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%I #43 Feb 27 2023 03:40:19
%S 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,
%T 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,
%U 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
%N Decimal expansion of Sum_{n>=1} 1/Fibonacci(2*n).
%H Richard André-Jeannin, <a href="https://fq.math.ca/Scanned/31-3/elementary31-3.pdf">Problem B-745</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 31, No. 3 (1993), p. 277; <a href="https://www.fq.math.ca/Scanned/33-1/elementary33-1.pdf">Fun with Unit Fractions</a>, Solution to Problem B-745 by Paul S. Bruckman, ibid., Vol. 33, No. 1 (1995), p. 86.
%H A. F. Horadam, <a href="https://www.fq.math.ca/Scanned/26-2/horadam.pdf">Elliptic Functions and Lambert Series in the Summation of Reciprocals in Certain Recurrence-Generated Sequences</a>, The Fibonacci Quarterly, Vol. 26, No. 2 (1988), pp. 98-114.
%F 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
%F From _Gleb Koloskov_, Sep 04 2021: (Start)
%F 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.
%F 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)
%F Equals 1 + Sum_{n>=1} 1/A065563(2*n-1) (André-Jeannin, 1993). - _Amiram Eldar_, Jan 15 2022
%F From _Peter Bala_, Aug 17 2022: (Start)
%F Equals 5/3 - 3*Sum_{n >= 1} 1/(F(2*n)*F(2*n+2)*F(2*n+4)), where F(n) = Fibonacci(n).
%F Conjecture: Equals 151/96 - 6*Sum_{n >= 1} 1/(F(2*n)*F(2*n+4)*F(2*n+6)). (End)
%F Equals A360928 * sqrt(5). - _Kevin Ryde_, Feb 27 2023
%e 1.5353705088362529850...
%t 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 *)
%t 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 *)
%o (PARI) sumpos(n=1, 1/fibonacci(2*n)) \\ _Michel Marcus_, Sep 04 2021
%Y Cf. A000045, A001906 (Fibonacci(2*n)), A065563.
%Y Cf. A000796, A001113, A002163, A002390, A016628, A079586, A153387, A256178, A265288, A360928.
%K nonn,cons
%O 1,2
%A _Eric W. Weisstein_, Dec 25 2008
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