Find a general form of a sequence and its sum I have a problem to find a general form of the sequence
\begin{align}
- \frac{{n\left( {n - 1} \right)}}{{2\left( {2n - 1} \right)}},\frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{2 \cdot 4 \cdot \left( {2n - 1} \right)\left( {2n - 3} \right)}}, - \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)}}{{2 \cdot 4 \cdot 6 \cdot \left( {2n - 1} \right)\left( {2n - 3} \right)\left( {2n - 5} \right)}}, \cdots :=a_n(k),\qquad n\ge 2
\end{align}
and then to find the sum $\sum |a_n(k)|^2$, $1\le k\le n$.
I have tried as follows:
$a_n(k)=\frac{(-1)^kP(n,2k)}{(2k)!!A_n(k)}$, where $P(n,2k)=\frac{n!}{(n-2k)!}$ and $\,\,A_n(k):=(2n-1)(2n-3)(2n-5)\cdots (2n-2k+1),\qquad k\ge1$
Is this true. If yes, can we write $A_n(k)$ in a closed form?! After all of that what is the sum of $|a_n(k)|^2$, $1\le k\le n$.

Here is a more compact representation as sum formula, most of it was already stated in the comment section.

Since
  \begin{align*}
a_n(k)=\frac{(-1)^kn(n-1)\cdots (n-2k+1)}{2\cdot4\cdots (2k)\cdot(2n-1)(2n-3)\cdots(2n-2k+1)}\qquad\qquad 1\leq k\leq n
\end{align*}
We obtain
  \begin{align*}
a_n(k)&=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{(2k)!!}\cdot\frac{(2n-2k-1)!!}{(2n-1)!!}\tag{1}\\
&=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{(2k)!!}\cdot\frac{(2n-2k)!}{(2n-2k)!!}\cdot\frac{(2n)!!}{(2n)!}\tag{2}\\
&=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{2^kk!}\cdot\frac{(2n-2k)!}{2^{n-k}(n-k)!}\cdot\frac{2^nn!}{(2n)!}\tag{3}\\
&=(-1)^k\frac{n!n!}{(2n)!}\cdot\frac{1}{k!(n-k)!}\cdot\frac{(2n-2k)!}{(n-2k)!}\\
&=\frac{(-1)^k\binom{n}{k}\binom{2n-2k}{n}}{\binom{2n}{n}}
\end{align*}

Comment:


*

*In (1) we use double factorials $(2n)!!=(2n)(2n-2)\cdots4\cdot 2$

*In (2) we use
$(2n)!=(2n)!!(2n-1)!!$

*In (3) we use
$(2n)!!=2^nn!$

I don't think that the series
  \begin{align*}
\sum_{k=1}^n\left|a_n(k)\right|^2=\binom{2n}{n}^{-2}\sum_{k=1}^{n}\binom{n}{k}^2\binom{2n-2k}{n}^2\qquad\qquad n\geq    1
\end{align*}
  has a nice closed formula. The first few terms are
  \begin{align*}
0,4,144,3636,82000,1764400,37164736,\ldots
\end{align*}
  but they are not known to OEIS.

I've also tried some standard techniques and checked for instance section 2.9 in Riordan Array Proofs of Identities in Gould’s Book which contains binomial identities of the type we need, but without success.

Wolfram Alpha provides following representation via hypergeometric series
  \begin{align*}
\sum_{k=1}^n\left|a_n(k)\right|^2
&={}_{4}F_{3}\left(\frac{1}{2}-\frac{n}{2},\frac{1}{2}-\frac{n}{2},-\frac{n}{2},-\frac{n}{2};1,\frac{1}{2}-n,\frac{1}{2}-n;1\right)-1
\end{align*}

