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A019670
Decimal expansion of Pi/3.
44
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, 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, 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
OFFSET
1,3
COMMENTS
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
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
60 degrees in radians. - M. F. Hasler, Jul 08 2016
Volume of a quarter sphere of radius 1. - Omar E. Pol, Aug 17 2019
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
The area of a circle circumscribing a unit-area regular dodecagon. - Amiram Eldar, Nov 05 2020
REFERENCES
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.3, p. 489.
LINKS
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Jean Desbois and Stéphane Ouvry, Algebraic and arithmetic area for m planar Brownian paths, arXiv:1101.4135 [math-ph], Jan 21, 2011.
FORMULA
A third of A000796, a sixth of A019692, the square root of A100044.
Sum_{k >= 0} (-1)^k/(6k+1) + (-1)^k/(6k+5). - Charles R Greathouse IV, Sep 08 2011
Product_{k >= 1}(1-(6k)^(-2))^(-1). - Fred Daniel Kline, May 30 2013
From Peter Bala, Feb 05 2015: (Start)
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.)
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)
Equals Sum_{k >= 1} arctan(sqrt(3)*L(2k)/L(4k)) where L=A000032. See also A005248 and A056854. - Michel Marcus, Mar 29 2016
Equals Product_{n >= 1} A016910(n) / A136017(n). - Fred Daniel Kline, Jun 09 2016
Equals Integral_{x=-oo..oo} sech(x)/3 dx. - Ilya Gutkovskiy, Jun 09 2016
From Peter Bala, Nov 16 2016: (Start)
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)) ).
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)
Equals Product_{p>=5, p prime} p/sqrt(p^2-1). - Dimitris Valianatos, May 13 2017
Equals A019699/4 or A019693/2. - Omar E. Pol, Aug 17 2019
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
Equals Integral_{x=0..oo} 1/(1 + x^6) dx. - Bernard Schott, Mar 12 2022
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
EXAMPLE
Pi/3 = 1.04719755119659774615421446109316762806572313312503527365831486...
From Peter Bala, Nov 16 2016: (Start)
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)) ).
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)
MATHEMATICA
RealDigits[N[Pi/3, 6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
PROG
(PARI) Pi/3 \\ Charles R Greathouse IV, Sep 08 2011
(PARI) N=150; default(realprecision, 3+N); digits(Pi*10^N\3) \\ M. F. Hasler, Jul 08 2016
CROSSREFS
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).
Sequence in context: A085508 A198347 A371333 * A093436 A343805 A082169
KEYWORD
nonn,cons
AUTHOR
N. J. A. Sloane, Dec 11 1996
STATUS
approved