OFFSET
1,1
COMMENTS
By using the formula of Binet one can easily show that all Fibonacci numbers greater than 3 are in the sequence. Furthermore the sequence a(n)/n converges to 1/5.
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
FORMULA
a(n) ~ 5n by the equidistribution theorem. - Charles R Greathouse IV, Oct 14 2016
EXAMPLE
47 is in the sequence because 1.6180339887*47 = 76.047 which is within .1 of an integer.
MATHEMATICA
Select[Range[300], Abs[ #*(1 + Sqrt[5])/2 - Round[ #*(1 + Sqrt[5])/2]] < 0.1 &] (* or *) grw1Q[n_]:=Module[{c=n*GoldenRatio}, Abs[c-Round[c]]<=.1]; Select[Range[ 300], grw1Q] (* Harvey P. Dale, Mar 20 2015 *)
PROG
(PARI) is(n)=my(phi=(sqrt(5)+1)/2); frac(n/phi + .1) < .2 \\ Charles R Greathouse IV, Oct 14 2016
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Ben Paul Thurston, Dec 15 2007
EXTENSIONS
Edited and extended by Stefan Steinerberger, Feb 20 2008
STATUS
approved