Let $H\_n=\sum\_{k=1}^n\frac 1 k$ be the $n$-th harmonic number with $H\_0=0.$


**Question:** Is the following true?
$$\det\left(H\_{i+j}\right)\_{i,j=0}^n=(-1)^n \frac{2H\_{n}}{n! \prod\_{j=1}^n \binom{2j}{j} \binom{2j-1}{j}}.$$


**Edit:**
Comparing with the orthogonal polynomials whose moments are the numbers $\frac{1}{n+1}$ it suffices to show the following identity:
$$\sum\_{j=0}^n (-1)^j\frac{\binom n j \binom{n+j} j}{\binom{2n} n} H\_j \prod\_{j=0}^{n-1}\frac{(j!)^3}{(n+j)!} = (-1)^n \frac{2H\_n}{n! \prod\_{j=1}^n \binom{2j}{j} \binom{2j-1}{j}}.$$



Hankel determinants of harmonic numbers
I prove your identity $$\sum\_{j=0}^n (-1)^j\binom{n}{j}\binom{n+j}{j} H\_j= 2(-1)^n H\_{n}$$
which you claim to imply the result.


The method is the same as [here](https://mathoverflow.net/questions/238978/a-combinatorial-identity-involving-harmonic-numbers). 


At first, use $(-1)^k\binom{n+k}k=\binom{-n-1}k$. Then $$F(y):=\sum\_k (-1)^k\binom{n}k\binom{n+k}ky^k=[x^n] (1+x)^n(1+xy)^{-n-1}.$$
Next, for any polynomial $F(y)=\sum c\_ky^k$ we have 
$$
\sum c\_kH\_k=\int\_0^1 \frac{F(y)-F(1)}{y-1}dy.
$$
Integration over $[0,1]$ and taking the coefficient of $x^n$ commute, thus we have to prove
$$
[x^n]\int\_0^1\frac{(\frac{x+1}{1+xy})^n\cdot \frac1{1+xy}-\frac1{1+x}}{y-1}dy=2(-1)^nH\_n.
$$
A natural change of variables here is $t=(1+x)/(1+xy)$, we get that our integral equals $$-\frac{1}{1+x}\int\_1^{1+x}\frac{1-t^{n+1}}{t(1-t)}dt=\frac{-\log(1+x)+H\_n}{1+x}-\sum\_{i=1}^n\frac{(1+x)^{i-1}}i.$$
A coefficient of $x^n$ indeed equals $2(-1)^nH\_n$.


