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Gourevitch's conjecture¹: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$

¹Jesús Guillera: About a New Kind of Ramanujan-Type Series; Experimental Mathematics (2003), Volume: 12, Issue: 4, page 507-510; DOI: 10.1080/10586458.2003.10504518, eudml

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UPDATE: As mentioned in a comment below, this conjecture has now been proved by K. C. Au, Wilf-Zeilberger seeds and non-trivial hypergeometric identities, arXiv:2312.14051, 26 Dec 2023.

Gourevitch's conjecture¹: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$

¹Jesús Guillera: About a New Kind of Ramanujan-Type Series; Experimental Mathematics (2003), Volume: 12, Issue: 4, page 507-510; DOI: 10.1080/10586458.2003.10504518, eudml

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UPDATE: As mentioned in a comment below, this conjecture has now been proved by K. C. Au, Wilf-Zeilberger seeds and non-trivial hypergeometric identities, arXiv:2312.14051, 26 Dec 2023. But Au mentions another identity that was still conjectural at the time of writing; if we define $(x)_n := x(x+1)(x+2)\cdots(x+n-1)$ then $$\sum_{n=0}^\infty {1920n^2 + 304n + 15 \over 7^{4n}n!^5}\biggl(\frac{1}{2}\biggr)_n \biggl(\frac{1}{8}\biggr)_n \biggl(\frac{3}{8}\biggr)_n \biggl(\frac{5}{8}\biggr)_n \biggl(\frac{7}{8}\biggr)_n = {56\sqrt{7}\over \pi^2}.$$

Gourevitch's conjecture¹: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$

¹Jesús Guillera: About a New Kind of Ramanujan-Type Series; Experimental Mathematics (2003), Volume: 12, Issue: 4, page 507-510; DOI: 10.1080/10586458.2003.10504518, eudml

Gourevitch's conjecture¹: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$

¹Jesús Guillera: About a New Kind of Ramanujan-Type Series; Experimental Mathematics (2003), Volume: 12, Issue: 4, page 507-510; DOI: 10.1080/10586458.2003.10504518, eudml

---

UPDATE: As mentioned in a comment below, this conjecture has now been proved by K. C. Au, Wilf-Zeilberger seeds and non-trivial hypergeometric identities, arXiv:2312.14051, 26 Dec 2023.

Gourevitch's conjecture: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$

Gourevitch's conjecture¹: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$

¹Jesús Guillera: About a New Kind of Ramanujan-Type Series; Experimental Mathematics (2003), Volume: 12, Issue: 4, page 507-510; DOI: 10.1080/10586458.2003.10504518, eudml