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%I #12 Mar 24 2024 14:55:11
%S 5,26,312,342,120,84,4912,6858,12166,280,61568,1368,240,162800,103822,
%T 303480,205378,226980,100254,357910,2664,998720,1157520,9320,368872,
%U 1030300,10608,1225042,2614040,13874,2048382,4530768,136,772880,3307948
%N Period of the Lucas 4-step sequence A073817 mod prime(n).
%C This sequence is the same as the period of Fibonacci 4-step sequence (A000078) mod prime(n) except for n=103, which corresponds to the prime 563 because the discriminant of the characteristic polynomial x^4-x^3-x^2-x-1 is -563. We have a(n) < prime(n) for primes 563 and A106280.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>
%F a(n) = A106295(prime(n)).
%t n=4; 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, 60}]
%Y Cf. A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1), A106280 (primes p such that x^4-x^3-x^2-x-1 mod p has 4 distinct zeros), A106295.
%K nonn
%O 1,1
%A _T. D. Noe_, May 02 2005
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