A319461 Decimal expansion of Pi^2/8 - (3/8)*log(2)^2.

1, 0, 5, 3, 5, 3, 0, 6, 6, 9, 9, 1, 6, 8, 4, 4, 2, 9, 3, 1, 0, 4, 1, 4, 7, 9, 2, 7, 6, 1, 2, 0, 1, 9, 5, 2, 7, 5, 1, 5, 2, 5, 5, 0, 6, 9, 0, 5, 7, 1, 4, 4, 2, 3, 3, 2, 2, 6, 5, 9, 4, 6, 2, 1, 9, 1, 8, 6, 3, 1, 0, 8, 4, 4, 5, 4, 9, 6, 2, 7, 2, 8, 3, 0, 0, 2, 4, 1

Offset

1,3

Comments

Ramanujan's question 642, part 2, in the Journal of the Indian Mathematical Society (VII, 80) asked: "Show that Sum_{n>=0} (Sum_{k=0..n} 1/(2*k + 1)) * 9^(-n)/(2*n + 1) = Pi^2/8 - (3/8)*log(2)^2".

Links

Table of n, a(n) for n=1..88.

B. C. Berndt, Y. S. Choi, and S. Y. Kang, The problems submitted by Ramanujan to the Journal of Indian Math. Soc., in: Continued fractions, Contemporary Math., 236 (1999), 15-56 (see Q642, JIMS VII).

B. C. Berndt, Y. S. Choi, and S. Y. Kang, The problems submitted by Ramanujan to the Journal of Indian Math. Soc., in: Continued fractions, Contemporary Math., 236 (1999), 15-56 (see Q642, JIMS VII).

Example

1.0535306699168442931041479276120195275152550690571442332...

Mathematica

RealDigits[Pi^2/8 - 3*Log[2]^2/8, 10, 120][[1]] (* Amiram Eldar, Jun 27 2023 *)

Prog

(PARI) Pi^2/8-(3/8)*log(2)^2
(PARI) suminf(n=0, sum(k=0, n, 1/(2*k+1))*9^(-n)/(2*n+1))

Crossrefs

Cf. A319460.

Sequence in context: A239805 A333236 A270915 * A201762 A153386 A112920

Adjacent sequences: A319458 A319459 A319460 * A319462 A319463 A319464

Keyword

nonn,cons

Author

Hugo Pfoertner, Sep 19 2018

Status

approved