This question is related to Hankel determinants of harmonic numbers.
Let $f(n)=\sum_{k=1}^n \frac{2^k}{k}$ and $r(n)=\sum_{j=0}^n (-2)^{n-j}\binom{n}{j}\binom{n+j}{j}f(j).$
In order to compute the Hankel determinants $\det\left(f(i+j)\right)_{i,j=0}^n$ I need the following identities:
1) $r(2n)=0$,
2) $r(2n+1)=(-1)^n \frac{(2n+1)!}{((2n+1)!!)^2}2^{2n+2}.$
Are these identities known?
Define the sequence $a_n(x):=\sum_{j=0}^n(-2)^{n-j}\binom{n}j\binom{n+j}jx^j$ so that $r(n)=\int_0^2\frac{a_n(x)-a_n(1)}{x-1}dx$.
We need the following fact which follows from the Vandermonde-Chu identity: for $n\geq1$, \begin{align}\sum_{j=0}^n(-1)^j\binom{n}j\binom{n+j}j\frac1{j+1} &=\sum_{j=0}^n\binom{n}{n-j}\binom{-n-1}j\frac1{j+1} \\ &=\frac1{n+1}\sum_{j=0}^n\binom{n+1}{n-j}\binom{-n-1}j=0. \tag1 \end{align} At present, the implication of (1) is that $$\int_0^2a_n(x)dx=2(-2)^n\sum_{j=0}^n(-1)^j\binom{n}j\binom{n+j}j\frac1{j+1}=0. \tag2$$ Next, we apply Zeilberger's algorithm to generate the two recursive relations \begin{align} (n+2)a_{n+2}(x)-2(x-1)(2n+3)a_{n+1}(x)+4(n+1)a_n(x)&=0, \\ (n+2)a_{n+2}(1)\qquad \qquad \qquad \qquad \qquad \qquad+4(n+1)a_n(1)&=0. \end{align} Subtract the $2^{nd}$ equation from the $1^{st}$, divide through by $x-1$ and integrate $\int_0^2(\cdot)dx$. So, $$(n+2)r(n+2)+4(n+1)r(n)=0\qquad \implies \qquad r(n+2)=-\frac{4(n+1)}{n+2}\,r(n);$$ where (2) has been utilized effectively. Initial conditions are $r(0)=0$ and $r(1)=4$. The case $n$ even is transparent. The case $n$ odd also follows from induction on $n$ and the fact that if we write the RHS of your claim for $r(n)$ as $$t(n):=\frac{(-1)^{\lfloor n/2\rfloor}4^n}{n\binom{n-1}{\lfloor n/2\rfloor}}\chi_{odd}(n)$$ then it is easy to check that $\frac{t(n+2)}{t(n)}=-\frac{4(n+1)}{2n+3}$ when $n$ is odd. The proof is now complete.
Note. Here $\chi_{odd}$ is understood as $\chi_{odd}(odd)=1$ and $\chi_{odd}(even)=0$.
