A352125 -id:A352125 - 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

A047522 Numbers that are congruent to {1, 7} mod 8.

Original entry on oeis.org

1, 7, 9, 15, 17, 23, 25, 31, 33, 39, 41, 47, 49, 55, 57, 63, 65, 71, 73, 79, 81, 87, 89, 95, 97, 103, 105, 111, 113, 119, 121, 127, 129, 135, 137, 143, 145, 151, 153, 159, 161, 167, 169, 175, 177, 183, 185, 191, 193, 199, 201, 207, 209, 215, 217, 223, 225, 231, 233

Offset: 1

N. J. A. Sloane

Also n such that Kronecker(2,n) = mu(gcd(2,n)). - Jon Perry and T. D. Noe, Jun 13 2003
Also n such that x^2 == 2 (mod n) has a solution. The primes are given in sequence A001132. - T. D. Noe, Jun 13 2003
As indicated in the formula, a(n) is related to the even triangular numbers. - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
Cf. property described by Gary Detlefs in A113801: more generally, these a(n) are of the form (2*h*n + (h-4)*(-1)^n-h)/4 (h,n natural numbers). Therefore a(n)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 8). Also a(n)^2 - 1 == 0 (mod 16). - Bruno Berselli, Nov 17 2010
A089911(3*a(n)) = 2. - Reinhard Zumkeller, Jul 05 2013
S(a(n+1)/2, 0) = (1/2)*(S(a(n+1), sqrt(2)) - S(a(n+1) - 2, sqrt(2)))  = T(a(n+1), sqrt(2)/2) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S is the Chebyshev polynomial (A049310) here extended to fractional n, evaluated at x = 0. (For T see A053120.) - Wolfdieter Lang, Jun 04 2023

L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 16.

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

Cf. A001132, A014494, A056575, A010709, A074378, A047336, A056020, A005408, A047209, A007310, A090771, A175885, A091998, A175886, A175887, A058529, A047621, A179260, A352125.
Cf. A077221 (partial sums).
Cf. A000217, A089911, A113801.
Cf. A010503, A049310, A053120.

a047522 n = a047522_list !! (n-1)
a047522_list = 1 : 7 : map (+ 8) a047522_list
-- Reinhard Zumkeller, Jan 07 2012

Select[Range[1, 191, 2], JacobiSymbol[2, # ]==1&]

a(n)=4*n-2+(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

a(n) = sqrt(8*A014494(n)+1) = sqrt(16*ceiling(n/2)*(2*n+1)+1) = sqrt(8*A056575(n)-8*(2n+1)*(-1)^n+1). - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
1 - 1/7 + 1/9 - 1/15 + 1/17 - ... = (Pi/8)*(1 + sqrt(2)). [Jolley] - Gary W. Adamson, Dec 16 2006
From R. J. Mathar, Feb 19 2009: (Start)
a(n) = 4n - 2 + (-1)^n = a(n-2) + 8.
G.f.: x(1+6x+x^2)/((1+x)(1-x)^2). (End)
a(n) = 8*n - a(n-1) - 8. - Vincenzo Librandi, Aug 06 2010
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 8*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
E.g.f.: 1 + (4*x - 1)*cosh(x) + (4*x - 3)*sinh(x). - Stefano Spezia, May 13 2021
E.g.f.: 1 + (4*x - 3)*exp(x) + 2*cosh(x). - David Lovler, Jul 16 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(2+sqrt(2)) (A179260).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/8)*cosec(Pi/8) (A352125). (End)

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

A094888 Decimal expansion of 2Piphi, where phi = (1+sqrt(5))/2.

Original entry on oeis.org

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

Offset: 2

N. J. A. Sloane, Jun 15 2004

			10.16640738463051963161901802648439768366367858644230824...

Ivan Panchenko, Table of n, a(n) for n = 2..1000
John Baez, Pi and the Golden Ratio, Azimuth, Mar 07 2017.
Index entries for transcendental numbers.

Cf. A000796, A001622, A090771, A094886, A135155.

Integral_{x>=0} 1/(1+x^m) dx: A019669 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), A019670 (m=6), A352125 (m=8), this sequence (m=10).

evalf(Pi*(1+sqrt(5)), 121);  # Alois P. Heinz, May 16 2022

RealDigits[2 * Pi * GoldenRatio, 10, 100][[1]] (* Amiram Eldar, May 18 2021 *)

From Peter Bala, Nov 03 2019: (Start)
Equals 10*Integral_{x >= 0} cosh(4*x)/cosh(5*x) dx = Integral_{x = 0..1} (1 + x^8)/(1 + x^10) dx .
Equals 100*Sum_{n >= 0} (-1)^n*(2*n + 1)/( (10*n + 1)*(10*n + 9) ). (End)
Equals 10 * Product_{k>=2} 2/sqrt(2 + sqrt(2 + ... sqrt(2 + phi)...)), with k nested radicals (Baez, 2017). - Amiram Eldar, May 18 2021
Equals Integral_{x>=0} 1/(1 + x^10) dx = (Pi/10) * csc(Pi/10). - Bernard Schott, May 15 2022
Equals Gamma(1/10)*Gamma(9/10). - Andrea Pinos, Jul 03 2023
Equals 10 * Product_{k >= 1} (10*k)^2/((10*k)^2 - 1). - Antonio Graciá Llorente, Mar 15 2024
Equals 10 * Product_{k>=2} (1 + (-1)^k/A090771(k)). - Amiram Eldar, Nov 23 2024
Equals 2*A094886 = 10*A135155/e. - Hugo Pfoertner, Nov 23 2024

A371858 Decimal expansion of Integral_{x=0..oo} 1 / (1 + x^7) dx.

Original entry on oeis.org

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

Offset: 1

Ilya Gutkovskiy, Apr 09 2024

			1.0343760552667964829453064065124887483642567...

Index entries for transcendental numbers.

Decimal expansion of Integral_{x=0..oo} 1 / (1 + x^k) dx: A019669 (k=2), A248897 (k=3), A093954 (k=4), A352324 (k=5), A019670 (k=6), this sequence (k=7), A352125 (k=8).

Cf. A019674, A047336, A121598.

RealDigits[(1/7) Pi Csc[Pi/7], 10, 102][[1]]

Equals (1/7) * Pi * csc(Pi/7).
Equals A019674 * A121598.
Equals Product_{k>=2} (1 + (-1)^k/A047336(k)). - Amiram Eldar, Nov 22 2024

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A019670 Decimal expansion of Pi/3.

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A047522 Numbers that are congruent to {1, 7} mod 8.

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

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A094888 Decimal expansion of 2*Pi*phi, where phi = (1+sqrt(5))/2.

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A371858 Decimal expansion of Integral_{x=0..oo} 1 / (1 + x^7) dx.

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

A094888 Decimal expansion of 2Piphi, where phi = (1+sqrt(5))/2.