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Period of the Lucas 3-step sequence A001644 mod prime(n).
3

%I #12 Feb 16 2025 08:32:57

%S 1,13,31,48,10,168,96,360,553,140,331,469,560,308,46,52,3541,1860,

%T 1519,5113,5328,3120,287,8011,3169,680,51,1272,990,12883,5376,5720,

%U 18907,3864,7400,2850,8269,162,9296,2494,32221,10981,36673,4656,3234,198,5565

%N Period of the Lucas 3-step sequence A001644 mod prime(n).

%C This sequence differs from the corresponding Fibonacci sequence (A106302) at n=1 and 5 because these correspond to the primes 2 and 11, which are the prime factors of -44, the discriminant of the characteristic polynomial x^3-x^2-x-1. We have a(n) < prime(n) for the primes 2, 11 and A106279.

%C For a prime p, the period depends on the zeros of x^3-x^2-x-1 mod p. If there are 3 zeros, then the period is < p. If there are no zeros, then the period is p^2+p+1 or a simple fraction of p^2+p+1. Also note that the period can be prime, as for p=3, 5, 31, 59, 71, 89, 97, 157, 223. When the period is prime, the orbits have a simple structure. [From _T. D. Noe_, Sep 18 2008]

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

%F a(n) = A106293(prime(n)).

%t n=3; 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, A106279, A106302.

%K nonn,changed

%O 1,2

%A _T. D. Noe_, May 02 2005