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A079586
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Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).
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40
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3, 3, 5, 9, 8, 8, 5, 6, 6, 6, 2, 4, 3, 1, 7, 7, 5, 5, 3, 1, 7, 2, 0, 1, 1, 3, 0, 2, 9, 1, 8, 9, 2, 7, 1, 7, 9, 6, 8, 8, 9, 0, 5, 1, 3, 3, 7, 3, 1, 9, 6, 8, 4, 8, 6, 4, 9, 5, 5, 5, 3, 8, 1, 5, 3, 2, 5, 1, 3, 0, 3, 1, 8, 9, 9, 6, 6, 8, 3, 3, 8, 3, 6, 1, 5, 4, 1, 6, 2, 1, 6, 4, 5, 6, 7, 9, 0, 0, 8, 7, 2, 9, 7, 0, 4
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OFFSET
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1,1
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COMMENTS
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André-Jeannin proved that this constant is irrational.
This constant does not belong to the quadratic number field Q(sqrt(5)) (Bundschuh and Väänänen, 1994). - Amiram Eldar, Oct 30 2020
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REFERENCES
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Daniel Duverney, Number Theory, World Scientific, 2010, 5.22, pp.75-76.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 358.
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LINKS
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Richard André-Jeannin, Problem H-450, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 29, No. 1 (1991), p. 89; Comparable, Solution to Problem H-450 by Paul S. Bruckman, ibid., Vol. 30, No. 2 (1992), p. 191-192.
Paul S. Bruckman, Problem B-602, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 25, No. 3 (1987), p. 279; Fibonacci Infinite Series, Solution to Problem B-602 by C. Georghiou, ibid., Vol. 26, No. 3 (1988), pp. 281-282.
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FORMULA
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Alternating series representation: 3 + Sum_{k >= 1} (-1)^(k+1)/(F(k)*F(k+1)*F(k+2)). - Peter Bala, Nov 30 2013
Equals sqrt(5) * Sum_{k>=0} (1/(phi^(2*k+1) - 1) - 2*phi^(2*k+1)/(phi^(4*(2*k+1)) - 1)), where phi is the golden ratio (A001622) (Greig, 1977).
Equals sqrt(5) * Sum_{k>=0} (-1)^k/(phi^(2*k+1) - (-1)^k) (Griffin, 1992).
Equals 1 + c1*(c2 + 32*Integral_{x=0..infinity} f(x) dx),
f(x) = sin(x)*(4+cos(2*x))/((exp(Pi*x/log(phi))-1)*(2*cos(2*x)+3)*(7-2*cos(2*x))) (End)
Equals 3 + 2 * Sum_{k>=1} 1/(F(2*k-1)*F(2*k+1)*F(2*k+2)) (Bruckman, 1987).
Equals 2 + Sum_{k>=1} 1/A350901(k) (André-Jeannin, Problem H-450, 1991).
Equals sqrt(5/4)*Sum_{j>=1} i^(1-j)/sin(j*c) where c = Pi/2 + i*arccsch(2). - Peter Luschny, Nov 15 2023
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EXAMPLE
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3.35988566624317755317201130291892717968890513373...
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MAPLE
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Digits := 120: c := Pi/2 + I*arccsch(2):
Jeannin := n -> sqrt(5/4)*add(I^(1-j)/sin(j*c), j = 1..n):
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MATHEMATICA
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digits = 105; Sqrt[5]*NSum[(-1)^n/(GoldenRatio^(2*n + 1) - (-1)^n), {n, 0, Infinity}, WorkingPrecision -> digits, NSumTerms -> digits] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Apr 09 2013 *)
First@RealDigits[Sqrt[5]/4 ((Log[5] + 2 QPolyGamma[1, 1/GoldenRatio^4] - 4 QPolyGamma[1, 1/GoldenRatio^2])/(2 Log[GoldenRatio]) + EllipticTheta[2, 0, 1/GoldenRatio^2]^2), 10, 105] (* Vladimir Reshetnikov, Nov 18 2015 *)
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PROG
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(PARI) /* Fast computation without splitting into even and odd indices, see the Arndt reference */
lambert2(x, a, S)=
{
/* Return G(x, a) = Sum_{n>=1} a*x^n/(1-a*x^n) (generalized Lambert series)
computed as Sum_{n=1..S} x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) )
As series in x correct up to order S^2.
We also have G(x, a) = Sum_{n>=1} a^n*x^n/(1-x^n) */
return( sum(n=1, S, x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) ) );
}
inv_fib_sum(p=1, q=1, S)=
{
/* Return Sum_{n>=1} 1/f(n) where f(0)=0, f(1)=1, f(n) = p*f(n-1) + q*f(n-1)
computed using generalized Lambert series.
Must have p^2+4*q > 0 */
my(al, be);
\\ Note: the q here is -q in the Horadam paper.
\\ The following numerical examples are for p=q=1:
al=1/2*(p+sqrt(p^2+4*q)); \\ == +1.6180339887498...
be=1/2*(p-sqrt(p^2+4*q)); \\ == -0.6180339887498...
return( (al-be)*( 1/(al-1) + lambert2(be/al, 1/al, S) ) ); \\ == 3.3598856...
}
default(realprecision, 100);
S = 1000; /* (be/al)^S == -0.381966^S == -1.05856*10^418 << 10^-100 */
inv_fib_sum(1, 1, S) /* 3.3598856... */ /* Joerg Arndt, Jan 30 2011 */
(PARI) suminf(k=1, 1/(fibonacci(k))) \\ Michel Marcus, Feb 19 2019
(Sage) m=120; numerical_approx(sum(1/fibonacci(k) for k in (1..10*m)), digits=m) # G. C. Greubel, Feb 20 2019
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CROSSREFS
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Cf. A000045, A000796, A001622, A002163, A002390, A084119, A093540, A016628, A105199, A153386, A153387.
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KEYWORD
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AUTHOR
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STATUS
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approved
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