cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A273816 Decimal expansion the Bessel moment c(3,0) = Integral_{0..inf} K_0(x)^3 dx, where K_0 is the modified Bessel function of the second kind.

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, May 31 2016

Keywords

Examples

			6.94882278107962978943643644547082975767485113260989173516238...
		

Crossrefs

Cf. A273817 (c(3,1)), A273818 (c(3,2)), A273819 (c(3,3)).

Programs

  • Mathematica
    c[3, 0] = 3*Gamma[1/3]^6/(32*Pi*2^(2/3));
    RealDigits[c[3, 0], 10, 103][[1]]

Formula

c(3, 0) = 3*Gamma(1/3)^6/(32*Pi*2^(2/3)).
Equals (1/2)*Pi*K[(1/4)*(2 - Sqrt[3])]*K[(1/4)*(2 + Sqrt[3])], where K(x) is the complete elliptic integral of the first kind.
Also equals sqrt(3) Pi^3/8 3F2(1/2, 1/2, 1/2; 1, 1; 1/4), where 3F2 is the generalized hypergeometric function A263490.

A271563 Decimal expansion of Sum_{j>=0} Sum_{i>=0} (-1/4)^i*(-1)^j*binomial(2i,i)/((2j+1)(i+2j+2)).

Original entry on oeis.org

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

Views

Author

John M. Campbell, Apr 09 2016

Keywords

Examples

			0.3445436367923706403320533879002043065894259746135921255085777...
		

Crossrefs

Programs

  • Maple
    evalf((Pi - 2*sqrt(1+sqrt(2)) * arctan(2*sqrt(2+10*sqrt(2))/7)) / sqrt(2), 120); # Vaclav Kotesovec, Apr 10 2016
  • Mathematica
    RealDigits[(Pi - 2*Sqrt[1 + Sqrt[2]] * ArcTan[(2/7)*Sqrt[2 + 10*Sqrt[2]]])/Sqrt[2], 10, 120][[1]]
    N[Sum[Sum[((-1)^(i + j) 4^-i Binomial[2 i, i])/((1 + 2 j) (2 + i + 2 j)), {i, 0, Infinity}], {j, 0, Infinity}]]
  • PARI
    (Pi - 2*sqrt(1+sqrt(2)) * atan(2*sqrt(2+10*sqrt(2))/7)) / sqrt(2)

Formula

Equals (Pi - 2*sqrt(1+sqrt(2)) * arctan(2*sqrt(2+10*sqrt(2))/7)) / sqrt(2). - Vaclav Kotesovec, Apr 10 2016
Showing 1-2 of 2 results.