OFFSET
0,1
COMMENTS
Original name: Decimal expansion of sum(1/(n*binomial(2*n,n)), n=1..infinity).
This appears to be Pi/sqrt(27). See A111510. - Marco Matosic, Feb 27 2008
Value of the Dirichlet L-series of the non-principal character modulo m=3 (A102283) at s=1. - R. J. Mathar, Oct 03 2011
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
REFERENCES
L. B. W. Jolley, Summation of Series, Dover, 1961, eq. (81), page 16.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jonathan M. Borwein and Roland Girgensohn, Evaluations of binomial series, Aequat. Math. 70 (2005), 25-36.
Étienne Fouvry, Claude Levesque, and Michel Waldschmidt, Representation of integers by cyclotomic binary forms, arXiv:1712.09019 [math.NT], 2017.
Alessandro Languasco, Pieter Moree, Sumaia Saad Eddin, and Alisa Sedunova, Computation of the Kummer ratio of the class number for prime cyclotomic fields, arXiv:1908.01152 [math.NT], 2019. See r(q) for q=3 in Table 1, p. 7.
D. H. Lehmer, Interesting Series Involving the Central Binomial Coefficient, Am. Math. Monthly 92 (1985) 449-457. See eq. (12).
Courtney Moen, Infinite series with binomial coefficients, Math. Mag. 64 (1) (1991), 53-55.
Paul J. Nahin, Inside interesting integrals, Undergrad. Lecture Notes in Physics, Springer (2020), (1.6.2)
Simon Plouffe, Sum(1/(n*binomial(2*n,n)), n=1..infinity), see p. 87.
Renzo Sprugnoli, Sums of reciprocals of the central binomial coefficients, El. J. Combin. Numb. Th. 6 (2006) # A27
Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
FORMULA
-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
Equals Integral_{0..oo} 2*x/((x^2+1)*(x^4+x^2+1)) dx. - Jean-François Alcover, Sep 10 2013
From Peter Bala, Feb 16 2015: (Start)
Pi/sqrt(27) = Sum_{n >= 0} 1/((3*n + 1)*(3*n + 2)) = 1 - 1/2 + 1/4 - 1/5 + 1/7 - 1/8 + ....
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 + ... ))))))).
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.
Pi/sqrt(27) = (1/4)*Sum_{n >= 0} (-1/8)^n * (9*n + 5)/((3*n + 1)*(3*n + 2)).
BBP-type formulas:
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.
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)
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
From Amiram Eldar, Aug 15 2020: (Start)
Equals Integral_{x=0..oo} 1/(exp(x) + exp(-x) + 1) dx.
Equals Integral_{x=0..oo} 1/(1 + x + x^2 + x^3 + x^4 + x^5) dx. (End)
Equals (3*S - 4)/8, where S = A248682. - Peter Luschny, Jul 22 2022
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
From Peter Bala, Dec 09 2023: (Start)
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).
Pi/sqrt(27) = (1/2)*Sum_{n >= 0} 1/((2*n + 1)*binomial(2*n,n)).
Pi/sqrt(27) = (9/4)*Sum_{n >= 3} (n - 1)*(n - 2)/binomial(2*n,n). (End)
Equals integral_{x=0..oo} 1/(1-x^3) dx [Nahin]. - R. J. Mathar, May 16 2024
EXAMPLE
0.60459978807807261686469275254738524409468...
MATHEMATICA
RealDigits[ N [Sum[1/(n*Binomial[2n, n]), {n, 1, Infinity}], 110]] [[1]]
RealDigits[Pi/(3*Sqrt[3]), 10, 105][[1]] (* T. D. Noe, Sep 11 2013 *)
PROG
(PARI) Pi/sqrt(27) \\ Charles R Greathouse IV, Sep 11 2013
(Magma) R:=RealField(106); SetDefaultRealField(R); n:=Pi(R)/Sqrt(27); Reverse(Intseq(Floor(10^105*n))); // Bruno Berselli, Mar 12 2018
CROSSREFS
KEYWORD
AUTHOR
Robert G. Wilson v, Aug 03 2002
STATUS
approved