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

Author

N. J. A. Sloane

Keywords

• nonn
• cons

Comments

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

Examples

			9.869604401089358618834490999876151135313699407240790626413349376220044...
		

References

• 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.

Links

• 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.

Crossrefs

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

Programs

Magma

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

Maple

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

Mathematica

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

PARI

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

Python

# 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

Formula

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

Extensions

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