ elements, but what if I want to find how many different equivalence relations with exactly two different equivalence classes are there ...
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New wrong recurrence formula for Bell numbers
Bell numbers are the numbers counting the total partitions on a set with $n$ distinct elements.
Explanation:
Consider a set like $A:=\left\{x_{1},x_{2},...,x_{n}\right\}$
A partial equivalence ...
user715522
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Number of preorder relations on a set related to the open problem about preorder relations
Consider a set $A=\left\{1,2,3\right\}$,I want to count the number of preorder relations on this set, so there is two cases two consider,either the relation is symmetric or it is not, if the relation ...
user715522
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Integral representation of Bell Polynomial?
From Wolfram Alpha: https://functions.wolfram.com/IntegerFunctions/BellB/07/01/0001/,
we have an integral representation for Bell numbers as:
$B_n = \frac{2n!}{\pi e} \displaystyle{\int_{0}^{\pi}} e^{...
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Closed form for $H(n,k)=\frac{1}{k!}\sum_{j=0}^\infty H(n-k, j)j^k$ where $H(1,k)=\frac{1}{k!}$
Let $H(n,k)$ be defined such that
$$H(1,k)=\frac{1}{k!}\text{, and }H(n,k)=\frac{1}{k!}\sum_{j=0}^\infty H(n-1,j)j^k$$
As pointed out in the comments, I should mention that we must define $0^0=1$ as ...
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Number of cycle partition of a set with repeating elements
We have a set $S$ with $E$ elements of which only $N$ are unique. We of course know how many repetitions of each of the $N$ elements are present: element $s_i$ is repeating $t_i$ times.
I would like ...
