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

It appears that for n>=2, a(n) is the number of distinct cycles that can be obtained when iterating x->f(x) where f(x) is defined in base n in the following way. In base n, take a number x with 2 nonzero digits, compute their sum modulo n-1; then f(x) is the concatenation of this digit and the unit digit of preceding term. See Nambi link. - Michel Marcus, Apr 13 2026

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.

Vineet Nambi, Cyclic Sequences of Two-Digit Numbers and Their Summation Invariants, 2026.

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.

For prime p not in A053032, a(p^n) = 1 + ((p+1)(p^n-1))/(A001175(p)) [except for p = 5].

For prime p in A053032, a(p^n) = 1 + ((p+1)(p^n-1)+n(p-1))/(A001175(p)). (End)

Conjecture: a(n) >= (n+1)/2. - Elliot Nutt, Dec 24 2025

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 A393147 A246796

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