A306617 Decimal expansion of a constant related to the asymptotics of A324425.

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

Offset: 0

Author

Vaclav Kotesovec, Feb 28 2019

Keywords

• nonn
• cons

Comments

Ulrich Neumann found a closed form, see the "Mathematica Stack Exchange" link.

Examples

			0.828859579669279286697229020775103026769105755977121145244040331795...
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..1000 [Terms beyond 79 were computed using Ulrich Neumann's program]
• Mathematica Stack Exchange, How to compute this integral with a better precision ?

Crossrefs

Cf. A324425.

Programs

Maple

evalf(exp(integrate(log(x^2 + y^2 + z^2), x = 0..1, y = 0..1, z = 0..1)), 20);
evalf(exp(integrate(-2 + 2*sqrt(y^2 + z^2) * arctan(1/sqrt(y^2 + z^2)) + log(1 + y^2 + z^2), y = 0..1, z = 0..1)), 20);

Mathematica

ixr = Exp[Integrate[1/3 (Log[1 + Sec[fi]^2] + (-7 + 3 Log[1 + Sec[fi]^2]) Sec[fi]^2 + 2 (Pi - 2 ArcTan[Sec[fi]]) Sec[fi]^3), {fi, 0, Pi/4}]]; Chop[N[ixr, 120]] (* A program by Ulrich Neumann added by Vaclav Kotesovec, Mar 03 2019. The calculation takes several minutes. *)

PARI

exp(intnum(z=0, 1 ,intnum(y=0, 1, intnum(x=0, 1, log(x^2 + y^2 + z^2)))))

PARI

exp(intnum(z=0, 1 ,intnum(y=0, 1, -2 + 2*sqrt(y^2 + z^2) * atan(1/sqrt(y^2 + z^2)) + log(1 + y^2 + z^2))))

Formula

Equals limit_{n->infinity} (A324425(n)^(1/n^3))/n^2.

Extensions

More terms computed by Ulrich Neumann added by Vaclav Kotesovec, Mar 03 2019