OFFSET
1,1
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
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).
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 20232: (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)
EXAMPLE
9.869604401089358618834490999876151135313699407240790626413349376220044...
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 *)
PROG
(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
(Magma) R:= RealField(100); Pi(R)^2; // G. C. Greubel, Mar 08 2018
(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
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
EXTENSIONS
More terms from Robert G. Wilson v, Dec 15 2005
STATUS
approved