Search: id:A106294 Results 1-1 of 1 results found. %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, Fibonacci n-Step %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 Search completed in 0.001 seconds