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A304656 Decimal expansion of Pi*sqrt(3).

%I A304656 #46 Jan 05 2025 19:51:41
%S A304656 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,
%T A304656 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,
%U A304656 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
%N A304656 Decimal expansion of Pi*sqrt(3).
%H A304656 C. Elsner, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/43-1/paper43-1-5.pdf">On recurrence formulas for sums involving binomial coefficients</a>, Fib. Q., 43,1 (2005), 31-45.
%H A304656 Melissa Larson, <a href="https://www.d.umn.edu/~jgreene/masters_reports/BBP%20Paper%20final.pdf">Verifying and discovering BBP-type formulas</a>, 2008.
%H A304656 Wikipedia, <a href="https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula">Bailey-Borwein-Plouffe formula</a>.
%H A304656 <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F A304656 Equals gamma(0, 1/6) - gamma(0, 5/6) where gamma(n,x) denotes the generalized Stieltjes constants.
%F A304656 Equals PolyGamma[0, 5/6] - PolyGamma[0, 1/6].
%F A304656 Equals 3*sqrt(2*zeta(2)).
%F A304656 Pi^2 = A304656 * A093602.
%F A304656 From _Amiram Eldar_, Aug 06 2020: (Start)
%F A304656 Equals Sum_{k>=0} 1/((k + 1/3)*(k + 2/3)).
%F A304656 Equals Integral_{x=0..oo} log(1 + 3/x^2) dx. (End)
%F A304656 Equals (27*S - 36)/8, where S = A248682. - _Peter Luschny_, Jul 22 2022
%F A304656 From _Peter Bala_, Oct 26 2023: (Start)
%F A304656 sqrt(3)*Pi = 9/2 + 9*Sum_{n >= 1} (-1)^(n+1)/(9*n^2 - 1);
%F A304656 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));
%F A304656 sqrt(3)*Pi = 1765/324 - (80/9)*Sum_{n >= 2} 1/((n^2 - 1)*(4*n^2 - 1)*(9*n^2 - 1)).
%F A304656 The following two series representations for the constant
%F A304656 sqrt(3)*Pi = 72 * Sum_{n >= 0} (2*n + 1)/((6*n + 1)*(6*n + 3)*(6*n + 5)) and
%F A304656 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:
%F A304656 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).
%F A304656 The first few values of c(k) for k >= 0 are [0, 8192/1485, 11341398016/2085060285, 62809601736704/11542783997745, 889063287831973723111424/ 163388820474305231710905, ...].
%F A304656 The following two series representations for the constant
%F A304656 sqrt(3)*Pi = 256/45 - 2560*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*...*(6*n + 11)) and
%F A304656 sqrt(3)*Pi = 337117184/62026965 + 2018508800*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*...*(6*n + 23)) appear to generalize as follows:
%F A304656 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).
%F A304656 The first few values of c(k) for k >= 0 are [256/45, 337117184/62026965, 1732370763874304/318357429615225, 733187044080753836032/134742553582636674675, 6361250411469779336874164224/1169047010493653932891525275, ...]. (End)
%F A304656 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
%e A304656 5.4413980927026535517822347729264671968521987442782217267096548061643695433790...
%p A304656 Pi*sqrt(3): evalf(%, 100);
%t A304656 RealDigits[N[StieltjesGamma[0,1/6]-StieltjesGamma[0,5/6],99]][[1]] (* corrected by _Harvey P. Dale_, Oct 13 2020 *)
%t A304656 RealDigits[Pi Sqrt[3],10,120][[1]] (* _Harvey P. Dale_, Oct 13 2020 *)
%o A304656 (Python)  # Use several guard digits when computing.
%o A304656 # BBP formula (9/32) P(1, 64, 6, (16, 8, 0, -2, -1, 0)).
%o A304656 from decimal import Decimal as dec, getcontext
%o A304656 def BBPpisqrt3(n: int) -> dec:
%o A304656     getcontext().prec = n
%o A304656     s = dec(0); f = dec(1); g = dec(64)
%o A304656     for k in range(int(n * 0.5536546824812272) + 1):
%o A304656         sixk = dec(6 * k)
%o A304656         s += f * ( dec(16) / (sixk + 1) + dec(8) / (sixk + 2)
%o A304656                  - dec(2)  / (sixk + 4) - dec(1) / (sixk + 5) )
%o A304656         f /= g
%o A304656     return (s * dec(9)) / dec(32)
%o A304656 print(BBPpisqrt3(200))  # _Peter Luschny_, Nov 03 2023
%Y A304656 Cf. A002388, A002391, A093602, A248682.
%K A304656 nonn,cons
%O A304656 1,1
%A A304656 _Peter Luschny_, May 16 2018