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Period of the Fibonacci 5-step sequence A001591 mod prime(n).
4

%I #28 Feb 16 2025 08:32:57

%S 6,104,781,2801,16105,30941,88741,13032,12166,70728,190861,1926221,

%T 2896405,79506,736,8042221,102689,3720,20151120,2863280,546120,

%U 39449441,48030024,3690720,29509760,104060400,37516960,132316201,28231632,6384,86714880,2248090,3128

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

%C This sequence is the same as the period of the Lucas 5-step sequence (A106298) mod prime(n) except for n=1 and 109, which correspond to the primes 2 and 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. We have a(n) < prime(n) for the primes in A106281.

%H Chai Wah Wu, <a href="/A106304/b106304.txt">Table of n, a(n) for n = 1..86</a>

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

%F a(n) = A106303(prime(n)).

%t n=5; Table[p=Prime[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, 40}]

%o (Python)

%o from itertools import count

%o from sympy import prime

%o def A106304(n):

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

%o s = 1 % p

%o for m in count(1):

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

%o if a == b:

%o return m # _Chai Wah Wu_, Feb 22-27 2022

%Y Cf. A106281 (primes p such that x^5-x^4-x^3-x^2-x-1 mod p has 5 distinct zeros).

%K nonn,changed

%O 1,1

%A _T. D. Noe_, May 02 2005, Nov 19 2006

%E a(31)-a(33) from _Chai Wah Wu_, Feb 27 2022