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A053004
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Decimal expansion of AGM(1,sqrt(2)).
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15
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1, 1, 9, 8, 1, 4, 0, 2, 3, 4, 7, 3, 5, 5, 9, 2, 2, 0, 7, 4, 3, 9, 9, 2, 2, 4, 9, 2, 2, 8, 0, 3, 2, 3, 8, 7, 8, 2, 2, 7, 2, 1, 2, 6, 6, 3, 2, 1, 5, 6, 5, 1, 5, 5, 8, 2, 6, 3, 6, 7, 4, 9, 5, 2, 9, 4, 6, 4, 0, 5, 2, 1, 4, 1, 4, 3, 9, 1, 5, 6, 7, 0, 8, 3, 5, 8, 8, 5, 5, 5, 6, 4, 8, 9, 7, 9, 3, 3, 8, 9, 3, 7, 5, 9, 0
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OFFSET
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1,3
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COMMENTS
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AGM(a,b) is the limit of the arithmetic-geometric mean iteration applied repeatedly starting with a and b: a_0=a, b_0=b, a_{n+1}=(a_n+b_n)/2, b_{n+1}=sqrt(a_n*b_n).
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REFERENCES
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George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press (2006), p. 195.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, page 5.
J. R. Goldman, The Queen of Mathematics, 1998, p. 92.
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LINKS
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FORMULA
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Equals Integral_{x=0..Pi/2} sqrt(sin(x)) or Integral_{x=0..1} sqrt(x/(1-x^2)). - Jean-François Alcover, Apr 29 2013 [cf. Boros & Moll p. 195]
Equals Product_{n>=1} (1+1/A033566(n)) and also 2*AGM(1, i)/(1+i)) where i is the imaginary unit. - Dimitris Valianatos, Oct 03 2016
Conjecturally equals 1/( Sum_{n = -infinity..infinity} exp(-Pi*(n+1/2)^2 ) )^2. Cf. A096427. - Peter Bala, Jun 10 2019
Equals Integral_{x=0..Pi} sin(x)^2/sqrt(1 + sin(x)^2) dx. (End)
Equals sqrt(2/Pi)*Gamma(3/4)^2 = Integral_{x = 0..1} 1/(1 - x^2)^(1/4) dx = Pi/Integral_{x = 0..1} 1/(1 - x^2)^(3/4) dx. - Peter Bala, Jan 05 2022
Equals 2*Integral_{x = 0..1} x^2/sqrt(1 - x^4) dx.
Equals 1 - Integral_{x = 0..1} (sqrt(1 - x^4) - 1)/x^2 dx.
Equals hypergeom([-1/2, -1/4], [3/4], 1) = 1 + Sum_{n >= 0} 1/(4*n + 3)*Catalan(n)*(1/2^(2*n+1)). Cf. A096427. (End)
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EXAMPLE
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1.19814023473559220743992249228...
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MAPLE
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MATHEMATICA
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RealDigits[N[(1+Sqrt[2])Pi/(4EllipticK[17-12Sqrt[2]]), 105]][[1]] (* Jean-François Alcover, Jun 02 2019 *)
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PROG
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(PARI) default(realprecision, 20080); x=agm(1, sqrt(2)); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b053004.txt", n, " ", d)) \\ Harry J. Smith, Apr 20 2009
(Python)
from mpmath import mp, agm, sqrt
mp.dps=106
print([int(z) for z in list(str(agm(1, sqrt(2))).replace('.', '')[:-1])]) # Indranil Ghosh, Jul 11 2017
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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