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A248178 Least k such that r - sum{1/F(n), h = 1..k} < 1/2^(n+1), where F(n) = A000045 (Fibonacci numbers) and r = sum{1/F(n), h = 1..infinity}.

%I A248178 #5 Oct 06 2014 22:59:49
%S A248178 6,7,9,10,12,13,15,16,18,19,20,22,23,25,26,28,29,31,32,33,35,36,38,39,
%T A248178 41,42,44,45,46,48,49,51,52,54,55,56,58,59,61,62,64,65,67,68,69,71,72,
%U A248178 74,75,77,78,80,81,82,84,85,87,88,90,91,92,94,95,97,98
%N A248178 Least k such that r - sum{1/F(n), h = 1..k} < 1/2^(n+1), where F(n) = A000045 (Fibonacci numbers) and r = sum{1/F(n), h = 1..infinity}.
%C A248178 This sequence gives a measure of the convergence rate of the sum of reciprocals of Fibonacci numbers.  It appears that a(n+1) - a(n) is in {1,2} for n >= 1.
%H A248178 Clark Kimberling, <a href="/A248178/b248178.txt">Table of n, a(n) for n = 1..100</a>
%e A248178 Let s(n) = sum{1/F(h), h = 1..n}.  Approximations are shown here:
%e A248178 n ... r - s(n) .... 1/2^(n+1)
%e A248178 1 ... 2.35989 ..... 0.25
%e A248178 2 ... 1.35989 ..... 0.125
%e A248178 3 ... 0.859886 .... 0.0625
%e A248178 4 ... 0.526552 .... 0.03125
%e A248178 5 ... 0.3265522 ... 0.015625
%e A248178 6 ... 0.201552 .... 0.0078125
%e A248178 a(1) = 6 because r - s(6) < 1/4 < r - s(5).
%t A248178 $MaxExtraPrecision = Infinity;
%t A248178 z = 100; p[k_] := p[k] = Sum[1/Fibonacci[h], {h, 1, k}] ;
%t A248178 r = Sum[1/Fibonacci[h], {h, 1, 1000}]; N[Table[r - p[n], {n, 1, z/10}]]
%t A248178 f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^(n + 1) &, 1]
%t A248178 u = Flatten[Table[f[n], {n, 1, z}]]  (* A248178 *)
%Y A248178 Cf. A079587, A248148.
%K A248178 nonn,easy
%O A248178 1,1
%A A248178 _Clark Kimberling_, Oct 03 2014