$$\sum_{n,m=0}^{\infty}\binom{2n}{n}\binom{2m}{m}\binom{2n+2m}{n+m}^3\left(\frac{1+6n+6m}{2^{10n+10m}}\right)=\frac{4}{\pi}$$
These are Ramanujan-Type Series for $1/\pi$ but based on a Double Sum Infinite Series.
Basically it's based on the following expansion of Complete Elliptic Integral of the First Kind $K$ with parameter $k$.
$$K=\frac{\pi}{2}\sum_{n,m=0}^{\infty}\binom{2n}{n}\binom{2m}{m}\binom{2n+2m}{n+m}^2\frac{k^{2n+2m}}{2^{6n+6m}}$$
From here on we can follow the similar process as deriving the normal $1/\pi$ series.
I haven't particularly seen these double sum ones, so was wondering whether anyone has worked upon them and looking for some reference on these.
Best Answer
We can rewrite it by substituting $k=m+n$, $$ \sum_{n\ge0}\frac{\left ( 6n+1 \right ) \left ( \frac12 \right )_n^3 }{2^{4n}(1)_n^3} \sum_{k=0}^{n} \binom{2k}{k} \binom{2n-2k}{n-k} =\frac{4}{\pi}. $$ The inner binomial sum is elementary and equals $2^{2n}$, giving the common type of Ramanujan's $\pi$ formula.
Z.W. Sun once conjectured lots of series identities in the paper. Some of them involve similar factors but seem much harder to handle.