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1 3 $begingroup$ I was looking at this Ramanujan phi-Function Let: $$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$ and this paper of the form, $$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and $$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$ By combining them together we have $$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and $$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$ The conjectured closed form for $(4)$ and $(5)$ , where $a=2$ are $$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and $$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$ Where G is the Catalan's constant. How can