cp's OEIS Frontend

A106297 Period of the Lucas 5-step sequence A074048 mod n.

%I A106297 #51 Aug 31 2025 22:43:40
%S A106297 1,1,104,6,781,104,2801,12,312,781,16105,312,30941,2801,81224,24,
%T A106297 88741,312,13032,4686,291304,16105,12166,312,3905,30941,936,16806,
%U A106297 70728,81224,190861,48,1674920,88741,2187581,312,1926221,13032,3217864,9372,2896405
%N A106297 Period of the Lucas 5-step sequence A074048 mod n.
%C A106297 This sequence can differ from the corresponding Fibonacci sequence (A106303) only when n is a multiple of 2 or 599 because 9584 is the discriminant of the characteristic polynomial x^5-x^4-x^3-x^2-x-1 and the prime factors of 9584 are 2 and 599. [corrected by _Avery Diep_, Aug 25 2025]
%C A106297 a(n) divides A106303(n). - _Avery Diep_, Aug 25 2025
%H A106297 Avery Diep, <a href="/A106297/b106297.txt">Table of n, a(n) for n = 1..388</a>
%H A106297 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>
%F A106297 Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).
%F A106297 Conjectures: a(5^k) = 781*5^(k-1) for k > 0. a(2^k) = 3*2^(k-1) for k > 1. If a(p) != a(p^2) for prime p > 2, then a(p^k) = p^(k-1)*a(p) for k > 0. - _Chai Wah Wu_, Feb 25 2022
%t A106297 n=5; Table[p=i; a=Join[Table[ -1, {n-1}], {n}]; a=Mod[a, p]; a0=a; k=0; While[k++; s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; a!=a0]; k, {i, 50}]
%o A106297 (Python)
%o A106297 from itertools import count
%o A106297 def A106297(n):
%o A106297     a = b = (5%n,1%n,7%n,3%n,15%n)
%o A106297     s = sum(b) % n
%o A106297     for m in count(1):
%o A106297         b, s = b[1:] + (s,), (s+s-b[0]) % n
%o A106297         if a == b:
%o A106297             return m # _Chai Wah Wu_, Feb 22-27 2022
%Y A106297 Cf. A074048, A106303 (period of Fibonacci 5-step sequence mod n), A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).
%K A106297 nonn,changed
%O A106297 1,3
%A A106297 _T. D. Noe_, May 02 2005, Nov 19 2006