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0 votes
1 answer
108 views

Stirling Numbers Exponential Generating Function Induction

I was reading the solution to a question written here, and it uses a fact which can be proved by induction. The question is to show that an EGF for Stirling Numbers of the second kind with fixed $k$, ...
Jeremy's user avatar
  • 5
8 votes
1 answer
311 views

Strange polynomial analog of the Bell numbers

Let $\vec{x} = (x_0, x_1, x_2, \dots)$ and $\vec{y}=(y_1,y_2,y_3, \dots)$ be two systems of parameters/variables. The Motzkin polynomials $P_n(\vec{x},\vec{y})$ for $n \geq 0$ are defined by the ...
Jeanne Scott's user avatar
4 votes
1 answer
308 views

Bell Polynomials

The complete Bell polynomials $B_n(x_1, x_2, \ldots, x_n)$ are defined through the relation $$\sum_{n=0}^{\infty} B_n(x_1, x_2, \ldots, x_n) \frac{t^n}{n!} =\exp\Big( \sum_{n=1}^{\infty} x_n \frac{t^n}...
Andrew's user avatar
  • 551
2 votes
1 answer
232 views

Exponential generating function with Stirling numbers

I want to prove in particular this result- $$ \newcommand{\gkpSII}[2]{{\genfrac{\lbrace}{\rbrace}{0pt}{}{#1}{#2}}} \sum_{k \geq 0} \gkpSII{2k}{j} \frac{\log(q)^k}{k!} = \frac{1}{\sqrt{2\pi}} \...
Jack's user avatar
  • 63
1 vote
2 answers
185 views

An approximation of the ordered Bell numbers

So my problem is the following: I have $n$ ice-cream flavors and I must rank them, allowing that I can place more than one flavor in some ranks. So for example if I have 4 flavors, I can put in the ...
PanYmermelada's user avatar
2 votes
1 answer
570 views

Generating function for the number of partitions of [n] without singletons.

I know the generating function for the total number of partitions of [n] is given by $$ B(x)=e^{e^x-1}$$ I am struggling to find $V(X)$, the exponential generating function for the number of ...
Mathsguy123's user avatar
7 votes
0 answers
155 views

Complicated recursion formula, seems similar to Bell numbers?

I came up with a recursive formula for a problem I was working on. It is as follows. $$a_n = \Big(\frac{1-q^{f \cdot n}}{1-q^n}\Big)\displaystyle\Big(1+\sum_{i=0}^{n-1}\binom{n}{i}p^{n-i}q^ia_i\Big)$$...
rabbitsLikeCarrots's user avatar
0 votes
2 answers
210 views

Remebering bell numbers

I read about Bell numbers and I'm looking for a way to generate these numbers quickly for tests and exams. I know there is a recursively relation but it is not useful for big numbers .
S.H.W's user avatar
  • 4,359