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A106303 Period of the Fibonacci 5-step sequence A001591 mod n.

%I A106303 #40 Aug 29 2025 00:22:24
%S A106303 1,6,104,12,781,312,2801,24,312,4686,16105,312,30941,16806,81224,48,
%T A106303 88741,312,13032,9372,291304,96630,12166,312,3905,185646,936,33612,
%U A106303 70728,243672,190861,96,1674920,532446,2187581,312,1926221,13032
%N A106303 Period of the Fibonacci 5-step sequence A001591 mod n.
%C A106303 This sequence can differ from the corresponding Lucas sequence (A106297) 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]
%H A106303 Chai Wah Wu, <a href="/A106303/b106303.txt">Table of n, a(n) for n = 1..388</a>
%H A106303 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.
%F A106303 Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).
%F A106303 Conjectures: a(5^k) = 781*5^(k-1) for k > 0. If a(p) != a(p^2) for p prime, then a(p^k) = p^(k-1)*a(p) for k > 0. - _Chai Wah Wu_, Feb 25 2022
%t A106303 n=5; Table[p=i; a=Join[{1}, Table[0, {n-1}]] 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 A106303 (Python)
%o A106303 from itertools import count
%o A106303 def A106303(n):
%o A106303     a = b = (0,)*4+(1 % n,)
%o A106303     s = 1 % n
%o A106303     for m in count(1):
%o A106303         b, s = b[1:] + (s,), (s+s-b[0]) % n
%o A106303         if a == b:
%o A106303             return m # _Chai Wah Wu_, Feb 21-27 2022
%Y A106303 Cf. A001591, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1), A106297 (period of Lucas 5-step sequence mod n).
%K A106303 nonn,changed
%O A106303 1,2
%A A106303 _T. D. Noe_, May 02 2005