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