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Period of the Fibonacci 5-step sequence A001591 mod n.
9

%I #36 Feb 16 2025 08:32:57

%S 1,6,104,12,781,312,2801,24,312,4686,16105,312,30941,16806,81224,48,

%T 88741,312,13032,9372,291304,96630,12166,312,3905,185646,936,33612,

%U 70728,243672,190861,96,1674920,532446,2187581,312,1926221,13032

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

%C This sequence differs from the corresponding Lucas sequence (A106297) at all n that are multiples 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.

%H Chai Wah Wu, <a href="/A106303/b106303.txt">Table of n, a(n) for n = 1..388</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.

%F Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).

%F 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 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 (Python)

%o from itertools import count

%o def A106303(n):

%o a = b = (0,)*4+(1 % n,)

%o s = 1 % n

%o for m in count(1):

%o b, s = b[1:] + (s,), (s+s-b[0]) % n

%o if a == b:

%o return m # _Chai Wah Wu_, Feb 21-27 2022

%Y Cf. A001591, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1), A106297 (period of Lucas 5-step sequence mod n).

%K nonn,changed

%O 1,2

%A _T. D. Noe_, May 02 2005