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

0,1

Comments

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

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 ?

Formula

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

Example

0.828859579669279286697229020775103026769105755977121145244040331795...

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. *)

Prog

(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))))

Crossrefs

Cf. A324425.

Sequence in context: A352473 A211269 A278809 * A278261 A296301 A019865

Adjacent sequences: A306614 A306615 A306616 * A306618 A306619 A306620

Keyword

nonn,cons

Author

Vaclav Kotesovec, Feb 28 2019

Extensions

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

Status

approved