A015134 - OEIS

%I #88 Sep 11 2024 00:54:57

%S 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,

%T 22,63,24,34,32,39,34,30,58,19,86,32,52,43,58,22,78,39,46,70,102,25,

%U 26,42,40,27,52,160,74,63,126,62,70,63,134,104,64,57,78,34,132,101,60,74,222

%N 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.

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

%H Peter McAndrew, <a href="/A015134/b015134.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from David Radcliffe)

%H B. Avila and T. Khovanova, <a href="http://arxiv.org/abs/1403.4614">Free Fibonacci Sequences</a>, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Avila/avila4.html">J. Int. Seq. 17 (2014) # 14.8.5</a>.

%H Jesse Fischer, <a href="https://jessefischer.github.io/numnec/">Number Necklace Generator</a>.

%H R. C. Johnson, <a href="https://web.archive.org/web/20100711074902/http://www.dur.ac.uk/bob.johnson/fibonacci/">Fibonacci Numbers and Resources</a>.

%F a(2^n) = 2^n. - _Thomas Anton_, Apr 16 2023

%F Conjectures from _Stephen Bartell_, Aug 20 2023: (Start)

%F a(3 * 2^n) = 3*(3 * 2^n - 14), n >= 3.

%F a(3^n) = (3^n + 1)/2, n >= 0.

%F a(2 * 3^n) = 2*(3^n - 1), n >= 1.

%F a(5 * 3^n) = (3^(n+2) - 3)/2, n >= 0.

%F a(10 * 3^n) = 6(3^(n+1) - 5), n >= 1.

%F a(5^n) = (5^n + 1)/2, n >= 0.

%F a(2 * 5^n) = 5^n + 1, n >= 0.

%F a(3 * 5^n) = (5^(n+1) - 1)/2, n >= 0.

%F a(4 * 5^n) = 3 * 5^n + 1, n >= 0.

%F a(6 * 5^n) = 5^(n+1) - 1, n >= 0.

%F a(7^n) = (7^n + 1)/2, n >= 0.

%F a(2 * 7^n) = 7^n + 1, n >= 0. (End)

%F Conjectures from _Stephen Bartell_, Aug 16 2024: (Start)

%F a(11^n) = (6*11^n - 1)/5 + n, n >= 0.

%F a(13^n) = (13^n + 1)/2, n >= 0.

%F a(17^n) = (17^n + 1)/2, n >= 0.

%F a(19^n) = (10*19^n - 1)/9 + n, n >= 0.

%F a(23^n) = (23^n + 1)/2, n >= 0.

%F a(29^n) = (15*29^n - 8)/7 + 2n, n >= 0.

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

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

%Y 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).

%K nonn

%O 1,2

%A _Phil Carmody_

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