A094888 -id:A094888 - cp's OEIS frontend

A019670 Decimal expansion of Pi/3.

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

N. J. A. Sloane, Dec 11 1996

With an offset of zero, also the decimal expansion of Pi/30 ~ 0.104719... which is the average arithmetic area  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

			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)

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.3, p. 489.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
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.
Index entries for transcendental numbers.

Cf. A000796 (Pi), A019669 (Pi/2), A019692 (2*Pi), A122952 (3*Pi), A003881(Pi/4), A137914, A238238, A002388, A091925, A093954, A016910, A019694, A136017, A352324.

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).

RealDigits[N[Pi/3,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

Pi/3 \\ Charles R Greathouse IV, Sep 08 2011

N=150; default(realprecision, 3+N); digits(Pi*10^N\3) \\ M. F. Hasler, Jul 08 2016

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
Equals Integral_{x >= 0} (2*x^2 + 1)/((x^2 + 1)*(4*x^2 + 1)) dx. - Peter Bala, Feb 12 2025

A090771 Numbers that are congruent to {1, 9} mod 10.

Original entry on oeis.org

1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69, 71, 79, 81, 89, 91, 99, 101, 109, 111, 119, 121, 129, 131, 139, 141, 149, 151, 159, 161, 169, 171, 179, 181, 189, 191, 199, 201, 209, 211, 219, 221, 229, 231, 239, 241, 249, 251, 259, 261, 269, 271, 279, 281

Offset: 1

Giovanni Teofilatto, Feb 07 2004

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 10). - Bruno Berselli, Nov 17 2010

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A000796, A090298, A005408, A047209, A007310, A019970, A047336, A047522, A091998, A094888, A113801, A175886, A175887, A188593.
Cf. A056020 (n = 1 or 8 mod 9), A175885 (n = 1 or 10 mod 11).
Cf. A045468 (primes), A195142 (partial sums).

a090771 n = a090771_list !! (n-1)
a090771_list = 1 : 9 : map (+ 10) a090771_list
-- Reinhard Zumkeller, Jan 07 2012

Flatten[Table[10n - {9, 1}, {n, 30}]] (* Alonso del Arte, Sep 02 2014 *)
LinearRecurrence[{1,1,-1},{1,9,11},60] (* Harvey P. Dale, Jul 05 2020 *)

a(n)=n\2*10-(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

a(n) = sqrt(40*A057569(n) + 1). - Gary Detlefs, Feb 22 2010
From Bruno Berselli, Sep 16 2010 - Nov 17 2010: (Start)
  G.f.: x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).
  a(n) = (10*n + 3*(-1)^n - 5)/2.
  a(n) = -a(-n + 1) = a(n-1) + a(n-2) - a(n-3) = a(n-2) + 10.
  a(n) = 10*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
a(n) = 10*n - a(n-1) - 10 (with a(1) = 1). - Vincenzo Librandi, Nov 16 2010
a(n) = sqrt(10*A132356(n-1) + 1). - Ivan N. Ianakiev, Nov 09 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/10)*cot(Pi/10) = A000796 * A019970 / 10 = sqrt(5 + 2*sqrt(5))*Pi/10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((10*x - 5)*exp(x) + 3*exp(-x))/2. - David Lovler, Sep 03 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(phi+2) (A188593).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi*phi/5 = A094888/10. (End)

Edited and extended by Ray Chandler, Feb 10 2004

A352324 Decimal expansion of 4Pi / (5sqrt(10-2*sqrt(5))).

Original entry on oeis.org

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

Offset: 1

Bernard Schott, Mar 12 2022

Cauchy's residue theorem implies that Integral_{x=0..oo} 1/(1 + x^m) dx = (Pi/m) * csc(Pi/m); this is the case m = 5.

The area of a circle circumscribing a unit-area regular decagon.

			1.0689593321155951134251843725068826399014509252665...

Jean-François Pabion, Éléments d'Analyse Complexe, licence de Mathématiques, page 111, Ellipses, 1995.

Index entries for transcendental numbers.

Cf. A019694, A047209, A121570, A371604, A377405.

Integral_{x=0..oo} 1/(1+x^m) dx: A019669 (m=2), A248897 (m=3), A093954 (m=4), this sequence (m=5), A019670 (m=6), A352125 (m=8), A094888 (m=10).

evalf(4*Pi / (5*(sqrt(10-2sqrt(5)))), 100);

First[RealDigits[N[4Pi/(5Sqrt[10-2Sqrt[5]]), 93]]] (* Stefano Spezia, Mar 12 2022 *)

Equals Integral_{x=0..oo} 1/(1 + x^5) dx.
Equals (Pi/5) *csc(Pi/5).
Equals (1/2) * A019694 * A121570.
Equals 1/Product_{k>=1} (1 - 1/(5*k)^2). - Amiram Eldar, Mar 12 2022
Equals Product_{k>=2} (1 + (-1)^k/A047209(k)). - Amiram Eldar, Nov 22 2024
Equals 1/A371604 = A377405/5. - Hugo Pfoertner, Nov 22 2024

A352125 Decimal expansion of Pisqrt(2)sqrt(2 + sqrt(2))/8.

Original entry on oeis.org

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

Offset: 1

Stefano Spezia, Mar 05 2022

			1.02617215297703088887146778087283197497962...

Jean-François Pabion, Éléments d'Analyse Complexe, licence de Mathématiques, page 111, Ellipses, 1995.

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, pp. 235-236.
Index entries for transcendental numbers.

Cf. A000796, A047522, A121601.

Integral_{x=0..oo} 1/(1+x^m) dx: A019669 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), A019670 (m=6), this sequence (m=8), A094888 (m=10).

First[RealDigits[N[Pi*Sqrt[2]Sqrt[2+Sqrt[2]]/8,95]]]

Pi*sqrt(4 + 2*sqrt(2))/8 \\ Michel Marcus, Mar 07 2022

Equals Integral_{x=0..oo} 1/(1 + x^8) dx.
Equals Pi*csc(Pi/8)/8.
Equals 1/Product_{k>=1} (1 - 1/(8*k)^2). - Amiram Eldar, Mar 12 2022
Equals Product_{k>=2} (1 + (-1)^k/A047522(k)). - Amiram Eldar, Nov 22 2024

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A019670 Decimal expansion of Pi/3.

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A090771 Numbers that are congruent to {1, 9} mod 10.

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A352324 Decimal expansion of 4*Pi / (5*sqrt(10-2*sqrt(5))).

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A352125 Decimal expansion of Pi*sqrt(2)*sqrt(2 + sqrt(2))/8.

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A352324 Decimal expansion of 4Pi / (5sqrt(10-2*sqrt(5))).

A352125 Decimal expansion of Pisqrt(2)sqrt(2 + sqrt(2))/8.