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0

Starting from the definition $$\frac{te^{tx}}{e^t-1} = \sum_{n\ge 0} B_n(x) \frac{t^n}{n!}$$ we obtain $$B_n(x) = \frac{n!}{2\pi i} \int_{|t|=\epsilon} \frac{1}{t^{n+1}} \frac{te^{tx}}{e^t-1} \; dt$$ and hence $$[x^q] B_n(x) = \frac{n!}{q!\times 2\pi i} \int_{|t|=\epsilon} \frac{1}{t^{n+1}} \frac{t^{q+1}}{e^t-1} \; dt \\ = \frac{n!}{q!\times 2\pi i} ...


0

You know that $$e^z=\sum_{n\ge 0}\frac1{n!}z^n$$ and that $$D(z)=\sum_{n\ge 0}\frac{d(n)}{n!}z^n\;.$$ Now take the Cauchy product: $$D(z)e^z=\left(\sum_{n\ge 0}\frac{d(n)}{n!}z^n\right)\left(\sum_{n\ge 0}\frac1{n!}z^n\right)=\sum_{n\ge 0}c_nz^n\;,$$ where ...


0

Denote the generating function for the Perrin sequence as $A(x) = \sum_{n=0}^\infty a_n x^n$, where $\{a_k\}_{k=0}^\infty$ is the Perrin sequence. We get $$\begin{align} A(x) &= \sum_{n=0}^\infty a_n x^n \\ &= \sum_{n=3}^\infty a_n x^n + a_2x^2+a_1x^1+a_0x^0\\ &= \sum_{n=3}^\infty \left(a_{n-2} + a_{n-3}\right)x^n+ a_2x^2+a_1x^1+a_0x^0\\ ...


1

As you started to say in your edit, this all comes down to plane partitions. An expression for the number of your partitions of $n$ is $$ \sum_{c=1}^n pp(n-c,c) $$ ($c$ for [core]) where $pp(i,j)$ is the number of plane partitions of $i$ with largest part $j$; this triangle of numbers is given in OEIS A242642 (with references back to MacMahon), which is ...


1

A closed for exists (attachment bellow) for the sum : $$ \sum_{m \ge 1} \frac{(xy)^m}{(1-y^m)} $$ thanks to the special function called q-digamma. See : http://mathworld.wolfram.com/q-PolygammaFunction.html By integration, a closed form for the sum : $$ \sum_{m \ge 1} \frac{(xy)^m}{m(1-y^m)} $$ is formally expressed. I cannot say if a simpler form can be ...


1

Hint: $\displaystyle\sum_{n\ge0}a_nx^n=\frac1x\cdot\sum_{n\ge0}\frac{x^{n+1}}{(n+1)!}=\frac1x\cdot\sum_{n\ge1}\frac{x^n}{n!}$.


0

$$e=\sum_{n=0}^{\infty }\frac{1}{n!}=1+\sum_{n=1}^{\infty }\frac{1}{(n)!}=1+\sum_{n=0}^{\infty }\frac{1}{(n+1)!}$$ hence $$\sum_{n=0}^{\infty }\frac{1}{(n+1)!}=e-1$$


0

I would do: $$\newcommand{\u}[2]{\underbrace{#1}_{#2}} \u{(x^0+x^1+x^2+...)}{x_1}\u{(x^2+x^3+x^4+...)}{x_2}\u{(x^0+x^4+x^8+...)}{x_3}\u{(x^1+x^2+x^3)}{x_4}$$ Or: $$\newcommand{\c}[2]{{}^{#1}{\mathbb ...


3

First, let's sort out what the generating function is. $x_1$ can be any non-negative integer: $$1+x+x^2+x^3+x^4+\ldots$$ $x_2$ is any integer greater than or equal to $2$: $$x^2+x^3+x^4+x^5+\ldots$$ $x_3$ is a non-negative multiple of four: $$1+x^4+x^8+x^{12}+\ldots$$ $x_4$ is in $\{1,2,3\}$: $$x+x^2+x^3$$ Multiplied together: ...


3

I think, you have no chance to get explicit formulas for a(k,i,j). From your differential equation you get $$ (-1)^k k! N^k B^{k+1} = p_{0,k} B +p_{1,k} B'+p_{2,k} B''+p_{3,k} B'''+...+p_{k,k} B^{(k)} $$ where $p_{j,k}=\sum_r a(j,k,r)x^r$. The recurrence relation is $$ p_{j,k+1}=(k+1) A p_{j,k}+B p_{j-1,k}+B p_{j,k}' $$ where $A=x-N$, $B=x$ and the ...


0

No: a summation is not a closed form. Roughly speaking, a closed form is a function of $n$ not involving a summation or product whose length depends on $n$. Here’s a simple example. The generating function $\frac1{1-2x}$ represents the series $$\sum_{n\ge 0}(2x)^n=\sum_{n\ge 0}2^nx^n\;,$$ whose associated sequence of coefficients is $\langle ...


0

Intersting question, but I think, only constant sequences verify this. take $(a_n)$ a sequence verifying a recurrence equation : $$a_n=\sum_{k=2}^r\lambda_ka_{\lfloor \frac{n}{k} \rfloor}$$ Then, evaluating it for $n=0$ we get : $$a_0=\sum_{k=2}^r\lambda_ka_0 $$ That is either $a_0=0$ or : $$\sum_{k=2}^r\lambda_k=1 $$ First, if $a_0=0$ then ...


5

As far as I can tell, the answer can only be written in terms of the Jacobi theta functions (or the like). Specifically, we have $$ \begin{align} \sum_{n=0}^\infty \lceil \sqrt{n} \rceil x^n & = x + 2x^2 + 2x^3 + 2x^4 + 3x^5 + 3x^6 + 3x^7 + 3x^8 + 3x^9 + \cdots \\ & = (x + x^2 + x^3 + \cdots) + (x^2 + x^3 + x^4 + \cdots) + (x^5 + x^6 ...


1

For a three-letter alphabet, the number of words of length $n$ is $3^n$. For a general alphabet $X=\{a_1\dots a_k\}$, the number of words of length $n$ is $k^n$ and the generating series is $$\sum_{n=0}^\infty k^n\, z^n=\frac{1}{1-kz}$$ Length (number of letters) is a particular case of ``length function'' : to each letter you attribute a length (or a ...


2

Suppose we have a sequence of values of length $k$ where the values are any non-negative integer. This is has the combinatorial specification $$\mathfrak{S}_{=k}\left(\sum_{q\ge 0} \mathcal{Z}^q\right).$$ This gives the generating function $$\frac{1}{(1-z)^k}.$$ On the other hand all such sequences can be obtained by stars-and-bars combining $n$ elements ...


2

It is quite easy to show by induction (or through other combinatorial arguments) that: $$ \sum_{n=0}^{N}\binom{n+k-1}{k-1}=\binom{N+k}{k},$$ hence if $$ f_k(x) = \sum_{n\geq 0}\binom{n+k-1}{k-1}x^n $$ it follows that $$ f_{k+1}(x) = \frac{f_k(x)}{1-x}=\frac{1}{(1-x)^{k+1}},$$ since if $$ f(z) = \sum_{n\geq 0} a_n z^n $$ we have: $$ ...


0

The following solution is supposed to be the simplest one. Let's consider a complex number $a$ such that $a^{4}=1$. So, denote $S_{n}=\sum_{s=0}{\binom{i}{m}}$. We know its generating function $A(s)=(1+s)^{m}$. So, the generating function for the sequence over all $i=4q, q\in \mathbb{Z}$ would look like this ...


0

Given the desired backward equations, the definitions of these generating functions appear to be slightly incorrect. Given a certain intermediate cell at time $\tau$, there will be, at any future time $t\ge \tau$, a certain number $I_1(t,\tau)$ of intermediate cells and $M_1(t,\tau)$ of malignant cells that descend from it via the intermediate cell dividing ...


0

Let's answer this question (even if it's not a very useful answer), the answer discuss the part how many information we can get from $G(z)$ We can use sets instead of sequences, given a set $A$ of sets $I$ such that :$|A|=m$ and $\forall I\in A |I|=n$ (The question imposes $n=m$ but the answer will be more general) we define: $$G_A(z)=\sum_{I\in ...


1

We have, if $\left|x\right|<1$ ...


3

Consider that, by the discrete Fourier transform: $$ 4\cdot\mathbb{1}_{n\equiv 0\pmod{4}} = 1^n+(-1)^n+i^n+(-i)^n \tag{1}$$ hence: $$ \mathbb{1}_{n\equiv 2\pmod{4}} = \frac{1}{4}\left(1^n + (-1)^n - (-i)^n - i^n\right)\tag{2} $$ and: $$\begin{eqnarray*} \sum_{n\equiv 2\pmod{4}}\binom{m}{n} &=& ...


2

Note: The following answer is divided in three steps. We start with a short Overview Step 1: A convenient representation We transform OPs expression by doing some index shifting and by adding one summand to both sides which considerably simplifies the RHS of OPs expression. We show OPs expression is equivalent to \begin{align*} ...


1

These are just a few with the generating functions or expansions. 1) Matthias Schork 2) Robert A. Sulanke 3) Sen-Peng Eu 4) Toufik Mansour 5) Ira M. Gessel 6) Eva Y. P. Deng


1

Square the Binet formula to get $$F_n^2=\frac{\phi^{2n}-2(-1)^n+\phi^{-2n}}{5},$$where $\phi=(1+\sqrt5)/2$. Now the series splits into three geometric series, where the common ratios are respectively $\phi^2s$, $-s$, and $\phi^{-2}s$.


3

Here is one approach that requires some technique but very little intuition: To begin with, we write down a recurrence relation. We have $a_{n+1}/a_n = (4n+1)/(n+1)$, or $(n+1)a_{n+1} = (4n+1)a_n$. Define $a_0 = 1$ for consistency. Since I posted this answer, the question was changed/merged to define $a_0 = 0$. The effect is to subtract $1$ from the ...


0

This is called generalized Binomial coefficient. Just expand $(1-4x)^{-\frac{1}{4}} = \sum_{k=0}^{\infty} \binom{-\frac{1}{4}}{k} (-4x)^k$ and apply the binomial expansion rule to the coefficient, this result will pop out. EDIT: the coefficient you have can be rewritten (after mulitplying by $(2n)!!$) as $ \frac{1}{2^n} \binom{2n}{n} $ EDIT 2: To get the ...


1

What you are looking for is a row in the table of Mahonian numbers. Here they are: Sloane's A008302. That link gives you lots and lots of other links and references, and it includes a recurrence for calculating them (which I added a couple of years ago). Here, $T(1,)$ represents $1$ and $T(2,)$ represents $(1+x)$, etc. $$T(1, 1) = 1,\qquad T(1, k\neq 1) = ...


2

This product gives the Poincare series for the symmetric group $S_{n+1}$ considered as a Coxeter group of type $A_n$. The answer is therefore the number of permutations in $S_{n+1}$ that can be minimally expressed as a product of $n$ adjacent transpositions. I think this is likely the best formula you will find for this number.



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