A153386 Decimal expansion of Sum_{n>=1} 1/Fibonacci(2*n).

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

Offset: 1

Author

Eric W. Weisstein, Dec 25 2008

Keywords

• nonn
• cons

Examples

			1.535370508836252985029852896651599006367...
		

References

• Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.14.1, p. 358.

Links

• Richard André-Jeannin, Problem B-745, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 31, No. 3 (1993), p. 277; Fun with Unit Fractions, Solution to Problem B-745 by Paul S. Bruckman, ibid., Vol. 33, No. 1 (1995), p. 86.
• A. F. Horadam, Elliptic Functions and Lambert Series in the Summation of Reciprocals in Certain Recurrence-Generated Sequences, The Fibonacci Quarterly, Vol. 26, No. 2 (1988), pp. 98-114.

Crossrefs

Cf. A000045, A001906 (Fibonacci(2*n)), A065563.

Cf. A000796, A001113, A002163, A002390, A016628, A079586, A153387, A256178, A265288, A360928.

Programs

Mathematica

rd[k_] := rd[k] = RealDigits[ N[ Sum[ 1/Fibonacci[2*n], {n, 1, 2^k}], 105]][[1]]; rd[k = 4]; While[ rd[k] != rd[k - 1], k++]; rd[k] (* Jean-François Alcover, Oct 29 2012 *)
RealDigits[Sqrt[5] * (Log[5] + 2*QPolyGamma[0, 1, 1/GoldenRatio^4] - 4*QPolyGamma[0, 1, 1/GoldenRatio^2]) / (8*ArcCsch[2]), 10, 105][[1]] (* Vaclav Kotesovec, Feb 26 2023 *)

PARI

sumpos(n=1, 1/fibonacci(2*n)) \\ Michel Marcus, Sep 04 2021

Formula

Equals sqrt(5) * (L((3-sqrt(5))/2) - L((7-3*sqrt(5))/2)), where L(x) = Sum_{k>=1} x^k/(1-x^k) (Horadam, 1988, equation (4.6)). - Amiram Eldar, Oct 04 2020

From Gleb Koloskov, Sep 04 2021: (Start)

Equals 1/2 + (sqrt(5)/log(phi))*(log(5)/8 + 3*Integral_{x=0..infinity} sin(x)/((4*sin(x)^2+5)*(exp(Pi*x/log(phi))-1)) dx), where phi = (1+sqrt(5))/2 = A001622.

Equals 1/2 + (A002163/A002390)*(A016628/8 + 3*Integral_{x=0..infinity} sin(x)/((4*sin(x)^2+5)*(A001113^(A000796*x/A002390)-1)) dx). (End)

Equals 1 + Sum_{n>=1} 1/A065563(2*n-1) (André-Jeannin, 1993). - Amiram Eldar, Jan 15 2022

From Peter Bala, Aug 17 2022: (Start)

Equals 5/3 - 3*Sum_{n >= 1} 1/(F(2*n)*F(2*n+2)*F(2*n+4)), where F(n) = Fibonacci(n).

Conjecture: Equals 151/96 - 6*Sum_{n >= 1} 1/(F(2*n)*F(2*n+4)*F(2*n+6)). (End)

Equals A360928 * sqrt(5). - Kevin Ryde, Feb 27 2023