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A106295 Period of the Lucas 4-step sequence A073817 mod n. 8
1, 5, 26, 10, 312, 130, 342, 20, 78, 1560, 120, 130, 84, 1710, 312, 40, 4912, 390, 6858, 1560, 4446, 120, 12166, 260, 1560, 420, 234, 1710, 280, 1560, 61568, 80, 1560, 24560, 17784, 390, 1368, 34290, 1092, 1560, 240, 22230, 162800, 120, 312, 60830, 103822 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
This sequence is the same as the period of Fibonacci 4-step sequence (A000078) mod n for n<563 because the discriminant of the characteristic polynomial x^4-x^3-x^2-x-1 is -563. The two sequences differ only at n that are multiples of 563.
LINKS
Marcellus E. Waddill, Some Properties of the Tetranacci Sequence Modulo m, The Fibonacci Quarterly, vol. 30, no. 3, 232-238 (1992).
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
FORMULA
Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).
a(2^k) = 5*2^(k-1) for k > 0. If a(p) != a(p^2) for p prime, then a(p^k) = p^(k-1)*a(p) for k > 0 [Waddill, 1992]. - Chai Wah Wu, Feb 25 2022
MATHEMATICA
n=4; Table[p=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}]
PROG
(Python)
from itertools import count
def A106295(n):
a = b = (4%n, 1%n, 3%n, 7%n)
s = sum(b) % n
for m in count(1):
b, s = b[1:] + (s, ), (s+s-b[0]) % n
if a == b:
return m # Chai Wah Wu, Feb 22-27 2022
CROSSREFS
Cf. A000078, A073817, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).
Sequence in context: A137113 A137115 A060063 * A057688 A259207 A300005
KEYWORD
nonn,changed
AUTHOR
T. D. Noe, May 02 2005
STATUS
approved

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Last modified March 29 08:53 EDT 2024. Contains 371268 sequences. (Running on oeis4.)