A304656 Decimal expansion of Pi*sqrt(3).

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

Offset: 1

Author

Peter Luschny, May 16 2018

Keywords

• nonn
• cons

Examples

			5.4413980927026535517822347729264671968521987442782217267096548061643695433790...
		

Links

• C. Elsner, On recurrence formulas for sums involving binomial coefficients, Fib. Q., 43,1 (2005), 31-45.
• Melissa Larson, Verifying and discovering BBP-type formulas, 2008.
• Wikipedia, Bailey-Borwein-Plouffe formula.
• Index entries for transcendental numbers

Crossrefs

Cf. A002388, A002391, A093602, A248682.

Programs

Maple

Pi*sqrt(3): evalf(%, 100);

Mathematica

RealDigits[N[StieltjesGamma[0,1/6]-StieltjesGamma[0,5/6],99]][[1]] (* corrected by Harvey P. Dale, Oct 13 2020 *)
RealDigits[Pi Sqrt[3],10,120][[1]] (* Harvey P. Dale, Oct 13 2020 *)

Python

# Use several guard digits when computing.
# BBP formula (9/32) P(1, 64, 6, (16, 8, 0, -2, -1, 0)).
from decimal import Decimal as dec, getcontext
def BBPpisqrt3(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(64)
    for k in range(int(n * 0.5536546824812272) + 1):
        sixk = dec(6 * k)
        s += f * ( dec(16) / (sixk + 1) + dec(8) / (sixk + 2)
                 - dec(2)  / (sixk + 4) - dec(1) / (sixk + 5) )
        f /= g
    return (s * dec(9)) / dec(32)
print(BBPpisqrt3(200))  # Peter Luschny, Nov 03 2023

Formula

Equals gamma(0, 1/6) - gamma(0, 5/6) where gamma(n,x) denotes the generalized Stieltjes constants.

Equals PolyGamma[0, 5/6] - PolyGamma[0, 1/6].

Equals 3*sqrt(2*zeta(2)).

Pi^2 = A304656 * A093602.

From Amiram Eldar, Aug 06 2020: (Start)

Equals Sum_{k>=0} 1/((k + 1/3)*(k + 2/3)).

Equals Integral_{x=0..oo} log(1 + 3/x^2) dx. (End)

Equals (27*S - 36)/8, where S = A248682. - Peter Luschny, Jul 22 2022

From Peter Bala, Oct 26 2023: (Start)

sqrt(3)*Pi = 9/2 + 9*Sum_{n >= 1} (-1)^(n+1)/(9*n^2 - 1);

sqrt(3)*Pi = 5 + 10*Sum_{n >= 1} 1/((4*n^2 - 1)*(9*n^2 - 1)) = 43/8 + 8*Sum_{n >= 2} (-1)^n/((n^2 - 1)*(9*n^2 - 1));

sqrt(3)*Pi = 1765/324 - (80/9)*Sum_{n >= 2} 1/((n^2 - 1)*(4*n^2 - 1)*(9*n^2 - 1)).

The following two series representations for the constant

sqrt(3)*Pi = 72 * Sum_{n >= 0} (2*n + 1)/((6*n + 1)*(6*n + 3)*(6*n + 5)) and

sqrt(3)*Pi = 8192/1485 - 860160 * Sum_{n >= 0} (2*n + 3)/((6*n + 1)*(6*n + 3)*...*(6*n + 17)) appear to generalize as follows:

for k >= 0, sqrt(3)*Pi = c(k) + (-1)^k*d(k)*Sum_{n >= 0} (2*n + 2*k + 1)/((6*n + 1)*(6*n + 3)*...*(6*n + 12*k + 5)), where c(k) is a rational number approximating sqrt(3)*Pi and d(k) = (6*k + 1)! * 2^(6*k+3) / 3^(3*k-2).

The first few values of c(k) for k >= 0 are [0, 8192/1485, 11341398016/2085060285, 62809601736704/11542783997745, 889063287831973723111424/ 163388820474305231710905, ...].

The following two series representations for the constant

sqrt(3)*Pi = 256/45 - 2560*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*...*(6*n + 11)) and

sqrt(3)*Pi = 337117184/62026965 + 2018508800*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*...*(6*n + 23)) appear to generalize as follows:

for k >= 0, sqrt(3)*Pi = c(k) - (-1)^k*d(k)*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*...*(6*n + 12*k + 11)), where c(k) is a rational number approximating sqrt(3)*Pi and d(k) = (6*k + 5)! * 2^(6*k+6) / 3^(3*k+1).

The first few values of c(k) for k >= 0 are [256/45, 337117184/62026965, 1732370763874304/318357429615225, 733187044080753836032/134742553582636674675, 6361250411469779336874164224/1169047010493653932891525275, ...]. (End)

For arbitrary integer k, Pi*sqrt(3) = Sum_{n >= 0} (1/(n - k + 1/6) - 1/(n + k + 5/6)) = Sum_{n >= 0} (1/(n + k + 7/6) - 1/(n - k - 1/6)). - Peter Bala, Jul 10 2024