A175576 Decimal expansion Pi^(3/2)/Gamma(3/4)^2.

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

Offset: 1

Author

R. J. Mathar, Jul 15 2010

Keywords

• nonn
• cons
• easy

Comments

Entry 34 e of chapter 11 of Ramanujan's second notebook.

In addition, Pi^(3/2) / Gamma(3/4)^2 is the area of the unit "squircle" as defined in MathWorld. (Note that 8*Gamma(5/4)^2 / sqrt(Pi) is the same constant.) - Jean-François Alcover, Feb 24 2011

Real period of the elliptic curve y^2 = x*(x - 1)*(x - 1/2). See Rouse. - Peter Bala, Dec 06 2024

Examples

			3.708149354602743836867700694390520092435197647...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000
• Bruce C. Berndt, Chapter 11 of Ramanujan's second notebook, Bull. Lond. Math. Soc. vol 15 no 4 (1983) 273-320.
• Jeremy Rouse, Hypergeometric functions and elliptic curves, Ramanujan Journal, Vol. 12 (2006), pp. 197-205.
• Eric Weisstein's World of Mathematics, Squircle

Crossrefs

Cf. A062539, A068465, A175476, A175575, A290570.

Programs

Maple

Pi^(3/2)/GAMMA(3/4)^2 ; evalf(%) ;

Mathematica

RealDigits[Pi*EllipticTheta[3, 0, Exp[-Pi]]^2, 10, 50][[1]]
RealDigits[Pi^(3/2)/(Gamma[3/4])^2, 10, 50][[1]] (* G. C. Greubel, Feb 12 2017 *)

PARI

Pi^1.5/gamma(3/4)^2 \\ Charles R Greathouse IV, Jun 06 2016

Formula

Equals A175476 / A068465^2 = 1/A175575.

Equals Integral_{-oo, oo} 1/(1+2*x^2)^(3/4) or Integral_{-oo, oo} 1/sqrt(1+x^4). - Jean-François Alcover, Jun 04 2013

Equals sqrt(2)*L, where L is the lemniscate constant A062539. - Jean-François Alcover, Aug 11 2014

From Peter Bala, Mar 01 2022 : (Start)

Equals 3*Sum_{n >= 0} (1/(4*n+1) + 1/(4*n-3))*binomial(1/2,n). Cf. A290570.

Equals hypergeom([-1/2, 3/4, -3/4], [-1/4, 5/4], -1).

Equals 2*hypergeom([1/4, 3/4], [5/4], 1) = (16/5)*hypergeom([-1/4, -3/4], [5/4], 1). (End)

Equals 2 * A093341. - R. J. Mathar, Dec 08 2023

From Peter Bala, Dec 06 2024: (Start)

Equals Pi*hypergeom([1/2, 1/2], [1], 1/2).

Equals 2*Integral_{x = 0..Pi/2} 1/sqrt(1 - (1/2)*sin^2(x)) dx. See Rouse. (End)