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A106297 Period of the Lucas 5-step sequence A074048 mod n. 4
1, 1, 104, 6, 781, 104, 2801, 12, 312, 781, 16105, 312, 30941, 2801, 81224, 24, 88741, 312, 13032, 4686, 291304, 16105, 12166, 312, 3905, 30941, 936, 16806, 70728, 81224, 190861, 48, 1674920, 88741, 2187581, 312, 1926221, 13032, 3217864, 9372, 2896405 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

This sequence differs from the corresponding Fibonacci sequence (A106303) at all n that are multiples of 2 or 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.

LINKS

Table of n, a(n) for n=1..41.

Eric Weisstein's World of Mathematics, Fibonacci n-Step

FORMULA

Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).

Conjectures: a(5^k) = 781*5^(k-1) for k > 0. a(2^k) = 3*2^(k-1) for k > 1. If a(p) != a(p^2) for prime p > 2, then a(p^k) = p^(k-1)*a(p) for k > 0. - Chai Wah Wu, Feb 25 2022

MATHEMATICA

n=5; 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, 50}]

PROG

(Python)

from itertools import count

def A106297(n):

    a = b = (5%n, 1%n, 7%n, 3%n, 15%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. A074048, A106303 (period of Fibonacci 5-step sequence mod n), A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).

Sequence in context: A088584 A238490 A097014 * A090849 A091025 A054904

Adjacent sequences:  A106294 A106295 A106296 * A106298 A106299 A106300

KEYWORD

nonn

AUTHOR

T. D. Noe, May 02 2005, Nov 19 2006

STATUS

approved

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Last modified August 13 16:07 EDT 2022. Contains 356107 sequences. (Running on oeis4.)