The Product of two rising factorials Let $(x)_{2 n}$ is a rising factorial of 2n terms ,$\left(\frac{x}{2}\right)_{n}$ and $\left(\frac{1+x}{2}\right)_{n}$ are rising factorial of n terms.
From https://mathworld.wolfram.com/PochhammerSymbol.html there is a following known formula as you see;
$(x)_{2 n}=2^{2 n}\left(\frac{x}{2}\right)_{n}\left(\frac{1+x}{2}\right)_{n}$
I want to ask for instance as a special case  $x=1$
$(1)_{2 n}=2^{2 n}\left(\frac{1}{2}\right)_{n}\left(1\right)_{n}$
But i want the left side to be a rising factorial in n terms. I mean the product of 2 rising factorial of n terms are equal a rising factorial in n terms ,too.
$(1)_{2 n}$=$(c)_{n}$ Is it possible? If it is So how can c be here?
 A: You're somewhat overlooking the definition and meaning of the symbols. Since
$$(1)_{2n} = 1 \cdot 2 \cdot \ldots \cdot 2n$$
we have that
$$(1)_{2n} = (2n)!$$
Hence, if we want a product of $n$ consecutive terms (that being $(c)_n$) starting at $c$ that equals this, we simply need $c=n+1$.
A: We can find $c$ for small $n$ and at least approximately for large $n$. We consider
\begin{align*}
(2n)!=(1)_{2n}=(c)_n=c\cdot (c+1)\cdots (c+n-1)\tag{1}
\end{align*}
where the right hand side is a polynomial in $c$ of degree $n$.

Solving (1) for small values of $n$ we obtain
\begin{align*}
\color{blue}{(1)_2}&=2!=2\color{blue}{=(2)_1}\\
\color{blue}{(1)_4}&=4!=24\color{blue}{=\left(\frac{\sqrt{97}-1}{2}\right)_2}=(4.4244\ldots)_2\\
\color{blue}{(1)_6}&=6!=720\color{blue}{=(8)_3}\\
\color{blue}{(1)_8}&=8!=40\,320=\color{blue}{\left(\frac{\sqrt{5+4\sqrt{40\,321}}-3}{2}\right)_4}=(12.714\ldots)_4\\
\color{blue}{(1)_{10}}&=10!=3\,628\,800\color{blue}{=(18.5582\ldots)_5}
\end{align*}
with some help of Wolfram Alpha.

For large $n$ we can employ Stirling's approximation formula
of $n!$ and we obtain a rough estimation by taking the $n$-th root of it.

We so get
\begin{align*}
\color{blue}{(1)_{2n}}&=(2n)!\sim \left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}\\
&\,\,\color{blue}{\simeq \left(\left(\frac{2n}{e}\right)^{2}\left(4\pi n\right)^{\frac{1}{2n}}\right)_n}\tag{2}
\end{align*}

Calculating for instance the value for $n=10$ we get $(1)_{10}\simeq(20.4\ldots)_5$.
