A069998 Decimal expansion of sqrt(Pi/2).

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

Offset: 1

Author

Benoit Cloitre, May 01 2002

Keywords

• cons
• easy
• nonn

Comments

This constant, sqrt(Pi/2), appears in one of the formulations of the Birthday Problem: An asymptotic expansion of the expected value for the average number of people required to find a pair having the same birthday out of k possible birthdays is sqrt(Pi/2)*sqrt(k) + 2/3 + 1/12*sqrt(Pi/2)*sqrt(1/k) - 4/135*1/k + ... found by the Indian mathematician Srinivasa Ramanujan (1887-1920). - Martin Renner, Sep 14 2016

Examples

			Sqrt(Pi/2) = 1.253314137315500251207882642... - _Wesley Ivan Hurt_, Sep 22 2016
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000
• P. Flajolet, P. J. Grabner, P. Kirschenhofer and H. Prodinger, On Ramanujan's Q-Function, Journal of Computational and Applied Mathematics 58 (1995), 103-116.
• I. S. Gradsteyn and I. M. Ryzhik, Table of integrals, series and products, (1980), page 420 (formulas 3.757.1, 3.757.2).
• MIT Integration Bee, 2024 MIT Integration Bee - Semifinals, Problem 3.
• Index entries for transcendental numbers

Crossrefs

Cf. A064619.

Programs

Maple

Digits:=100: evalf(sqrt(Pi/2)); # Wesley Ivan Hurt, Sep 22 2016

Mathematica

RealDigits[Sqrt[Pi/2], 10, 120][[1]] (* Harvey P. Dale, Jul 24 2012 *)

PARI

sqrt(Pi/2) \\ G. C. Greubel, Jan 09 2017

PARI

intnum(x=0, [oo, -I], sin(x)/sqrt(x)) \\ Gheorghe Coserea, Sep 23 2018

PARI

intnum(x=[0, -1/2], [oo, I], cos(x)/sqrt(x)) \\ Gheorghe Coserea, Sep 23 2018

Formula

From A.H.M. Smeets, Sep 22 2018: (Start)

Equals Integral_{x >= 0} sin(x)/sqrt(x) dx [see Gradsteyn and Ryzhik].

Equals Integral_{x >= 0} cos(x)/sqrt(x) dx [see Gradsteyn and Ryzhik]. (End)

Equals Integral_{x>=0} (sin(x)-x*cos(x))/x^(3/2) dx. - Amiram Eldar, May 08 2021

Equals Integral_{x=0..Pi/2} sin(cot(x)^2)*sec(x)^2 dx. See MIT link. - Kritsada Moomuang, Jun 23 2025