Pochhammer symbol is the notation used for rising factorial and falling factorial: $(x)^n=x(x+1)\dots(x+n-1)$ and $(x)_n=x(x-1)\dots(x-n+1)$.

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Can this sum over the q-Pochhammer symbol be simplified?

While considering the problem of the expected value of a dice fixing strategy on a two-sided die that comes up as $1$ with a probability of $\alpha$ and $0$ otherwise. I was studying the strategy ...
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Falling Factorial Notation

If $(x)_{n}$ refers to $$x(x-1)\cdots(x-n+1)$$ then what does $(xy;x)_{n}$ refer to? Is it $$xy(xy-1)\ldots(xy-n+1)?$$ Thanks. The notation in question is used on page two of this paper.
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Understanding falling factorial notation with semicolons.

I have a question about the following notation: $$\prod_{j=1}^{k/d}(z^dt^{dj}q^d;q^d)_{\infty}$$ I'm confused as to what I do with $q$ and the other variables especially when the semicolon is ...
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Improper integral involving sinc function and Pochhammer symbol

Can anyone please advise me how to integrate expressions of the form $\text{sinc}\,(x) / (1-x)_n$ along the real axis? Using a CAS, one could suggest that $$ n! \int_{-\infty}^\infty \frac{\sin \pi ...
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show $\sum_{j=0}^n (-1)^j {n \brack j}_q =0$ for n odd

I would like to show $\sum_{j=0}^n (-1)^j {n \brack j}_q =0$ for n odd, or preferably even more generally that $\sum_{j=0}^n (-1)^j {n \brack j }_q =\frac{1}{2}((-1)^n+1)(q;q)_{\frac{n}{2}}$. Using ...
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Product of a modified/generalized geometric progression

What is the solution to the following? It's sort of a modified/generalized geometric progression (or is there a known name for this kind?), $$\lim_{N\to\infty}\prod_{n=0}^{N}(1+a^n), a\in(0,1)$$
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Coefficient of $x^{50}$ in the expansion of $\prod_{n=1}^{52}{(x+n)}$

Find the coefficient of $x^{50}$ in the expansion of $$\prod_{n=1}^{52}{(x+n)}$$ I can't find a way out.
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Prove that ${}_2F_1(0,b;c;z)=1$

I do not know how I could prove that ${}_2F_1(0,\beta;\gamma;t)=1$ because when I apply the definition I get $0$, namely.. $$ \sum_{n=0}^{\infty}\frac{(0)_n(\beta)_n}{n!(\gamma)_n}t^n=0$$ someone ...
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How to derive Such infinite sum representation for Hypergeometric function?

I was reading a paper $[1]$ in which authors claimed that we can simplify below Gauss function to finite series if $m $ and $v$ are positive integers. $$ _2F_{1}(v,m+v;m+1;x)=\psi\sum_{c=0}^{v-1} ...
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Mathematical expressions for binomial coefficient and Pochhammer’s Symbol with negative values

I have two questions regarding the binomial coefficient and Pochhammer’s Symbol when they contain negative value; In the following example $\sum\limits_{k=0}^{-n} \binom{-n}{k} \left(a\right)_{-n}$. ...
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Integral of binomial coefficients

Let the integral in question be given by \begin{align} f_{n}(x) = \int_{1}^{x} \binom{t-1}{n} \, dt. \end{align} The integral can also be seen in the form \begin{align} f_{n}(x) = \frac{1}{n!} \, ...
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Falling power of a sum in terms of falling powers of the terms

I am trying to come up with an expression for $(x+y)^{\underline{n}}$ in terms of $x^{\underline{r}}$ and $y^{\underline{r}}$. I tried for $n=2$ and $n=3$ and it looks like binomial expansion holds, ...
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How can I show that $\left(a-n-1\right)!/\left(a-1\right)!=\left(-a\right)!\left(-1\right)^n/\left(-a+n\right)!$?

Is it possible to show that \begin{align}\frac{\left(a-n-1\right)!}{\left(a-1\right)!}\stackrel{?}{=}\frac{\left(-a\right)!\left(-1\right)^n}{\left(-a+n\right)!}\tag{1},\end{align} or, more ...
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Q Pochammer Symbol Product Identities

Consider the expression $$G(x,a) = \frac{1}{((1-a)x;a)_{\infty}}$$ Based on: Infinite sum involving ascending powers It follows that in the limit as $a \rightarrow 1$ ...
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Infinite series with a rising factorial?

Is there a closed-form expression for the infinite series $\sum_{i=0}^\infty (-\pi)^i\alpha^{(i)}$ For known $\pi,\alpha\in [0,1)$ where $\alpha^{(i)}$ is the rising factorial or Pochhammer symbol ...
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Is $\frac{(\alpha)^n (\beta)^n} {(\delta)^n} > \frac{(\alpha+1)^n (\beta+1)^n} {(\delta+1)^n}$ for any $n$ ?(in this specific case)

Let $\alpha=(K-1)a$, $\beta=K$ and $\delta=Ka$, where $K>a\ge 1$ ($\delta>\alpha>\beta$). Can we claim that $\frac{(\alpha)^n (\beta)^n} {(\delta)^n} > \frac{(\alpha+1)^n (\beta+1)^n} ...
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is there anyone able to develop this series in order to get the following equality?

$\sum_{i=1}^\infty (1-\alpha)_{(i-1)}*\frac{\varepsilon^i}{i!}$ = $\frac{1-(1-\varepsilon)^{\alpha}}{\alpha}$ where $(1-\alpha)_{(i-1)}$ is the Pochammer symbol or rising\ascending factorial. Can ...
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Product identity for $n^n$

I came across the rather nice identity \begin{align} &&\frac{(-n)^{n-1} \Gamma (n+1)}{(1-n)_{n-1}}&&\tag{1}&\\ \\ &=&\prod _{k=1}^{n-1} \frac{(k+1) n^2}{n^2-k ...
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Another question about ratios of Pochhammer symbols

My question is similar to this question. Can $$\frac{(11/6)_n (7/6)_n (3/2)_n}{(3)_n}$$ be expressed 'nicely' in terms of factorials just like $(1/6)_n (1/2)_n (5/6)_n$ in the aforementioned question? ...
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Strange claim by WA involving nth derivative

Playing a bit around with WA i found this Namely: $$\frac {d^n}{d^nx} \left(\frac x{f(x)}\right)^{n+1}=x\left(\frac 1{f(x)}\right)^{n+1}(2)_n$$ For $n\in\mathbb{N_0}$ and $n+1\ne x$ and $x \ne 0$ ...
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Understanding relation between Product and Summation Notation

So I am given the following: $n = \sum_{i=1}^{k}m_{i}$ I am also given $x = \sum_{i=1}^{k}log(m_{i}) = log\prod_{i=1}^{k}m_{i}$ I was only given the first part, however I believe that is a ...
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Coefficients in Pochhammer Expansion

Can anyone tell me if there is a formula for finding the coefficient of $x^3$ in the expansion of $(3x+5)_{6}$, where $(a)_n$ denotes the Pochhammer symbol, i.e. $(a)_{n}=a\cdot(a+1)\cdots(a+n-1)$? ...
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An identity involving the Pochhammer symbol

I need help proving the following identity: $$\frac{(6n)!}{(3n)!} = 1728^n \left(\frac{1}{6}\right)_n \left(\frac{1}{2}\right)_n \left(\frac{5}{6}\right)_n.$$ Here, $$(a)_n = a(a + 1)(a + 2) \cdots (a ...
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Combinatorial Identity $ \sum_{k=1}^n (-1)^{k-1} \cdot q^{\frac{k(k-1)}{2}} \cdot \frac{\prod_{i=n-k+1}^n(1-q^i)}{\prod_{i=1}^k(1-q^i)} = 1 $

I have to validate the following identity which is defined: $$ \sum_{k=1}^n (-1)^{k-1} \cdot q^{\frac{k(k-1)}{2}} \cdot \frac{\prod_{i=n-k+1}^n(1-q^i)}{\prod_{i=1}^k(1-q^i)} = 1 $$ where ...
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Unimodality of q-binomial coefficients

The q-Pochhammer symbol $[n]_q!$ is defined as $$[n]_q! = \frac{(1-q^n)(1-q^{n-1})\cdots(1-q)}{(1-q)^n} = (1+q) (1+q+q^2) \cdots (1+q+\cdots+q^{n-1})$$ It can be easily shown that $[n]_q!$ (function ...
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Identity using q-Pochhammer symbols

Prove - $$∑_{n=0}^{∞} \frac{(a;q)_n}{(q;q)_n} q^{n\choose 2} q^n={(−q;q)_∞}{(aq;q^2)_∞}.$$ where $(a;q)$ are the q-Pochhammer symbols. I know that the RHS is the product of generating functions of ...
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Identity involving the rising factorial

I am reading a book about hypergeometric functions and in a proof of a transformation they use the supposedly obvious fact $$ \displaystyle\frac{(c-a-b)_{n-r}}{(n-r)!} = \frac{(c-a-b)_{n} ...
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The multinomial formula as three Pochhammer rising factorials

I need to describe: $${n \choose k,0,l,0,m}$$ as three rising factorials. How can I do this? As far as I know I can delete zero's, so it would be: $${n \choose k,l,m}=\frac{n!}{k!l!m!},$$ where ...
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Binomial-like sum involving falling factorials

We know that $\sum_{k=0}^n a^k \frac{n^{\underline k}}{k!} = (1+a)^n$. Is there a known (preferably closed) form for $\sum_{k=0}^n a^k n^{\underline k}$?
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Trying to prove that $\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$

How could one prove that: $$\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$$ This is about as far as I got: $$\prod_{k=1}^j \frac{2 k}{j+k-1} = \frac{2^j j!}{(j)_j} \implies$$ ...
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Definition domains of the pochhammer symbols?

What are the definition domains for $n$ and $x$ that gives $x^{(n)}$ (upper pochhammer symbol) and $(x)_n$ (lower pochhammer symbol) in $\mathbb{R}$ ?
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factoring infinite products of $q$-series with constant term equal to 1

I was thinking about the following infinite product: $$\prod_{n=0}^{\infty} \frac{ae^{-2n}+be^{-n}+c}{c}$$ The right way of generalizing it is to think in terms of $q$-Pochhammer symbols. If $r_{1}$ ...