A073010 Decimal expansion of Pi/sqrt(27).

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, 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, 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

Offset: 0

Author

Robert G. Wilson v, Aug 03 2002

Keywords

• nonn
• cons
• easy

Comments

Original name: Decimal expansion of Sum_{n>0} 1/(n*binomial(2*n,n)).

This appears to be Pi/sqrt(27). See A111510. - Marco Matosic, Feb 27 2008

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

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

Examples

			0.60459978807807261686469275254738524409468...
		

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.
• Michael I. Shamos, A catalog of the real numbers, (2007). See p. 501.
• 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.
• Index entries for transcendental numbers.

Crossrefs

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

Programs

Magma

R:=RealField(106); SetDefaultRealField(R); n:=Pi(R)/Sqrt(27); Reverse(Intseq(Floor(10^105*n))); // Bruno Berselli, Mar 12 2018

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 *)

PARI

Pi/sqrt(27) \\ Charles R Greathouse IV, Sep 11 2013

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

From Peter Bala, Mar 05 2025: (Start)

Equals 2*Integral_{x = 0..1} 1/(3 + x^2) dx = Integral_{x = 0..1} (4 - x)/(sqrt(x)*(12 + x*(1 - x))) dx.

Equals Sum_{n >= 1} (-1/3)^n * (3 - 14*n)/(n*(2*n-1)*binomial(4*n, 2*n)). The series terms are O(7*sqrt(2*Pi/n)/48^n). (End)

Equals Integral_{x=0..oo} (x^3)/(x^6 + 1) dx. - Kritsada Moomuang, Jun 04 2025