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A106298
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Period of the Lucas 5-step sequence A074048 mod prime(n).
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5
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1, 104, 781, 2801, 16105, 30941, 88741, 13032, 12166, 70728, 190861, 1926221, 2896405, 79506, 736, 8042221, 102689, 3720, 20151120, 2863280, 546120, 39449441, 48030024, 3690720, 29509760, 104060400, 37516960, 132316201, 28231632, 6384, 86714880, 2248090, 3128
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OFFSET
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1,2
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COMMENTS
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This sequence is the same as the period of Fibonacci 5-step sequence (A106304) mod prime(n) except for n=1 and 109, which correspond to the primes 2 and 599 because 9584 is the discriminant of the characteristic polynomial x^5-x^4-x^3-x^2-x-1 and the prime factors of 9584 are 2 and 599. We have a(n) < prime(n) for the primes 2, 599 and A106281.
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LINKS
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Chai Wah Wu, Table of n, a(n) for n = 1..82
Eric Weisstein's World of Mathematics, Fibonacci n-Step
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FORMULA
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a(n) = A106297(prime(n)).
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MATHEMATICA
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n=5; 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, 40}]
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PROG
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(Python)
from itertools import count
from sympy import prime
def A106298(n):
a = b = (5%(p:=prime(n)), 1%p, 7%p, 3%p, 15%p)
s = sum(b) % p
for m in count(1):
b, s = b[1:] + (s, ), (s+s-b[0]) % p
if a == b:
return m # Chai Wah Wu, Feb 22-27 2022
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CROSSREFS
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Cf. A106281 (primes p such that x^5-x^4-x^3-x^2-x-1 mod p has 5 distinct zeros).
Sequence in context: A250677 A206021 A206014 * A337145 A230028 A340898
Adjacent sequences: A106295 A106296 A106297 * A106299 A106300 A106301
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe, May 02 2005
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EXTENSIONS
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a(31)-a(33) from Chai Wah Wu, Feb 27 2022
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STATUS
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approved
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