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A106294
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Period of the Lucas 3-step sequence A001644 mod prime(n).
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3
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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
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1,2
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COMMENTS
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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.
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]
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LINKS
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FORMULA
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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nonn,changed
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AUTHOR
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STATUS
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approved
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