

A106294


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


3



1, 13, 31, 48, 10, 168, 96, 360, 553, 140, 331, 469, 560, 308, 46, 52, 3541, 1860, 1519, 5113, 5328, 3120, 287, 8011, 3169, 680, 51, 1272, 990, 12883, 5376, 5720, 18907, 3864, 7400, 2850, 8269, 162, 9296, 2494, 32221, 10981, 36673, 4656, 3234, 198, 5565
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OFFSET

1,2


COMMENTS

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^3x^2x1. We have a(n) < prime(n) for the primes 2, 11 and A106279.
For a prime p, the period depends on the zeros of x^3x^2x1 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]


LINKS

Table of n, a(n) for n=1..47.
Eric Weisstein's World of Mathematics, Fibonacci nStep


FORMULA

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


MATHEMATICA

n=3; Table[p=Prime[i]; a=Join[Table[ 1, {n1}], {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}]


CROSSREFS

Cf. A106273, A106279, A106302.
Sequence in context: A080387 A100589 A111489 * A101649 A063305 A301622
Adjacent sequences: A106291 A106292 A106293 * A106295 A106296 A106297


KEYWORD

nonn


AUTHOR

T. D. Noe, May 02 2005


STATUS

approved



