This question is related to [Hankel determinants of harmonic numbers](https://mathoverflow.net/questions/262177/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?



An identity related to Hankel determinants of $\sum_{k=1}^n \frac{2^k}{k}$
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](https://en.wikipedia.org/wiki/Vandermonde's_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](https://en.wikipedia.org/wiki/Zeilberger%27s_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 $\chi\_{odd}(even)=0$.


