|
%I #120 Mar 21 2024 08:34:56
%S 1,0,4,7,1,9,7,5,5,1,1,9,6,5,9,7,7,4,6,1,5,4,2,1,4,4,6,1,0,9,3,1,6,7,
%T 6,2,8,0,6,5,7,2,3,1,3,3,1,2,5,0,3,5,2,7,3,6,5,8,3,1,4,8,6,4,1,0,2,6,
%U 0,5,4,6,8,7,6,2,0,6,9,6,6,6,2,0,9,3,4,4,9,4,1,7,8,0,7,0,5,6,8
%N Decimal expansion of Pi/3.
%C With an offset of zero, also the decimal expansion of Pi/30 ~ 0.104719... which is the average arithmetic area <S_0> of the 0-winding sectors enclosed by a closed Brownian planar path, of a given length t, according to Desbois, p. 1. - _Jonathan Vos Post_, Jan 23 2011
%C Polar angle (or apex angle) of the cone that subtends exactly one quarter of the full solid angle. See comments in A238238. - _Stanislav Sykora_, Jun 07 2014
%C 60 degrees in radians. - _M. F. Hasler_, Jul 08 2016
%C Volume of a quarter sphere of radius 1. - _Omar E. Pol_, Aug 17 2019
%C Also smallest positive zero of Sum_{k>=1} cos(k*x)/k = -log(2*|sin(x/2)|). Proof of this identity: Sum_{k>=1} cos(k*x)/k = Re(Sum_{k>=1} exp(k*x*i)/k) = Re(-log(1-exp(x*i))) = -log(2*|sin(x/2)|), x != 2*m*Pi, where i = sqrt(-1). - _Jianing Song_, Nov 09 2019
%C The area of a circle circumscribing a unit-area regular dodecagon. - _Amiram Eldar_, Nov 05 2020
%H Ivan Panchenko, <a href="/A019670/b019670.txt">Table of n, a(n) for n = 1..1000</a>
%H Kunle Adegoke, <a href="http://arxiv.org/abs/1603.08097">Infinite arctangent sums involving Fibonacci and Lucas numbers</a>, arXiv:1603.08097 [math.NT], 2016.
%H J. M. Borwein, P. B. Borwein, and K. Dilcher, <a href="http://www.jstor.org/stable/2324715">Pi, Euler numbers and asymptotic expansions</a>, Amer. Math. Monthly, 96 (1989), 681-687.
%H Jean Desbois and Stéphane Ouvry, <a href="http://arxiv.org/abs/1101.4135">Algebraic and arithmetic area for m planar Brownian paths</a>, arXiv:1101.4135 [math-ph], Jan 21, 2011.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F A third of A000796, a sixth of A019692, the square root of A100044.
%F Sum_{k >= 0} (-1)^k/(6k+1) + (-1)^k/(6k+5). - _Charles R Greathouse IV_, Sep 08 2011
%F Product_{k >= 1}(1-(6k)^(-2))^(-1). - _Fred Daniel Kline_, May 30 2013
%F From _Peter Bala_, Feb 05 2015: (Start)
%F Pi/3 = Sum {k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k = 2F1(1/2,1/2;3/2;1/4). Similar series expansions hold for Pi^2 (A002388), Pi^3 (A091925) and Pi/(2*sqrt(2)) (A093954.)
%F The integer sequences A(n) := 4^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k ) both satisfy the second-order recurrence equation u(n) = (20*n^2 + 4*n + 1)*u(n-1) - 8*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/3 = 1 + 1/(24 - 8*3^3/(89 - 8*2*5^3/(193 - 8*3*7^3/(337 - ... - 8*(n - 1)*(2*n - 1)^3/((20*n^2 + 4*n + 1) - ... ))))). Cf. A002388 and A093954. (End)
%F Equals Sum_{k >= 1} arctan(sqrt(3)*L(2k)/L(4k)) where L=A000032. See also A005248 and A056854. - _Michel Marcus_, Mar 29 2016
%F Equals Product_{n >= 1} A016910(n) / A136017(n). - _Fred Daniel Kline_, Jun 09 2016
%F Equals Integral_{x=-oo..oo} sech(x)/3 dx. - _Ilya Gutkovskiy_, Jun 09 2016
%F From _Peter Bala_, Nov 16 2016: (Start)
%F Euler's series transformation applied to the series representation Pi/3 = Sum_{k >= 0} (-1)^k/(6*k + 1) + (-1)^k/(6*k + 5) given above by Greathouse produces the faster converging series Pi/3 = (1/2) * Sum_{n >= 0} 3^n*n!*( 1/(Product_{k = 0..n} (6*k + 1)) + 1/(Product_{k = 0..n} (6*k + 5)) ).
%F The series given above by Greathouse is the case n = 0 of the more general result Pi/3 = 9^n*(2*n)! * Sum_{k >= 0} (-1)^(k+n)*( 1/(Product_{j = -n..n} (6*k + 1 + 6*j)) + 1/(Product_{j = -n..n} (6*k + 5 + 6*j)) ) for n = 0,1,2,.... Cf. A003881. See the example section for notes on the case n = 1.(End)
%F Equals Product_{p>=5, p prime} p/sqrt(p^2-1). - _Dimitris Valianatos_, May 13 2017
%F Equals A019699/4 or A019693/2. - _Omar E. Pol_, Aug 17 2019
%F Equals Integral_{x >= 0} (sin(x)/x)^4 = 1/2 + Sum_{n >= 0} (sin(n)/n)^4, by the Abel-Plana formula. - _Peter Bala_, Nov 05 2019
%F Equals Integral_{x=0..oo} 1/(1 + x^6) dx. - _Bernard Schott_, Mar 12 2022
%F Pi/3 = -Sum_{n >= 1} i/(n*P(n, 1/sqrt(-3))*P(n-1, 1/sqrt(-3))), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The first twenty terms of the series gives the approximation Pi/3 = 1.04719755(06...) correct to 8 decimal places. - _Peter Bala_, Mar 16 2024
%e Pi/3 = 1.04719755119659774615421446109316762806572313312503527365831486...
%e From _Peter Bala_, Nov 16 2016: (Start)
%e Case n = 1. Pi/3 = 18 * Sum_{k >= 0} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ).
%e Using the methods of Borwein et al. we can find the following asymptotic expansion for the tails of this series: for N divisible by 6 there holds Sum_{k >= N/6} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ) ~ 1/N^3 + 6/N^5 + 1671/N ^7 - 241604/N^9 + ..., where the sequence [1, 0, 6, 0, 1671, 0, -241604, 0, ...] is the sequence of coefficients in the expansion of ((1/18)*cosh(2*x)/cosh(3*x)) * sinh(3*x)^2 = x^2/2! + 6*x^4/4! + 1671*x^6/6! - 241604*x^8/8! + .... Cf. A024235, A278080 and A278195. (End)
%t RealDigits[N[Pi/3,6! ]] (* _Vladimir Joseph Stephan Orlovsky_, Dec 02 2009 *)
%o (PARI) Pi/3 \\ _Charles R Greathouse IV_, Sep 08 2011
%o (PARI) N=150; default(realprecision, 3+N); digits(Pi*10^N\3) \\ _M. F. Hasler_, Jul 08 2016
%Y Cf. A000796 (Pi), A019669 (Pi/2), A019692 (2*Pi), A122952 (3*Pi), A003881(Pi/4), A137914, A238238, A002388, A091925, A093954, A016910, A019694, A136017, A352324.
%Y Integral_{x=0..oo} 1/(1+x^m) dx: A013661 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), this sequence (m=6), A352125 (m=8), A094888 (m=10).
%K nonn,cons
%O 1,3
%A _N. J. A. Sloane_, Dec 11 1996
|
