OFFSET
1,1
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.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..100
EXAMPLE
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).
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 03 2014
STATUS
approved