A015134 Consider Fibonacci-type sequences b(0)=X, b(1)=Y, b(k)=b(k-1)+b(k-2) mod n; all are periodic; sequence gives number of distinct periods.
1, 2, 2, 4, 3, 4, 4, 8, 5, 6, 14, 10, 7, 8, 12, 16, 9, 16, 22, 16, 29, 28, 12, 30, 13, 14, 14, 22, 63, 24, 34, 32, 39, 34, 30, 58, 19, 86, 32, 52, 43, 58, 22, 78, 39, 46, 70, 102, 25, 26, 42, 40, 27, 52, 160, 74, 63, 126, 62, 70, 63, 134, 104, 64, 57, 78, 34, 132, 101, 60, 74, 222
Offset: 1
Keywords
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.
Crossrefs
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.
Extensions
More terms from Larry Reeves (larryr(AT)acm.org), Jan 06 2005
Comments