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A twisted hypergeometric series $\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2$
Question. I was given that $$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2=\frac{32}\pi G\ln2+\frac{64}\pi\Im\operatorname{Li}_3\left(\frac{1+i}2\right)-2\ln^22-\frac53\pi^2$$ ...
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