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Recall that the Apéry numbers are given by $$A_n:=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2\ \ \ \ (n=0,1,2,\dotsc).$$ In 2002 T. Sato discovered the following series for $1/\pi$ involving Apéry numbers: $$\sum_{k=0}^\infty(20k+10-3\sqrt5)\frac{A_k}{((\sqrt5+1)/2)^{12k}}=\frac{20\sqrt3+9\sqrt{15}}{6\pi}.$$ In a 2009 paper, H. H. Chan and H. Verrill deduced six more such series. In a 2025 preprint, R. Hemmecke, P. Paule and C.-S. Radu gave five more such series for $1/\pi$.

Here I report two new series for $1/\pi$ involving Apéry numbers that I have found.

Conjecture. We have $$\sum_{k=0}^\infty(6\sqrt{11}(2k+1)-13)\frac{(-1)^kA_k}{(199+60\sqrt{11})^k}=\frac{10+3\sqrt{11}}{\pi}\tag{1}\label{495096_1}$$ and $$\sum_{k=0}^\infty(38\sqrt{3}(2k+1)-49)\frac{(-1)^kA_k}{(1351+780\sqrt{3})^k}=\frac{26+15\sqrt{3}}{\pi}.\tag{2}\label{495096_2}$$

Remark. As the two series in \eqref{495096_1} and \eqref{495096_2} converge fast, it is easy to check \eqref{495096_1} and \eqref{495096_2} via Mathematica.

QUESTION. Are \eqref{495096_1} and \eqref{495096_2} missing from public references? Can one prove them via the theory of modular forms?

Your comments are welcome!

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    $\begingroup$ Where you write "check", can we assume that you mean "check numerically"? $\endgroup$ Commented May 24 at 14:44
  • $\begingroup$ Yes, "check numerically" is more accurate. $\endgroup$ Commented May 24 at 22:13

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