A093954 -id:A093954 - cp's OEIS frontend

A002388 Decimal expansion of Pi^2.

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

Offset: 1

N. J. A. Sloane

Also equals the volume of revolution of the sine or cosine curve for one full period, Integral_{x=0..2*Pi} sin(x)^2 dx. - Robert G. Wilson v, Dec 15 2005

Equals Sum_{n>0} 20/A026424(n)^2 where A026424 are the integers such that the number of prime divisors (counted with multiplicity) is odd. - Michel Lagneau, Oct 23 2015

			9.869604401089358618834490999876151135313699407240790626413349376220044...

W. E. Mansell, Tables of Natural and Common Logarithms. Royal Society Mathematical Tables, Vol. 8, Cambridge Univ. Press, 1964, p. XVIII.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 76.

Harry J. Smith, Table of n, a(n) for n = 1..20000
Mohammad K. Azarian, Al-Risala al-Muhitiyya: A Summary (The Treatise on the Circumference), Missouri Journal of Mathematical Sciences, Vol. 22, No. 2, 2010, pp. 64-85.
D. H. Bailey and J. M. Borwein, Experimental Mathematics: Examples, Methods and Implications, Notices AMS, 52 (No. 5 2005), 502-514.
David H. Bailey, Jonathan M. Borwein, Andrew Mattingly, and Glenn Wightwick, The Computation of Previously Inaccessible Digits of (Pi)^2 and Catalan's Constant, Notices AMS, 60 (No. 7 2013), 844-854.
N. D. Elkies, Why is (Pi)^2 so close to 10?
Melissa Larson, Verifying and discovering BBP-type formulas, 2008.
Simon Plouffe, Pi^2 to 10000 digits.
Simon Plouffe, Plouffe's Inverter, Pi^2 to 10000 digits.
Wikipedia, Bailey-Borwein-Plouffe formula.
Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science, Vol 3, No 4 (1999).
Index entries for sequences related to the number Pi.
Index entries for transcendental numbers.

Cf. A102753, A058284, A019670, A091925, A093954, A093602, A304656.

R:= RealField(100); Pi(R)^2; // G. C. Greubel, Mar 08 2018

Digits:=100: evalf(Pi^2); # Wesley Ivan Hurt, Jul 13 2014

RealDigits[Pi^2, 10, 111][[1]] (* Robert G. Wilson v, Dec 15 2005 *)

default(realprecision, 20080); x=Pi^2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002388.txt", n, " ", d)); \\ Harry J. Smith, May 31 2009

# Use some guard digits when computing.
# BBP formula (9 / 8) P(2, 64, 6, (16, -24, -8, -6,  1, 0)).
from decimal import Decimal as dec, getcontext
def BBPpi2(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(64)
    for k in range(int(n * 0.5536546824812272) + 1):
        sixk = dec(6 * k)
        s += f * ( dec(16) / (sixk + 1) ** 2 - dec(24) / (sixk + 2) ** 2
                 - dec(8)  / (sixk + 3) ** 2 - dec(6)  / (sixk + 4) ** 2
                 + dec(1)  / (sixk + 5) ** 2 )
        f /= g
    return (s * dec(9)) / dec(8)
print(BBPpi2(200))  # Peter Luschny, Nov 03 2023

Pi^2 = 11/2 + 16 * Sum_{k>=2} (1+k-k^3)/(1-k^2)^3. - Alexander R. Povolotsky, May 04 2009
Pi^2 = 3*(Sum_{n>=1} ((2*n+1)^2/Sum_{k=1..n} k^3)/4 - 1). - Alexander R. Povolotsky, Jan 14 2011
Pi^2 = (3/2)*(Sum_{n>=1} ((7*n^2+2*n-2)/(2*n^2-1)/(n+1)^5) - zeta(3) - 3*zeta(5) + 22 - 7*polygamma(0,1-1/sqrt(2)) + 5*sqrt(2)*polygamma(0,1-1/sqrt(2)) - 7*polygamma(0,1+1/sqrt(2)) - 5*sqrt(2)*polygamma(0,1+1/sqrt(2)) - 14*EulerGamma). - Alexander R. Povolotsky, Aug 13 2011
Also equals 32*Integral_{x=0..1} arctan(x)/(1+x^2) dx. - Jean-François Alcover, Mar 25 2013
From Peter Bala, Feb 05 2015: (Start)
Pi^2 = 20 * Integral_{x = 0 .. log(phi)} x*coth(x) dx, where phi = (1/2)*(1 + sqrt(5)) is the golden ratio.
Pi^2 = 10 * Sum_{k >= 0} binomial(2*k,k)*(1/(2*k + 1)^2)*(-1/16)^k. Similar series expansions hold for Pi/3 (see A019670) and (7*/216)*Pi^3 (see A091925).
The integer sequences A(n) := 2^n*(2*n + 1)!^2/n! and B(n) := A(n)*( Sum_{k = 0..n} binomial(2*k,k)*1/(2*k + 1)^2*(-1/16)^k ) both satisfy the second order recurrence equation u(n) = (24*n^3 + 44*n^2 + 2*n + 1)*u(n-1) + 8*(n - 1)*(2*n - 1)^5*u(n-2). From this observation we can obtain the continued fraction expansion Pi^2/10 = 1 - 1/(72 + 8*3^5/(373 + 8*2*5^5/(1051 + ... + 8*(n - 1)*(2*n - 1)^5/((24*n^3 + 44*n^2 + 2*n + 1) + ... )))). Cf. A093954. (End)
Pi^2 = A304656 * A093602 = (gamma(0, 1/6) - gamma(0, 5/6))*(gamma(0, 2/6) - gamma(0, 4/6)), where gamma(n,x) are the generalized Stieltjes constants. This formula can also be expressed by the polygamma function. - Peter Luschny, May 16 2018
Equals 8 + Sum_{k>=1} 1/(k^2 - 1/4)^2 = -8 + Sum_{k>=0} 1/(k^2 - 1/4)^2. - Amiram Eldar, Aug 21 2020
From Peter Bala, Dec 10 2021: (Start)
Pi^2 = (2^6)*Sum_{n >= 1} n^2/(4*n^2 - 1)^2 = (2^11)*Sum_{n >= 1} n^2/ ((4*n^2 - 1)^2*(4*n^2 - 3^2)^2) = ((2^19)*(3^2)/7) * Sum_{n >= 1} n^2/((4*n^2 - 1)^2*(4*n^2 - 3^2)^2*(4*n^2 - 5^2)^2).
More generally, it appears that for k >= 0 we have Pi^2 = (2*k+1)*2^(4*k+6) * (2*k)!^4/(4*k)! * Sum_{n >= 1} n^2/((4*n^2 - 1)^2*...*(4*n^2 - (2*k+1)^2)^2).
It also appears that for k >= 0 we have Pi^2 = (-1)^k * 2^(6*k+8)*(2*k+1)^3/(6*k+1) * ((2*k)!^6 * (3*k)!)/(k!^3 * (6*k)!) * Sum_{n >= 1} n^2/((4*n^2 - 1)^3*...*(4*n^2 - (2*k+1)^2)^3). (End)
From Peter Bala, Oct 27 2023: (Start)
Pi^2 = 10 - Sum_{n >= 1} 1/(n*(n + 1))^3.
Pi^2 = 6217/630 + (648/35)*Sum_{n >= 1} 1/(n*(n + 1)*(n + 2)*(n + 3))^3.
The general result (verified using the WZ method - see Wilf) is : for n >= 0,
Pi^2 = A(n) + (-1)^(n+1) * B(n)*Sum_{k >= 1} 1/(k*(k + 1)*...*(k + 2*n + 1))^3, where A(n) = 10 - Sum_{i = 1..n} (-1)^(i+1) * (56*i^2 + 24*i + 3)*(2*i)!^3*(3*i)!/(2*i^2*(2*i + 1)*(6*i + 1)!*i!^3) and B(n) = (2*n + 1)!^6 * (3*n)! / ( (2*n + 1)*(6*n + 1)!*n!^3 ).
Letting n -> oo gives the fast converging alternating series
Pi^2 = 10 - Sum_{i >= 1} (-1)^(i+1) * (56*i^2 + 24*i + 3)*(2*i)!^3 * (3*i)!/(2*i^2*(2*i + 1)*(6*i + 1)!*i!^3). The i-th summand of the series is asymptotic to (14/3) * 1/(i^2 * 27^i) so taking 70 terms of the series gives a value for Pi^2 accurate to more than 100 decimal places.
The series representation Pi^2 = 3*Sum_{k >= 1} (2*k)/k^3 can be accelerated to give the faster converging series
Pi^2 = 99/10 - (8/5)*Sum_{k >= 1} (2*k + 2)/(k*(k + 1)*(k + 2))^3 and
Pi^2 = 54715/5544 + (41472/385)*Sum_{k >= 1} (2*k + 4)/(k*(k + 1)*(k + 2)*(k + 3)*(k + 4))^3.
The general result is: for n >= 1, Pi^2 = C(n) + (-1)^n * D(n)*Sum_{k >= 1} (2*k + 2*n)/(k*(k + 1)*...*(k + 2*n))^3, where C(n) = A(n) - 10*(-1)^n*(3*n)!*(2*n)!^3/((2*n + 1)*n!^3*(6*n + 1)!) and D(n) = (2*n)!^6 * (3*n)! / ( 2*n*(6*n - 1)!*n!^3 ). (End)
Equals 9 + 3*Sum_{n>=1} 1/((n^2*(n+1)^2)). - Davide Rotondo, May 29 2025

More terms from Robert G. Wilson v, Dec 15 2005

A019670 Decimal expansion of Pi/3.

Original entry on oeis.org

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

A002325 Glaisher's J numbers.

Original entry on oeis.org

1, 1, 2, 1, 0, 2, 0, 1, 3, 0, 2, 2, 0, 0, 0, 1, 2, 3, 2, 0, 0, 2, 0, 2, 1, 0, 4, 0, 0, 0, 0, 1, 4, 2, 0, 3, 0, 2, 0, 0, 2, 0, 2, 2, 0, 0, 0, 2, 1, 1, 4, 0, 0, 4, 0, 0, 4, 0, 2, 0, 0, 0, 0, 1, 0, 4, 2, 2, 0, 0, 0, 3, 2, 0, 2, 2, 0, 0, 0, 0, 5, 2, 2, 0, 0, 2, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 2, 1, 6, 1, 0, 4, 0, 0, 0

Offset: 1

N. J. A. Sloane

Number of integer solutions to the equation x^2 + 2*y^2 = n when (-x, -y) and (x, y) are counted as the same solution.
For n nonzero, a(n) is nonzero if and only if n is in A002479. - Michael Somos, Dec 15 2011
Coefficients of Dedekind zeta function for the quadratic number field of discriminant -8. See A002324 for formula and Maple code. - N. J. A. Sloane, Mar 22 2022

			x + x^2 + 2*x^3 + x^4 + 2*x^6 + x^8 + 3*x^9 + 2*x^11 + 2*x^12 + x^16 + ...

B. C. Berndt, Ramanujan's Notebooks Part III, Springer-Verlag, see p. 114 Entry 8(iii).
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 3, p. 19.
N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 78, Eq. (32.24).
J. W. L. Glaisher, Table of the excess of the number of (8k+1)- and (8k+3)-divisors of a number over the number of (8k+5)- and (8k+7)-divisors, Messenger Math., 31 (1901), 82-91.
D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, pp. 7-10.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..10000
J. W. L. Glaisher, Table of the excess of the number of (8k+1)- and (8k+3)-divisors of a number over the number of (8k+5)- and (8k+7)-divisors, Messenger Math., 31 (1901), 82-91. [Incomplete annotated scanned copy]
Michael D. Hirschhorn, The number of representations of a number by various forms, Discrete Mathematics 298 (2005), 205-211.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 5.
N. J. A. Sloane, Transforms.
Index entries for sequences related to Glaisher's numbers.

Cf. A033715, A093954.
Dedekind zeta functions for imaginary quadratic number fields of discriminants -3, -4, -7, -8, -11, -15, -19, -20 are A002324, A002654, A035182, A002325, A035179, A035175, A035171, A035170, respectively.
Dedekind zeta functions for real quadratic number fields of discriminants 5, 8, 12, 13, 17, 21, 24, 28, 29, 33, 37, 40 are A035187, A035185, A035194, A035195, A035199, A035203, A035188, A035210, A035211, A035215, A035219, A035192, respectively.

S:= series( (JacobiTheta3(0,q)*JacobiTheta3(0,q^2)-1)/2, q, 1001):
seq(coeff(S,q,j), j=1..1000); # Robert Israel, Dec 01 2015

a[n_] := Total[ KroneckerSymbol[-8, #] & /@ Divisors[n]]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Nov 25 2011, after Michael Somos *)
QP = QPochhammer; s = ((QP[q^2]^3*QP[q^4]^3)/(QP[q]^2*QP[q^8]^2)-1)/(2q) + O[q]^105; CoefficientList[s, q] (* Jean-François Alcover, Dec 01 2015, adapted from PARI *)

a(n) = if( n<1, 0, issquare(n)-issquare(2*n) + 2*sum(i=1,sqrtint(n\2), issquare(n-2*i^2)))

{a(n) = if( n<1, 0, qfrep([ 1, 0; 0, 2],n)[n])} \\ Michael Somos, Jun 05 2005

{a(n) = if( n<1, 0, direuler(p=2, n, 1 / (1 - X) / (1 - kronecker( -2, p) * X))[n])} \\ Michael Somos, Jun 05 2005

{a(n) = if( n<1, 0, sumdiv(n, d, kronecker( -2, d)))} \\ Michael Somos, Aug 23 2005

{a(n) = local(A, p, e); if( n<1, 0, A = factor(n); prod( k=1, matsize(A)[1], if( p=A[k, 1], e=A[k, 2]; if( p==2, 1, if( p%8<4, e+1, !(e%2))))))} \\ Michael Somos, Oct 23 2006

{a(n) = local(A); if( n<1, 0, A = x * O(x^n); polcoeff( eta(x + A)^-2 * eta(x^2 + A)^3 * eta(x^4 + A)^3 * eta(x^8 + A)^-2, n) / 2)}

a(n) = my(f=factor(n>>valuation(n,2)), e); prod(i=1, #f~, e=f[i, 2]; if( f[i, 1]%8<4, e+1, 1 - e%2)) \\ Charles R Greathouse IV, Sep 09 2014

Coefficients in expansion of Dirichlet series Product_p (1-(Kronecker(m, p)+1)*p^(-s) + Kronecker(m, p)*p^(-2s))^(-1) for m = -2.
Moebius transform is period 8 sequence [ 1, 0, 1, 0, -1, 0, -1, 0, ...]. - Michael Somos, Aug 23 2005
G.f.: (theta_3(q) * theta_3(q^2) - 1) / 2 = Sum_{k>0} Kronecker( -2, n) * x^k / (1 - x^k) = Sum_{k>0} (x^k + x^(3*k)) / (1 + x^(4*k)).
Multiplicative with a(2^e) = 1, a(p^e) = e+1 if p == 1, 3 (mod 8), a(p^e) = (1+(-1)^e)/2 if p == 5, 7 (mod 8). - Michael Somos, Oct 23 2006
A033715(n) = 2 * a(n) unless n=0.
a(n) = A188169(n) + A188170(n) - A188171(n) - A188172(n) [Hirschhorn]. - R. J. Mathar, Mar 23 2011
G.f.: A(x) = 2*(1+x^2)/(G(0)-2*x*(1+x^2)); G(k) = 1+x+x^(2*k)*(1+x^3+x^(2*k+1)+x^(2*k+4)+x^(4*k+3)+x^(4*k+4)) - x*(1+x^(2*k))*(1+x^(2*k+4))*(1+x^(4*k+4))^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 03 2012
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi/(2*sqrt(2)) = 1.110720... (A093954). - Amiram Eldar, Oct 11 2022

A091925 Decimal expansion of Pi^3.

Original entry on oeis.org

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

Offset: 2

Mohammad K. Azarian, Mar 16 2004

Surface area of the 6-dimensional unit sphere. - Stanislav Sykora, Nov 08 2013
Surface area of a sphere of diameter Pi equals the volume of the circumscribed cube. - Omar E. Pol, Dec 25 2013
Area of a circle of radius Pi. - Omar E. Pol, Jan 31 2016

			31.00627668029982017547631506710139520222528856588510769414453810380639...

Harry J. Smith, Table of n, a(n) for n = 2..20000
G. L. Honaker, Jr. and Chris Caldwell, Prime Curios! 31
Jean-Christophe Pain, Series representations for Pi^3 involving the golden ratio, arXiv:2206.15281 [math.NT], 2022.
Khodabakhsh Hessami Pilehrood, Tatiana Hessami Pilehrood, Series acceleration formulas for beta values, Discrete Mathematics & Theoretical Computer Science 12:2 (2010), pp. 223-236.
Stanislav Sykora, Surface Integrals over n-Dimensional Spheres.
Index entries for transcendental numbers

Cf. A000796, A002388, A058285 (continued fraction), A019670, A093954, A092731, A092735.

R:= RealField(100); (Pi(R))^3; // G. C. Greubel, Mar 09 2018

First@ RealDigits@ N[Pi^3, 120] (* Michael De Vlieger, Jan 31 2016 *)

default(realprecision, 20080); x=Pi^3/10; for (n=2, 20000, d=floor(x); x=(x-d)*10; write("b091925.txt", n, " ", d)); \\ Harry J. Smith, Jun 22 2009

Sum_{k >= 0} binomial(2*k,k)/((2*k + 1)^3*16^k) = 7*Pi^3/216. (Kh. Hessami Pilehrood and T. Hessami Pilehrood).
From Peter Bala, Feb 05 2015: (Start)
The integer sequences A(n) := 2^n*(2*n + 1)!^3/n!^2 and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)^3*(1/16)^k ) both satisfy the second order recurrence equation u(n) = (160*n^4 + 128*n^3 + 144*n^2 + 2)*u(n-1) - 32*(n - 1)*(2*n - 1)^7*u(n-2). From this observation we can obtain the continued fraction expansion 7/216*Pi^3 = 1 + 2/(432 - 32*3^7/(4162 - 32*2*5^7/(17714 - ... - 32*(n - 1)*(2*n - 1)^7/((160*n^4 + 128*n^3 + 144*n^2 + 2) - ... )))). Cf. A002388, A019670 and A093954. (End)
From Peter Bala, Oct 31 2019: (Start)
Pi^3 = (1/7) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/6)^3 + 1/(n + 5/6)^3 ).
Pi^3 = (1/31) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/10)^3 - 1/(n + 3/10)^3 - 1/(n + 7/10)^3 + 1/(n + 9/10)^3 ). Cf. A019692, A092731 and A092735. (End)
Equals Integral_{x=-oo..oo} x^2/(exp(x/2) + exp(-x/2)) dx. - Amiram Eldar, May 21 2021

A000281 Expansion of cos(x)/cos(2x).

Original entry on oeis.org

1, 3, 57, 2763, 250737, 36581523, 7828053417, 2309644635483, 898621108880097, 445777636063460643, 274613643571568682777, 205676334188681975553003, 184053312545818735778213457, 193944394596325636374396208563

Offset: 0

N. J. A. Sloane

a(n) is (2n)! times the coefficient of x^(2n) in the Taylor series for cos(x)/cos(2x).

			cos x / cos 2*x = 1 + 3*x^2/2 + 19*x^4/8 + 307*x^6/80 + ...

J. W. L. Glaisher, "On the coefficients in the expansions of cos x / cos 2x and sin x / cos 2x", Quart. J. Pure and Applied Math., 45 (1914), 187-222.
I. J. Schwatt, Intro. to Operations with Series, Chelsea, p. 278.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..215 (terms 0..50 from T. D. Noe)
P. Bala, A triangle for calculating A000281
P. Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
Kwang-Wu Chen, An Interesting Lemma for Regular C-fractions, J. Integer Seqs., Vol. 6, 2003.
D. Dumont, Further triangles of Seidel-Arnold type and continued fractions related to Euler and Springer numbers, Adv. Appl. Math., 16 (1995), 275-296.
Matthieu Josuat-Vergès and Jang Soo Kim, Touchard-Riordan formulas, T-fractions, and Jacobi's triple product identity, arXiv:1101.5608 [math.CO] (2011).
D. Shanks, Generalized Euler and class numbers. Math. Comp. 21 (1967) 689-694, sequence c(2,n).
D. Shanks, Corrigenda to: "Generalized Euler and class numbers", Math. Comp. 22 (1968), 699.
D. Shanks, Generalized Euler and class numbers, Math. Comp. 21 (1967), 689-694; 22 (1968), 699. [Annotated scanned copy]

Cf. A000364 A086646, A002437.
Cf. A064069. Bisection of A000825, A001586.
Cf. A098432. Cf. A093954, A188458, A212435, A235605, A255883.

a := n -> (-1)^n*2^(6*n+1)*(Zeta(0,-2*n,1/8)-Zeta(0,-2*n,5/8)):
seq(a(n), n=0..13); # Peter Luschny, Mar 11 2015

With[{nn=30},Take[CoefficientList[Series[Cos[x]/Cos[2x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Oct 06 2011 *)

{a(n) = if( n<0, 0, n*=2; n! * polcoeff( cos(x + x * O(x^n)) / cos(2*x + x * O(x^n)), n))}; /* Michael Somos, Feb 09 2006 */

a(n) = Sum_{k=0..n} (-1)^k*binomial(2n, 2k)*A000364(n-k)*4^(n-k). - Philippe Deléham, Jan 26 2004
E.g.f.: Sum_{k>=0} a(k)x^(2k)/(2k)! = cos(x)/cos(2x).
a(n-1) is approximately 2^(4*n-3)*(2*n-1)!*sqrt(2)/((Pi^(2*n-1))*(2*n-1)). The approximation is quite good a(250) is of the order of 10^1181 and this formula is accurate to 238 digits. - Simon Plouffe, Jan 31 2007
G.f.: 1 / (1 - 1*3*x / (1 - 4*4*x / (1 - 5*7*x / (1 - 8*8*x / (1 - 9*11*x / ... ))))). - Michael Somos, May 12 2012
G.f.: 1/E(0) where E(k) = 1 - 3*x - 16*x*k*(2*k+1) - 16*x^2*(k+1)^2*(4*k+1)*(4*k+3)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
G.f.: T(0)/(1-3*x), where T(k) = 1 - 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2/( 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2 - (32*x*k^2+16*x*k+3*x-1 )*(32*x*k^2+80*x*k+51*x -1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
From Peter Bala, Mar 09 2015: (Start)
a(n) = (-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial.
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(4*k + 1)^2) = 1 + 3*x + 57*x^2 + 2763*x^3 + ....
We appear to have the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + .... See A093954.
Bisection of A001586. See also A188458 and A212435. Second row of A235605 (read as a square array).
The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255883. (End)
From Peter Luschny, Mar 11 2015: (Start)
a(n) = ((-64)^n/((n+1/2)))*(B(2*n+1,7/8)-B(2*n+1,3/8)), B(n,x) Bernoulli polynomials.
a(n) = 2*(-16)^n*LerchPhi(-1, -2*n, 1/4).
a(n) = (-1)^n*Sum_{0..2*n} 2^k*C(2*n,k)*E(k), E(n) the Euler secant numbers A122045.
a(n) = (-4)^n*SKP(2*n,1/2) where SKP are the Swiss-Knife polynomials A153641.
a(n) = (-1)^n*2^(6*n+1)*(Zeta(-2*n,1/8) - Zeta(-2*n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. (End)
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + x - 4*x/(1 - 12*x/(1 + x - 40*x/(1 - 56*x/(1 + x - ... - 4*n(4*n - 3)*x/(1 - 4*n(4*n - 1)*x/(1 + x - ...
G.f.: 1/(1 + 9*x - 12*x/(1 - 4*x/(1 + 9*x - 56*x/(1 - 40*x/(1 + 9*x - ... - 4*n(4*n - 1)*x/(1 - 4*n(4*n - 3)*x/(1 + 9*x - .... (End)
From Peter Bala, Nov 08 2019: (Start)
a(n) = sqrt(2)*4^n*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/2)/cosh(Pi*x) dx. Cf. A002437.
The L-series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)! (see Shanks), which gives a(n) ~ (1/sqrt(2))*(2*n)!*(4/Pi)^(2*n+1). (End)

A063448 Decimal expansion of Pi * sqrt(2).

Original entry on oeis.org

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

Offset: 1

Jason Earls, Jul 24 2001

Hypotenuse of the right triangle with legs Pi and Pi. - Zak Seidov, May 04 2005
Circumference of the circumcircle of the unit square. - Jonathan Sondow, Nov 23 2017
Half-perimeter of the closed curve with implicit Cartesian equation x^2 + y^2 = abs(x) + abs(y). - Stefano Spezia, Oct 20 2020

			4.4428829381583662470158809900606936986146216893756902230853...

Harry J. Smith, Table of n, a(n) for n = 1..20000
Peter Bala, Series for sqrt(2)*Pi
Eric Weisstein's World of Mathematics, Gauss Multiplication Formula.
Wikipedia, Bailey-Borwein-Plouffe formula.
Index entries for transcendental numbers

Cf. A063447 (continued fraction), A093954, A153799, A193887, A244976, A247719.

RealDigits[N[Pi*Sqrt[2], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Mar 21 2011*)

PARI
```
\p 400; Pi * sqrt(2)
```

default(realprecision, 20080); x=Pi*sqrt(2); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b063448.txt", n, " ", d)) \\ Harry J. Smith, Aug 21 2009

# Use some guard digits when computing.
# BBP formula (1/8) P(1, 64, 12, (32, 0, 8, 0, 8, 0, -4, 0, -1, 0, -1, 0))
from decimal import Decimal as dec, getcontext
def BBPpisqrt2(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(64)
    for k in range(int(n * 0.5536546824812272) + 1):
        twk = dec(12 * k)
        s += f * ( dec(32) / (twk + 1) + dec(8)  / (twk + 3)
                 + dec(8)  / (twk + 5) - dec(4)  / (twk + 7)
                 - dec(1)  / (twk + 9) - dec(1)  / (twk + 11))
        f /= g
    return s / dec(8)
print(BBPpisqrt2(200))  # Peter Luschny, Nov 03 2023

Equals Gamma(1/4)*Gamma(3/4). - Jean-François Alcover, Nov 24 2014
From Amiram Eldar, Aug 06 2020: (Start)
Equals Integral_{x=0..oo} log(1 + 1/x^4) dx.
Equals Integral_{x=0..oo} log(1 + 2/x^2) dx.
Equals Integral_{x=-oo..oo} exp(x/4)/(exp(x) + 1) dx.
Equals Integral_{x=0..2*Pi} 1/(cos(x)^2 + 1) dx = Integral_{x=0..2*Pi} 1/(sin(x)^2 + 1) dx. (End)
From Andrea Pinos, Jul 03 2023: (Start)
Equals (Product_{k=1..4} Gamma(k/8)*Gamma(1 - k/8))^(1/4).
General result: 2*Pi/(4*y)^(1/(2*y)) = (Product_{k=1..y} Gamma(k/(2*y))*Gamma(1 - k/(2*y)) )^(1/y). (End)
From Peter Bala, Oct 22 2023: (Start)
sqrt(2)*Pi = 4 + 8*Sum_{n >= 0} (-1)^n/(16*n^2 + 32*n + 15). See A141759.
In the following the Eisenstein summation convention is assumed: that is,
Sum_{n = -oo..oo} means Limit_{N -> oo} Sum_{n = -N..N}:
sqrt(2)*Pi = 4*Sum_{n = -oo..oo} (-1)^n/(4*n + 1).
More generally, it appears that for k >= 0, k not of the form 4*m + 1,
sqrt(2)*Pi = -sign(cos(Pi*(k - 3)/4)) * 4*(2^floor(k/2))*k! * Sum_{n = -oo..oo} (-1)^n/((4*n + 1)*(4*n + 3)*...*(4*n + 2*k + 1)) (verified up to k = 50).
sqrt(2)*Pi = (2^4)*Sum_{n >= 0} (-1)^n * (2*n + 1)/((4*n + 1)*(4*n + 3)) = 512/105 - (2^6)*4!*Sum_{n >= 0} (-1)^n * (2*n + 3)/((4*n + 1)*(4*n + 3)*...*(4*n + 11)).
sqrt(2)*Pi = 4 + (2^3)*Sum_{n >= 0} (-1)^n * (4*n + 1)/((4*n + 1)*(4*n + 3)*(4*n + 5)) = 1408/315 - (2^5)*5!*Sum_{n >= 0} (-1)^n * (4*n + 1)/((4*n + 1)*(4*n + 3)*...*(4*n + 13)).
sqrt(2)*Pi = 16/3 - (2^4)*3!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 3)*(4*n + 5)*(4*n + 7)) = 14848/3465 + (2^6)*7!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 3)*...*(4*n + 15)). (End)
From Peter Bala, Nov 19 2023: (Start)
sqrt(2)*Pi = 512*Sum_{k >= 1} (-1)^(k+1) * k^2/((16*k^2 - 1)*(16*k^2 - 9)).
This is the case n = 1 of the more general result: for n >= 1,
sqrt(2)*Pi = (-1)^(n+1) * 2^(n+7) * (2*n)!/(2*n - 1) * Sum_{k >= 1} (-1)^(k+1) * k^2/( Product_{i = 0..n} (16*k^2 - (2*i+1)^2) ). Cf. A334422. (End)
Equals Integral_{x=-oo..oo} (x^2 + 1)/(x^4 + 1) dx. - Kritsada Moomuang, Jun 04 2025

Edited by N. J. A. Sloane, May 05 2007

Corrected by Neven Juric, Apr 24 2008

A113411 Excess of number of divisors of 2n+1 of form 8k+1, 8k+3 over those of form 8k+5, 8k+7.

Original entry on oeis.org

1, 2, 0, 0, 3, 2, 0, 0, 2, 2, 0, 0, 1, 4, 0, 0, 4, 0, 0, 0, 2, 2, 0, 0, 1, 4, 0, 0, 4, 2, 0, 0, 0, 2, 0, 0, 2, 2, 0, 0, 5, 2, 0, 0, 2, 0, 0, 0, 2, 6, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 3, 4, 0, 0, 4, 2, 0, 0, 2, 2, 0, 0, 0, 2, 0, 0, 6, 0, 0, 0, 0, 2, 0, 0, 1, 6, 0, 0, 4, 2, 0, 0, 0, 4, 0, 0, 2, 0, 0, 0, 4, 0, 0, 0, 4

Offset: 0

Michael Somos, Oct 29 2005

Ramanujan theta functions: f(q) := Prod_{k>=1} (1-(-q)^k) (see A121373), phi(q) := theta_3(q) := Sum_{k=-oo..oo} q^(k^2) (A000122), psi(q) := Sum_{k=0..oo} q^(k*(k+1)/2) (A010054), chi(q) := Prod_{k>=0} (1+q^(2k+1)) (A000700).

Bisection of A002325. Number of ways to write n as a sum of a square plus four times a triangular number [Hirschhorn]. - R. J. Mathar, Mar 23 2011

			1 + 2*x + 3*x^4 + 2*x^5 + 2*x^8 + 2*x^9 + x^12 + 4*x^13 + 4*x^16 + ...
q + 2*q^3 + 3*q^9 + 2*q^11 + 2*q^17 + 2*q^19 + q^25 + 4*q^27 + 4*q^33 + ...

Nathan J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 82, Eq. (32.55).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael D. Hirschhorn, The number of representations of a number by various forms, Discrete Mathematics 298 (2005), 205-211.
Michael Somos, Introduction to Ramanujan theta functions, 2019.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

Cf. A033761, A093954, A112603, A133692.

Cf. A000122, A000700, A010054, A121373.

a[n_] := DivisorSum[2n+1, Switch[Mod[#, 8], 1|3, 1, 5|7, -1]&]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Dec 04 2015 *)

a(n) = if( n<0, 0, n = 2*n + 1; sumdiv(n, d, (-1)^(d%8>3)))

a(n) = local(n1); if( n<0, 0, n1 = sqrtint(n); polcoeff( sum(k=1,n1, 2*x^k^2, 1 + x*O(x^n)) * sum(k=0,n1, x^(2*k^2 + 2*k)), n))

a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^5 * eta(x^8 + A)^2 / (eta(x + A)^2 * eta(x^4 + A)^3), n))

a(n) = local(A, p, e); if( n<0, 0, n = 2*n + 1; A = factor(n); prod(k=1, matsize(A)[1], if( p=A[k,1], e=A[k,2]; if( p==2, 0, if( abs(p%8-6)==1, (1+(-1)^e)/2, e+1)))))

Expansion of phi(q) * psi(q^4) in powers of q where psi(), phi() are Ramanujan theta functions.
Expansion of q^(-1) * (eta(q^4)^5 * eta(q^16)^2) / (eta(q^2)^2 * eta(q^8)^3) in powers of q^2.
a(n) = b(2*n + 1) where b(n) is multiplicative and b(2^e) = 0^e, b(p^e) = e+1 if p == 1, 3 (mod 8), b(p^e) = (1+(-1)^e)/2 if p == 5, 7 (mod 8).
Euler transform of period 8 sequence [ 2, -3, 2, 0, 2, -3, 2, -2, ...].
G.f. is a period 1 Fourier series which satisfies f(-1 / (16 t)) = 2^(1/2) (t/i) g(t) where q = exp(2 Pi i t) and g() is g.f. for A133692. - Michael Somos, Mar 16 2011
G.f.: (Sum_{k} x^k^2) * (Sum_{k>=0} x^(2*k^2 + 2*k)).
G.f.: Sum_{k>=0} a(k) * x^(2*k + 1) = Sum_{k>=0} F(x^(2*k + 1), x^(3*(2*k + 1))) where F(x, y) = (x + y) / (1 + x*y).
a(4*n + 2) = a(4*n + 3) = 0. a(4*n) = A112603(n). a(4*n + 1) = 2 * A033761(n).
From Peter Bala, Jan 07 2021: (Start)
Conjectural g.f.s: A(x) = Sum_{n >= 0} (-1)^(n*(n-1)/2)*x^n/(1 - x^(2*n+1)).
A(x) = Sum_{n = -oo..oo} (-1)^n*x^(2*n)/(1 - x^(4*n+1)) = Sum_{n = -oo..oo} (-1)^n*x^(2*n+1)/(1 - x^(4*n+3)). (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi/(2*sqrt(2)) = 1.1107207... (A093954). - Amiram Eldar, Dec 28 2023

A247719 Decimal expansion of Integral_{t=0..Pi/2} sqrt(tan(t)) dt.

Original entry on oeis.org

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

Offset: 1

Jean-François Alcover, Sep 23 2014

			2.22144146907918312350794049503034684930731...

G. C. Greubel, Table of n, a(n) for n = 1..10000
D. H. Bailey and J. M. Borwein, Highly Parallel, High-Precision Numerical Integration p. 7. (2005) Lawrence Berkeley National Laboratory.
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, pp. 235-236.

Cf. A063448, A093954, A193887, A244976, A181048, A181049, A186706.

SetDefaultRealField(RealField(100)); R:= RealField(); Pi(R)/Sqrt(2); // G. C. Greubel, Sep 07 2018

RealDigits[Pi/Sqrt[2], 10, 104] // First

default(realprecision, 100); Pi/sqrt(2) \\ G. C. Greubel, Sep 07 2018

Equals Pi/sqrt(2).
Equals A063448/2.
c = 2*( Sum_{k >= 0} (-1)^k/(4*k + 1) + Sum_{k >= 0} (-1)^k/(4*k + 3) ) = 2*(A181048 + A181049). - Peter Bala, Sep 21 2016
From Amiram Eldar, Aug 07 2020: (Start)
Equals Integral_{x=0..Pi} 1/(cos(x)^2 + 1) dx = Integral_{x=0..Pi} 1/(sin(x)^2 + 1) dx.
Equals Integral_{x=-oo..oo} 1/(x^4 + 1) dx.
Equals Integral_{x=-oo..oo} x^2/(x^4 + 1) dx.
Equals Integral_{x=0..oo} log(1 + 1/(2 * x^2)) dx. (End)
Equals Integral_{x=0..2*Pi} 1/(3 + sin(x)) dx; since for a>1: Integral_{x=0..2*Pi} 1/(a + sin(x)) dx = 2*Pi/sqrt(a^2-1). - Bernard Schott, Feb 19 2023
Equals 20/9 - 160*Sum_{n >= 1} 1/((64*n^2 - 1)*(64*n^2 - 4)*(64*n^2 - 9)). - Peter Bala, Nov 09 2023

A244644 Consider the method used by Archimedes to determine the value of Pi (A000796). This sequence denotes the number of iterations of his algorithm which would result in a difference of less than 1/10^n from that of Pi.

Original entry on oeis.org

0, 1, 3, 5, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 25, 26, 28, 29, 31, 33, 34, 36, 38, 39, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 58, 59, 61, 63, 64, 66, 68, 69, 71, 73, 74, 76, 78, 79, 81, 83, 84, 86, 88, 89, 91, 93, 94, 96, 98, 99, 101, 103, 104, 106, 108, 109, 111, 113, 114

Offset: 0

William H. Richardson and Robert G. Wilson v, Jul 03 2014

It takes on average 5/3 iterations to yield another digit in the decimal expansion of Pi.
The side of a 96-gon inscribed in a unit circle is equal to sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(3))))). This is the size of one of the two polygons that Archimedes used to derive that 3 + 10/70 < Pi < 3 + 10/71.
In the Mathematica program, we started with an inscribed triangle and a circumscribed triangle of a unit circle and used decimal precision to just over a 1000 places.
The perimeter of the circumscribed 3*2^n-polygon exceeds Pi by more than the deficit of the perimeter of the inscribed 3*2^n-polygon. If we were to give twice the weight of the inscribed 3*2^n-polygon to that of the circumscribed 3*2^n-polygon, then the convergence would be twice as fast!
From A.H.M. Smeets, Jul 12 2018: (Start)
Archimedes's scheme: set upp(0) = 2*sqrt(3), low(0) = 3 (hexagons); upp(n+1) = 2*upp(n)*low(n)/(upp(n)+low(n)) (harmonic mean, i.e., 1/upp(n+1) = (1/upp(n) + 1/low(n))/2), low(n+1) = sqrt(upp(n+1)*low(n)) (geometric mean, i.e., log(low(n+1)) = (log(upp(n+1)) + log(low(n)))/2), for n >= 0. Invariant: low(n) < Pi < upp(n); variant function: upp(n)-low(n) tends to zero for n -> inf. The error of low(n) and upp(n) decreases by a factor of approximately 4 each iteration, i.e., approximately 2 bits are gained by each iteration.
From Archimedes's scheme, set r(n) = (2*low(n) + upp(n))/3, r(n) > Pi and the error decreases by a factor of approximately 16 for each iteration, i.e., approximately 4 bits are gained by each iteration. This is often called "Snellius acceleration".
For similar schemes see also A014549 (in this case with quadratically convergence), A093954, A129187, A129200, A188615, A195621, A202541.
Note that replacing "5/3" by "log(10)/log(4)" would be better in the first comment. (End)

			Just averaging the initial two triangles (3.89711) would yield Pi to one place of accuracy, i.e., the single digit '3'. Therefore a(0) = 0.
The first iteration yields, as the perimeters of the two hexagons, 4*sqrt(3) and 6. Their average is ~ 3.2320508 which is within 1/10 of the true value of Pi. Therefore a(1) = 1.
a(3) = 5 since it takes 5 iterations of Archimedes's algorithm to drive the averaged value of the circumscribed 96-gon and the inscribed 96-gon to yield a value within 0.001 of the correct value of Pi.
a(4) = 6 since it takes 6 iterations of Archimedes's algorithm to drive the averaged value of the circumscribed 3*2^6-gon and the inscribed 3*2^6-gon to yield a value within 0.0001 of the correct value of Pi.

Petr Beckmann, A History of Pi, 5th Ed. Boulder, Colorado: The Golem Press (1982).
Jonathan Borwein and David Bailey, Mathematics by Experiment, Second Edition, A. K. Peters Ltd., Wellesley, Massachusetts 2008.
Jonathan Borwein & Keith Devlin, The Computer As Crucible, An Introduction To Experimental Mathematics, A. K. Peters, Ltd., Wellesley, MA, Chapter 7, 'Calculating [Pi]' pp. 71-79, 2009.
Eli Maor, The Pythagorean Theorem, Princeton Science Library, Table 4.1, page 55.
Daniel Zwillinger, Editor-in-Chief, CRC Standard Mathematical Tables and Formulae, 31st Edition, Chapman & Hall/CRC, Boca Raton, London, New York & Washington, D.C., 2003, §4.5 Polygons, page 324.

Mike Bertrand, Ex Libris, Archimedes and Pi
Frits Beukers and Weia Reinboud, Snellius versneld, (text in English), preprint.
Frits Beukers and Weia Reinboud, Snellius versneld, (text in English), NAW 5/3 no. 1, pp. 60-63 (2002).
Lee Fook Loong Eugene, The Computation of [Pi] And Its History
Kyutae Paul Han, Pi and Archimedes Polygon Method
Eric Weisstein's World of Mathematics, Archimedes' Recurrence Formula
Eric Weisstein's World of Mathematics, Regular Polygon
Michael Woltermann Ph.D., Washington & Jefferson College, 38. Archimedes' Determination of [Pi].

Cf. A000796.

a[n_] := a[n] = N[2 a[n - 1] b[n - 1]/(a[n - 1] + b[n - 1]), 2^10]; b[n_] := b[n] = N[ Sqrt[ b[n - 1] a[n]], 2^10]; a[-1] = 2Sqrt[27]; b[-1] = a[-1]/2; f[n_] := Block[{k = 0}, While[ 10^n*((a[k] + b[k])/4 -Pi) > 1, k++]; k]; Array[f, 70]

Conjecture: There exists a c such that a(n) = floor(n*log(10)/log(4)+c); where c is in the range [0.08554,0.10264]. Critical values to narrow the range are believed to be at a(74), a(133), a(192), a(251), a(310), a(366), a(425), a(484). - A.H.M. Smeets, Jul 23 2018

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 06 2010

			0.86697298733991103757399516388287071365217536734524490433....
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) = 1.7339(3)5974(5)7982(5)075(25)79(846)27(404)7... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181048. The numbers 1, 1, -3, -11, 57, 361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) ) = 1.7339(4)5974(6)7982(2)075(14)79(903)27(765)7.... - _Peter Bala_, Sep 23 2016

Jolley, Summation of Series, Dover (1961) eq 82 page 16.
Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.17

Ivan Panchenko, Table of n, a(n) for n = 0..1000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric W. Weisstein, Euler's Series Transformation.
Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).

Cf. A001586, A003881, A024396, A091648, A093954, A113476, A181049, A181122, A188458, A193534, A262246.

RealDigits[(Pi Sqrt[2])/8 + (Sqrt[2] Log[1 + Sqrt[2]])/4, 10, 100][[1]] (* Alonso del Arte, Aug 11 2011 *)

(log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) \\ G. C. Greubel, Jul 05 2017

(asinh(1)+Pi/2)/sqrt(8) \\ Charles R Greathouse IV, Jul 06 2017

Equals (A093954 + A091648/sqrt(2))/2.
Integral_{x = 0..1} 1/(1+x^4) = Sum_{k >= 0} (-1)^k/(4*k+1) = (log(1+sqrt(2)) + Pi/2)/(2*sqrt(2)).
1 - 1/5 + 1/9 - 1/13 + 1/17 - ... = (Pi*sqrt(2))/8 + (sqrt(2)*log(1 + sqrt(2)))/4 = (Pi + 2*log(1 + sqrt(2)))/(4 sqrt(2)). The first two are the formulas as given in Spiegel et al., the third is how Mathematica rewrites the infinite sum. - Alonso del Arte, Aug 11 2011
Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) ) ~ 1/N + 1/N^2 - 3/N^3 - 11/N^4 + 57/N^5 + 361/N^6 - ..., where the sequence of coefficients [1, 1, -3, -11, 57, 361, ...] is A188458. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 1/4. An example is given below. Cf. A181049. - Peter Bala, Sep 23 2016
Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 1) = Sum_{n >= 0} 2^(n-1)*n!/A007696(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k + 1)). - Peter Bala, Dec 01 2021
From Peter Bala, Oct 23 2023: (Start)
The slowly converging series representation Sum_{n >= 0} (-1)^n/(4*n + 1) for the constant can be accelerated to give the following faster converging series:
1/2 + 2*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5));
7/10 + 8*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9));
71/90 + 48*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13));
971/1170 + 384*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13)*(4*n + 17)).
These results may be easily verified by taking the partial fraction expansions of the summands. The general result appears to be that for r >= 0, the constant equals
C(r) + (2^r)*r!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 5)*...*(4*n + 4*r + 1)), where C(r) is the rational number Sum_{k = 0..r-1} 2^(k-1)*k!/(1*5*9*...*(4*k + 1)). [added 19 Feb 2024: the general result can be proved by the WZ method as described in Wilf.]
In the limit as r -> oo we find that the constant equals Sum_{k >= 0} 2^(k-1)*k!/(Product_{i = 0..k} 4*i + 1) as noted above. (End)
From Peter Bala, Mar 03 2024: (Start)
Continued fraction: 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler.
Equals hypergeom([1/4, 1], [5/4], -1).
Gauss's continued fraction: 1/(1 + 1^2/(5 + 4^2/(9 + 5^2/(13 + 8^2/(17 + 9^2/(21 + 12^2/(25 + 13^2/(29 + 16^2/(33 + 17^2/(37 + ... )))))))))). (End)

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A002388 Decimal expansion of Pi^2.

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A019670 Decimal expansion of Pi/3.

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A002325 Glaisher's J numbers.

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A091925 Decimal expansion of Pi^3.

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A000281 Expansion of cos(x)/cos(2x).

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A063448 Decimal expansion of Pi * sqrt(2).

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).