%I A106294
%S A106294 1,13,31,48,10,168,96,360,553,140,331,469,560,308,46,52,3541,1860,1519,
%T A106294 5113,5328,3120,287,8011,3169,680,51,1272,990,12883,5376,5720,18907,
%U A106294 3864,7400,2850,8269,162,9296,2494,32221,10981,36673,4656,3234,198,5565
%N A106294 Period of the Lucas 3-step sequence A001644 mod prime(n).
%C A106294 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 A106294 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 (noe(AT)sspectra.com), Sep 18 2008]
%H A106294 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
Fibonaccin-Step.html">Fibonacci n-Step</a>
%t A106294 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 A106294 Cf. A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1), A106279
(primes p such that x^3-x^2-x-1 mod p has 3 distinct zeros), A106302
(period of Fibonacci 3-step sequence mod prime(n)).
%Y A106294 Sequence in context: A097443 A100589 A111489 this_sequence A101649 A063305
A166143
%Y A106294 Adjacent sequences: A106291 A106292 A106293 this_sequence A106295 A106296
A106297
%K A106294 nonn
%O A106294 1,2
%A A106294 T. D. Noe (noe(AT)sspectra.com), May 02 2005
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