A186642 Decimal expansion of the "squircle" perimeter.

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

Offset: 1

Author

Jean-François Alcover, Feb 25 2011

Keywords

• cons
• easy
• nonn

Comments

This squircle constant can also be computed as a series in terms of incomplete beta function with coefficients from sequences A002596 and A120777:

a(n) = (-1)^(n+1) numerator((2n-3)!!/n!) ( sequence A002596);

b(n) = denominator(binomial(2n+2, n+1)/2^(2n+1)) ( sequence A120777).

Generic term:

u(n) = (a(n)/b(n-1))*beta(1/2, (6n+1)/4, 1-(3/2)*n).

Here is the series computed up to 5 terms:

4*2^(3/4) + sum(u(n), {n, 1, 5}) =

4*2^(3/4) + beta(1/2, 7/4, -1/2) - (1/4)*beta(1/2, 13/4, -2) + (1/8)* beta(1/2, 19/4, -7/2) - (5/64)*beta(1/2, 25/4, -5) + (7/128)*beta(1/2, 31/4, -13/2).

It evaluates to 7.018901897260651...

Numeric check with 10000 terms:

4*2^(3/4) + sum(u(n), {n, 1, 10000}) = 7.017697943556135...

Examples

			7.01769794356404...
		

Links

• Eric Weisstein's World of Mathematics, Squircle

Crossrefs

Cf. A175576 (unit squircle area).

Programs

Mathematica

First @ RealDigits[N[2*Integrate[Sqrt[1 + x^(3/2)/(1 - x)^(3/2)]/x^(3/4), {x, 0, 1/2}], 100]]
(* This other series formula gives 100 correct digits: *)
First @ RealDigits[1/Sqrt[Pi]*NSum[(-1)^(n+1)*Gamma[n - 1/2]*Beta[1/2, (6n + 1)/4, 1 - (3/2)n] / n!, {n, 0, Infinity},WorkingPrecision -> 100, Method -> "AlternatingSigns"], 10, 100]

Formula

-((3^(1/4) MeijerG[{{1/3, 2/3, 5/6, 1, 4/3}, {}}, {{1/12, 5/12, 7/12, 3/4, 13/12}, {}}, 1])/(16 Sqrt[2] Pi^(7/2) Gamma[5/4])). - Eric W. Weisstein, Oct 25 2011