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}.

6, 7, 9, 10, 12, 13, 15, 16, 18, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 46, 48, 49, 51, 52, 54, 55, 56, 58, 59, 61, 62, 64, 65, 67, 68, 69, 71, 72, 74, 75, 77, 78, 80, 81, 82, 84, 85, 87, 88, 90, 91, 92, 94, 95, 97, 98

Offset: 1

Author

Clark Kimberling, Oct 03 2014

Keywords

• nonn
• easy

Comments

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.

Examples

			Let s(n) = sum{1/F(h), h = 1..n}.  Approximations are shown here:
n ... r - s(n) .... 1/2^(n+1)
1 ... 2.35989 ..... 0.25
2 ... 1.35989 ..... 0.125
3 ... 0.859886 .... 0.0625
4 ... 0.526552 .... 0.03125
5 ... 0.3265522 ... 0.015625
6 ... 0.201552 .... 0.0078125
a(1) = 6 because r - s(6) < 1/4 < r - s(5).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A079587, A248148.

Programs

Mathematica

$MaxExtraPrecision = Infinity;
z = 100; p[k_] := p[k] = Sum[1/Fibonacci[h], {h, 1, k}] ;
r = Sum[1/Fibonacci[h], {h, 1, 1000}]; N[Table[r - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^(n + 1) &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248178 *)