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A244979 Decimal expansion of Pi/(2*sqrt(5)).

%I A244979 #29 Feb 16 2025 08:33:23
%S A244979 7,0,2,4,8,1,4,7,3,1,0,4,0,7,2,6,3,9,3,1,5,6,3,7,4,6,4,3,2,0,4,8,9,4,
%T A244979 7,9,9,4,6,6,5,0,9,1,8,7,0,6,7,2,0,2,4,1,9,9,8,9,7,2,1,0,2,6,1,9,2,1,
%U A244979 4,1,8,8,0,6,1,9,1,8,8,2,0,5,1,0,4,1,4,2,4,1,5,3,6,5,7,6,7,2,4,0,2,1,5,0,7
%N A244979 Decimal expansion of Pi/(2*sqrt(5)).
%D A244979 George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press (2006), Chapter 13 A Master Formula, p. 250.
%H A244979 Vincenzo Librandi, <a href="/A244979/b244979.txt">Table of n, a(n) for n = 0..10000</a>
%H A244979 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BetaFunction.html">Beta Function</a>
%H A244979 <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F A244979 Equals Integral_(0..1) (1 + x^2)/(1 + 3*x^2 + x^4) dx.
%F A244979 From _Peter Bala_, Feb 16 2015: (Start)
%F A244979 Also equals beta(1/2, 1/2)/(2*sqrt(5)), where 'beta' is Euler's beta function.
%F A244979 Pi/(2*sqrt(5)) = Integral_{t = 0..a} (1 + t^2)*(1 + t^6)/(1 + t^10) dt = a + a^3/3 + a^7/7 + a^9/9 - a^11/11 - a^13/13 - a^17/17 - a^19/19 + ..., where a = 1/2(sqrt(5) - 1). Hint: differentiate atan( sqrt(5)*(t - t^3)/(1 - 3*t^2 + t^4) ). (End)
%F A244979 Equals (1/2)*Sum_{n >= 0} (-1)^n*( 1/(10*n + 1) + 1/(10*n + 3) + 1/(10*n + 7) + 1/(10*n + 9) ). Cf. A019692. - _Peter Bala_, Oct 30 2019
%F A244979 From _Amiram Eldar_, Aug 06 2020: (Start)
%F A244979 Equals Integral_{x=0..oo} 1/(x^2 + 5) dx.
%F A244979 Equals 0.1 * Integral_{x=0..oo} log(1 + 5/x^2) dx. (End)
%F A244979 Equals Integral_{x = 0..1} 2/(4*x^2 + 5*(1 - x)^2) dx. - _Peter Bala_, Jul 22 2022
%e A244979 0.702481473104072639315637464320489479946650918706720241998972102619214188...
%t A244979 RealDigits[Pi/(2*Sqrt[5]), 10, 105] // First
%o A244979 (PARI) Pi/sqrt(20) \\ _Charles R Greathouse IV_, Sep 30 2022
%Y A244979 Cf. A244976, A244977, A244978, A019692.
%K A244979 nonn,cons,easy
%O A244979 0,1
%A A244979 _Jean-François Alcover_, Jul 09 2014