OFFSET
1,4
COMMENTS
Ratio a(n)/A095354(n) (i.e. average number of 1-fibits in Zeckendorf-expansions of primes p which Fib(n+1) <= p < Fib(n+2)) grows as: 1, 1, 1, 1.5, 2., 2.333333, 2.333333, 2.8, 3.285714, 3.181818, 3.5, 3.916667, 4.189189, 4.418182, 4.785714, 4.873016, 5.358586, 5.575758, 5.871179, 6.100852, 6.382705, 6.676225, 6.954266, 7.223132, 7.489542, 7.773978, 8.045173, 8.331323, 8.598659, 8.886546, 9.161734, 9.440489, 9.71936, 9.995484, 10.266207, 10.54327, 10.820602, 11.096084, 11.374267.
Ratio of that average compared to A010049(n)/A000045(n) (the expected value of that same sum computed for all integers in the same range) converges as: 1, 1, 0.666667, 0.9, 1, 1.037037, 0.919192, 0.99661, 1.063946, 0.945946, 0.96142, 1, 0.999059, 0.988519, 1.008389, 0.970278, 1.011305, 1.000122, 1.003368, 0.995592, 0.996635, 0.999338, 0.999601, 0.998575, 0.997298, 0.998427, 0.997837, 0.999078, 0.998056, 0.99941, 0.999296, 0.999567, 0.999834, 0.999811, 0.999265, 0.999347, 0.999451, 0.999382, 0.999555.
LINKS
A. Karttunen and J. Moyer: C-program for computing the initial terms of this sequence
EXAMPLE
a(1) = a(2) = 0, as there are no primes in ranges [1,2[ and [2,3[. a(3)=1 as in [3,5[ there is prime 3 with Fibonacci-representation 100. a(4)=3, as in [5,8[ there are primes 5 and 7, whose Fibonacci-representations are 1000 and 1010 respectively and we have three 1-fibits in total. a(5)=2, as in [8,13[ there is only one prime 11, with Zeckendorf-representation 10100.
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jun 04 2004
EXTENSIONS
a(2) corrected by Chai Wah Wu, Jan 16 2020
STATUS
approved