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Decimal expansion of Pi/sqrt(27).
48

%I #108 Jul 14 2024 08:10:22

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

%T 4,4,0,9,4,6,8,8,7,4,9,3,6,4,2,4,6,8,5,8,5,2,3,2,9,4,9,7,8,4,6,2,7,0,

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

%N Decimal expansion of Pi/sqrt(27).

%C Original name: Decimal expansion of sum(1/(n*binomial(2*n,n)), n=1..infinity).

%C This appears to be Pi/sqrt(27). See A111510. - _Marco Matosic_, Feb 27 2008

%C This is Pi*sqrt(3)/9 = A019676*A002194, see eq. (12) in Lehmer link. - _R. J. Mathar_, Mar 04 2009

%C Value of the Dirichlet L-series of the non-principal character modulo m=3 (A102283) at s=1. - _R. J. Mathar_, Oct 03 2011

%C Construct the largest possible circle inside a given equilateral triangle. This constant is the ratio of the area of the circle to the area of the triangle (A245670 is analogous square in triangle). - _Rick L. Shepherd_, Jul 29 2014

%D L. B. W. Jolley, Summation of Series, Dover, 1961, eq. (81), page 16.

%H Vincenzo Librandi, <a href="/A073010/b073010.txt">Table of n, a(n) for n = 0..1000</a>

%H Jonathan M. Borwein and Roland Girgensohn, <a href="http://dx.doi.org/10.1007/s00010-005-2774-x">Evaluations of binomial series</a>, Aequat. Math. 70 (2005), 25-36.

%H Étienne Fouvry, Claude Levesque, and Michel Waldschmidt, <a href="https://arxiv.org/abs/1712.09019">Representation of integers by cyclotomic binary forms</a>, arXiv:1712.09019 [math.NT], 2017.

%H Alessandro Languasco, Pieter Moree, Sumaia Saad Eddin, and Alisa Sedunova, <a href="https://arxiv.org/abs/1908.01152">Computation of the Kummer ratio of the class number for prime cyclotomic fields</a>, arXiv:1908.01152 [math.NT], 2019. See r(q) for q=3 in Table 1, p. 7.

%H D. H. Lehmer, <a href="http://www.jstor.org/stable/2322496">Interesting Series Involving the Central Binomial Coefficient</a>, Am. Math. Monthly 92 (1985) 449-457. See eq. (12).

%H Courtney Moen, <a href="https://www.jstor.org/stable/2690456">Infinite series with binomial coefficients</a>, Math. Mag. 64 (1) (1991), 53-55.

%H Paul J. Nahin, <a href="https://doi.org/10.1007/978-3-030-43788-6">Inside interesting integrals</a>, Undergrad. Lecture Notes in Physics, Springer (2020), (1.6.2)

%H Simon Plouffe, <a href="http://vixra.org/pdf/1409.0110v1.pdf">Sum(1/(n*binomial(2*n,n)), n=1..infinity)</a>, see p. 87.

%H Renzo Sprugnoli, <a href="https://www.emis.de/journals/INTEGERS/papers/g27/g27.Abstract.html">Sums of reciprocals of the central binomial coefficients</a>, El. J. Combin. Numb. Th. 6 (2006) # A27

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CentralBinomialCoefficient.html">Central Binomial Coefficient</a>.

%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>.

%F -Pi/(3*sqrt(3)) = Sum_{n>=0} (1/(6*n+1) - 2/(6*n+2) - 3/(6*n+3) - 1/(6*n+4) + 2/(6*n+5) + 3/(6*n+6)). - _Mats Granvik_, Sep 08 2013

%F Equals Integral_{0..oo} 2*x/((x^2+1)*(x^4+x^2+1)) dx. - _Jean-François Alcover_, Sep 10 2013

%F From _Peter Bala_, Feb 16 2015: (Start)

%F Pi/sqrt(27) = Sum_{n >= 0} 1/((3*n + 1)*(3*n + 2)) = 1 - 1/2 + 1/4 - 1/5 + 1/7 - 1/8 + ....

%F Continued fraction: 1/(1 + 1^2/(1 + 2^2/(2 + 4^2/(1 + 5^2/(2 + ... + (3*n + 1)^2/(1 + (3*n + 2)^2/(2 + ... ))))))).

%F Pi/sqrt(27) = Integral_{t = 0..1/2} 1/(t^2 - t + 1) dt = Integral_{t = 0..1/2} (1 + t - t^3 - t^4)/(1 - t^6) dt.

%F Pi/sqrt(27) = (1/4)*Sum_{n >= 0} (-1/8)^n * (9*n + 5)/((3*n + 1)*(3*n + 2)).

%F BBP-type formulas:

%F Pi/sqrt(27) = Sum_{n >= 0} (1/64)^(n+1)*( 32/(6*n + 1) + 16/(6*n + 2) - 4/(6*n + 4) - 2/(6*n + 5) ) follows from the above integral representation.

%F Pi/sqrt(27) = Sum_{n >= 0} (-1)^n*(1/27)^(n+1)*( 9/(6*n + 1) + 9/(6*n + 2) + 6/(6*n + 3) + 3/(6*n + 4) + 1/(6*n + 5) ) follows from the result: Pi/3 = Integral_{t = 0..1/sqrt(3)} 1/(1 - sqrt(3)*t + t^2) dt. (End)

%F Equals Integral_{x=0..oo} x*I_0(x)*K_0(x)^2 dx over a triple product of modified Bessel functions. - _R. J. Mathar_, Oct 14 2015

%F From _Amiram Eldar_, Aug 15 2020: (Start)

%F Equals Integral_{x=0..oo} 1/(exp(x) + exp(-x) + 1) dx.

%F Equals Integral_{x=0..oo} 1/(1 + x + x^2 + x^3 + x^4 + x^5) dx. (End)

%F Equals (3*S - 4)/8, where S = A248682. - _Peter Luschny_, Jul 22 2022

%F Equals Product_{p prime} (1 - Kronecker(-3, p)/p)^(-1) = Product_{p prime != 3} (1 + (-1)^(p mod 3)/p)^(-1). - _Amiram Eldar_, Nov 06 2023

%F From _Peter Bala_, Dec 09 2023: (Start)

%F Pi/sqrt(27) = Sum_{n >= 1} 1/(n*binomial(2*n,n)) = (1/6)*Sum_{n >= 1} 3^n/(n*binomial(2*n,n)) (see Lehmer, equation 12, and also p. 456).

%F Pi/sqrt(27) = (1/2)*Sum_{n >= 0} 1/((2*n + 1)*binomial(2*n,n)).

%F Pi/sqrt(27) = (9/4)*Sum_{n >= 3} (n - 1)*(n - 2)/binomial(2*n,n). (End)

%F Equals integral_{x=0..oo} 1/(1-x^3) dx [Nahin]. - _R. J. Mathar_, May 16 2024

%e 0.60459978807807261686469275254738524409468...

%t RealDigits[ N [Sum[1/(n*Binomial[2n, n]), {n, 1, Infinity}], 110]] [[1]]

%t RealDigits[Pi/(3*Sqrt[3]), 10, 105][[1]] (* _T. D. Noe_, Sep 11 2013 *)

%o (PARI) Pi/sqrt(27) \\ _Charles R Greathouse IV_, Sep 11 2013

%o (Magma) R:=RealField(106); SetDefaultRealField(R); n:=Pi(R)/Sqrt(27); Reverse(Intseq(Floor(10^105*n))); // _Bruno Berselli_, Mar 12 2018

%Y Cf. A002194, A019676, A111510, A245670, A248682.

%K nonn,cons,easy

%O 0,1

%A _Robert G. Wilson v_, Aug 03 2002