A106295 Period of the Lucas 4-step sequence A073817 mod n.

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

Offset: 1

Author

T. D. Noe, May 02 2005

Keywords

• nonn

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

• Chai Wah Wu, Table of n, a(n) for n = 1..1486
• 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

Crossrefs

Cf. A000078, A073817, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).

Programs

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}]

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

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