A106306 -id:A106306 - cp's OEIS frontend

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

Phil Carmody

nonn

b(k) >= k/4 (by counting zeros). - R. C. Johnson (bob.johnson(AT)dur.ac.uk), Nov 20 2003

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.

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).

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)

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

A015135 Consider Fibonacci-type sequences f(0)=X, f(1)=Y, f(k)=f(k-1)+f(k-2) mod n; all are periodic; sequence gives number of distinct periods.

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 2, 4, 3, 6, 3, 5, 2, 4, 5, 5, 2, 4, 3, 7, 3, 6, 2, 6, 4, 4, 4, 5, 3, 10, 3, 6, 5, 3, 5, 5, 2, 4, 4, 7, 2, 6, 2, 7, 7, 3, 2, 6, 3, 8, 4, 5, 2, 5, 5, 6, 5, 6, 3, 11, 2, 4, 5, 7, 5, 10, 2, 4, 3, 10, 3, 6, 2, 4, 7, 5, 5, 8, 3, 9, 5, 4, 2, 7, 5, 4, 5, 9, 2, 10, 4, 4, 5, 4, 7, 7, 2, 6, 7, 9, 3, 6, 2

Offset: 1

Phil Carmody

nonn

Consider the 2-step recursion f(k)=f(k-1)+f(k-2) mod n. For any of the n^2 initial conditions f(1) and f(2) in Zn, the recursion has a finite period. Each of these n^2 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths for each n. For n=8, there are 4 different lengths: 1, 3, 6 and 12. The maximum possible length of an orbit is A001175(n), the period of the Fibonacci 2-step sequence mod n. - T. D. Noe, May 02 2005

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
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Cf. A015134 (orbits of 2-step sequences), A106306 (primes that yield a simple orbit structure in 2-step recursions).

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

cp's OEIS Frontend

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.

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