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%I #99 Mar 14 2024 13:00:21
%S 1,2,0,9,1,9,9,5,7,6,1,5,6,1,4,5,2,3,3,7,2,9,3,8,5,5,0,5,0,9,4,7,7,0,
%T 4,8,8,1,8,9,3,7,7,4,9,8,7,2,8,4,9,3,7,1,7,0,4,6,5,8,9,9,5,6,9,2,5,4,
%U 1,5,4,5,4,0,8,4,2,3,5,9,2,2,4,5,6,0,8
%N Decimal expansion of Sum_{i >= 0} (i!)^2/(2*i+1)!.
%C Value of the Borwein-Borwein function I_3(a,b) for a = b = 1. - _Stanislav Sykora_, Apr 16 2015
%C The area of a circle circumscribing a unit-area regular hexagon. - _Amiram Eldar_, Nov 05 2020
%D George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press (2006), pp. 120-121.
%D L. B. W. Jolley, Summation of Series, Dover (1961), No. 261, pp. 48, 49, (and No. 275).
%H Xavier Gourdon and Pascal Sebah, <a href="http://numbers.computation.free.fr/Constants/Pi/piSeries.html">Collection of series for Pi</a> (see paragraph 7).
%H Su Hu and Min-Soo Kim, <a href="https://arxiv.org/abs/2201.09674">A generalization of Wallis' formula</a>, arXiv:2201.09674 [math.NT], 2022.
%H Richard Kershner, <a href="http://www.jstor.org/stable/2371320">The Number of Circles Covering a Set</a>, American Journal of Mathematics, 61(3), 665-671.
%H MIT Integration Bee, <a href="https://www.youtube.com/watch?v=MfI1gA910Cw">2023 MIT Integration Bee - Finals</a>, Problem 1.
%H László Fejes Tóth, <a href="http://dx.doi.org/10.1090/S0002-9904-1948-08969-8">An Inequality concerning polyhedra</a>, Bull. Amer. Math. Soc. 54 (1948), 139-146. See (9) p. 146.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Arithmetic-GeometricMean.html">Arithmetic-Geometric Mean</a>, equations 26-32.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F Equals 2*sqrt(3)*Pi/9 = 1 + 1/6 + 1/30 + 1/140 + 1/630 + 1/2772 + 1/12012 + ...
%F Equals m*I_3(m,m) = m*Integral_{x>=0} (x/(m^3+x^3)), for any m>0. - _Stanislav Sykora_, Apr 16 2015
%F Equals Integral_{x>=0} (1/(1+x^3)) dx. - _Robert FERREOL_, Dec 23 2016
%F From _Peter Bala_, Oct 27 2019: (Start)
%F Equals 3/4*Sum_{n >= 0} (n+1)!*(n+2)!/(2*n+3)!.
%F Equals Sum_{n >= 1} 3^(n-1)/(n*binomial(2*n,n)).
%F Equals 2*Sum_{n >= 1} 1/(n*binomial(2*n,n)). See Boros and Moll, pp. 120-121.
%F Equals Integral_{x = 0..1} 1/(1 - x^3)^(1/3) dx = Sum_{n >= 0} (-1)^n*binomial(-1/3,n) /(3*n + 1).
%F Equals 2*Sum_{n >= 1} 1/((3*n-1)*(3*n-2)) = 2*(1 - 1/2 + 1/4 - 1/5 + 1/7 - 1/8 + ...) (added Oct 30 2019). (End)
%F Equals Product_{k>=1} 9*k^2/(9*k^2 - 1). - _Amiram Eldar_, Aug 04 2020
%F From _Peter Bala_, Dec 13 2021: (Start)
%F Equals (2/3)*A093602.
%F Conjecture: for k >= 0, 2*sqrt(3)*Pi/9 = (3/2)^k * k!*Sum_{n = -oo..oo} (-1)^n/ Product_{j = 0..k} (3*n + 3*j + 1). (End)
%F Equals (3/4)*S - 1, where S = A248682. - _Peter Luschny_, Jul 22 2022
%F Equals Integral_{x=0..Pi/2} tan(x)^(1/3)/(sin(2*x) + 1) dx. See MIT Link. - _Joost de Winter_, Aug 26 2023
%F Continued fraction: 1/(1 - 1/(7 - 12/(12 - 30/(17 - ... - 2*n*(2*n - 1)/((5*n + 2) - ... ))))). See A000407. - _Peter Bala_, Feb 20 2024
%e 1.2091995761561452337293855050947704881893774987284937170465899569254...
%t RealDigits[2 Sqrt[3] Pi/9, 10, 100][[1]]
%o (PARI) a = 2*Pi/(3*sqrt(3)) \\ _Stanislav Sykora_, Apr 16 2015
%Y Cf. A000796, A002194, A093602, A248181, A257096, A257097, A186706.
%Y Cf. A091682 (Sum_{i >= 0} (i!)^2/(2*i)!).
%K nonn,cons,easy
%O 1,2
%A _Bruno Berselli_, Mar 06 2015
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