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

1,2

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)

Crossrefs

Cf. A015135 (number of different orbit lengths of the 2-step recursion mod n), A106306 (primes that yield a simple orbit structure in 2-step recursions).

Sequence in context: A046701 A140472 A109168 * A171580 A246796 A177235

Adjacent sequences: A015131 A015132 A015133 * A015135 A015136 A015137

Keyword

nonn

Author

Phil Carmody

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jan 06 2005

Status

approved