OFFSET
1,2
COMMENTS
b(k) >= k/4 (by counting zeros). - R. C. Johnson (bob.johnson(AT)dur.ac.uk), Nov 20 2003
LINKS
Peter McAndrew, Table of n, a(n) for n = 1..10000 (terms 1..1000 from David Radcliffe)
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.
Jesse Fischer, Number Necklace Generator.
R. C. Johnson, Fibonacci Numbers and Resources.
FORMULA
a(2^n) = 2^n. - Thomas Anton, Apr 16 2023
Conjectures from Stephen Bartell, Aug 20 2023: (Start)
a(3 * 2^n) = 3*(3 * 2^n - 14), n >= 3.
a(3^n) = (3^n + 1)/2, n >= 0.
a(2 * 3^n) = 2*(3^n - 1), n >= 1.
a(5 * 3^n) = (3^(n+2) - 3)/2, n >= 0.
a(10 * 3^n) = 6(3^(n+1) - 5), n >= 1.
a(5^n) = (5^n + 1)/2, n >= 0.
a(2 * 5^n) = 5^n + 1, n >= 0.
a(3 * 5^n) = (5^(n+1) - 1)/2, n >= 0.
a(4 * 5^n) = 3 * 5^n + 1, n >= 0.
a(6 * 5^n) = 5^(n+1) - 1, n >= 0.
a(7^n) = (7^n + 1)/2, n >= 0.
a(2 * 7^n) = 7^n + 1, n >= 0. (End)
Conjectures from Stephen Bartell, Aug 16 2024: (Start)
a(11^n) = (6*11^n - 1)/5 + n, n >= 0.
a(13^n) = (13^n + 1)/2, n >= 0.
a(17^n) = (17^n + 1)/2, n >= 0.
a(19^n) = (10*19^n - 1)/9 + n, n >= 0.
a(23^n) = (23^n + 1)/2, n >= 0.
a(29^n) = (15*29^n - 8)/7 + 2n, n >= 0.
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
More terms from Larry Reeves (larryr(AT)acm.org), Jan 06 2005
STATUS
approved