A375594 Decimal expansion of Pi*(Pi^2*log(2) + 4*log(2)^3 + 6*zeta(3))/48.

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

Offset: 1

Author

R. J. Mathar, Aug 20 2024

Keywords

• nonn
• cons

Comments

Apart from a factor sqrt(Pi)/16 the same as Adamchik's generalized Stirling number [1/2,4].

Examples

			1.006980484962515...
		

Links

• V. S. Adamchik, On Stirling numbers and Euler sums, J. Comput. Appl. Math. 79 (1) (1997) 119-130.
• R. J. Mathar, Chebyshev approximation of x^m*(-log x)^l in the interval 0<=x<=1, arXiv:2408.15212 [math.CA] (2024).

Crossrefs

Cf. A019669 (2F1), A173623 (3F2), A318741 (4F3).

Programs

Maple

1/48*Pi*(Pi^2*log(2)+4*log(2)^3+6*Zeta(3)) ; evalf(%) ;

Mathematica

First[RealDigits[Pi*(Pi^2*Log[2] + 4*Log[2]^3 + 6*Zeta[3])/48, 10, 100]] (* Paolo Xausa, Aug 23 2024 *)

Formula

Equals 5F4(1/2,1/2,1/2,1/2,1/2; 3/2,3/2,3/2,3/2; 1) = Sum_{k>= 0} binomial(2k,k)/[2^(2k)*(2k+1)^4].

Equals A196878/6. - R. J. Mathar, Aug 23 2024