Simplify the following expression $\sum_{j=0}^n 2^n \binom{1/2}{n-j}\binom{j-1/2}{j}$ Simplify the following expression $\sum_{j=0}^n 2^n \binom{1/2}{n-j}\binom{j-1/2}{j}$ to something without a summutation mark (and probably elementary).
The question originates from the $n$th term of the polynomial expansion of $\sqrt{\frac{1+2x}{1-2x}}$, i.e. the coefficient of $x^n$ in its Maclaurin series.
We define $\binom{r}{n}=\frac{a(a-1)(a-2)...(a-b+1)}{b!}$ where $r$ is any real number and $n$ is a natural number.
 A: We already know for $n\geq 0$:
\begin{align*}
\sum_{j=0}^n 2^n \binom{1/2}{n-j}\binom{j-1/2}{j}=[x^n]\sqrt{\frac{1+2x}{1-2x}}\tag{1}
\end{align*}

We obtain from (1)
  \begin{align*}
\color{blue}{\sum_{j=0}^n}&\color{blue}{ 2^n \binom{1/2}{n-j}\binom{j-1/2}{j}}\\
&=[x^n]\sqrt{\frac{1+2x}{1-2x}}\\
&=[x^n]\frac{1+2x}{\sqrt{1-4x^2}}\tag{2}\\
&=[x^n]\left(1+2x\right)\sum_{j=0}^\infty\binom{-\frac{1}{2}}{j}\left(-4x^2\right)^j\tag{3}\\
&=\left([x^n]+2[x^{n-1}]\right)\sum_{j=0}^\infty\binom{-\frac{1}{2}}{j}(-4)^jx^{2j}\tag{4}\\
&\,\,\color{blue}{=}\begin{cases}
\color{blue}{\binom{-\frac{1}{2}}{m}(-4)^m}&\qquad\color{blue}{ n=2m}\\
\tag{5}\\
\color{blue}{2\binom{-\frac{1}{2}}{m}(-4)^m}&\qquad\color{blue}{ n=2m+1}
\end{cases}
\end{align*}

Comment:


*

*In (2) we  expand   the fraction with $1+2x$.

*In (3) we apply the binomial series expansion.

*In (4) we use $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

*In (5) we select the coefficient of $x^n$ for even and odd $n$.
A: Looks like $$2\binom{n-1}{\lfloor(n-1)/2\rfloor}-[n=0].$$  See OEIS A063886.
