OFFSET
1,2
COMMENTS
This sequence differs from the Fibonacci periods (A001175) only when n is a multiple of 5, which can be traced to 5 being the discriminant of the characteristic polynomial x^2-x-1.
This sequence coincides with the Fibonacci periods (A001175) if n is a multiple of 5^j and the following conditions apply: n contains at least one prime factor of the form p = 10*k+1 (A030430) which occurs in Fibonacci(m) or Lucas(m) as prime factor, where m must be the smallest possible index containing p and a factor 5^i and j <= i. If n contains several prime factors from A030430 that satisfy the above conditions, the largest applicable i is decisive. - Klaus Purath, Apr 26 2019
REFERENCES
S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989. See p. 89. - From N. J. A. Sloane, Feb 20 2013
LINKS
G. C. Greubel and D. Turner, Table of n, a(n) for n = 1..10000
Brennan Benfield and Oliver Lippard, Fixed points of K-Fibonacci sequences, arXiv:2404.08194 [math.NT], 2024. See p. 11.
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)).
EXAMPLE
From Klaus Purath, Jul 10 2019: (Start)
n = 3*5*31 = 465, j = 1; L(15) is the smallest Lucas number with prime factor 31; 15 = 3*5, i = 1 = j. Hence Lucas period (mod 465) = Fibonacci period (mod 465) = 120, but if n = 3*5^2*31 = 2325, j = 2 > i. Hence Lucas period (mod 2325) = 120 < Fibonacci period (mod 2325) = 600.
n = 5*701 = 3505, j = 1; F(175) is the smallest Fibonacci number with prime factor 701; 175 = 7*5^2, i = 2 > j. Therefore Lucas period (mod 3505) = Fibonacci period (mod 3505) = 700, but if n = 5^3*701 = 87625, j = 3 > i. Therefore Lucas period (mod 87625) = 700 < Fibonacci period (mod 87625) = 3500.
n = 5^2*11*101 = 27775, j =2; L(5) is the smallest Lucas number with prime factor 11, i = 1; L(25) = is the smallest Lucas number with prime factor 101; 25 = 5^2, i = 2 ( decisive); j = i. Hence Lucas period (mod 27775) = Fibonacci period (mod 27775) = 100, but if n = 5^3*11*101 = 138875, j = 3 > i. Hence Lucas period (mod 138875) = 100 < Fibonacci period (mod 138875) = 500. (End)
MATHEMATICA
n=2; 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, 70}]
PROG
(Sage)
def a(n): return BinaryRecurrenceSequence(1, 1, 2, 1).period(n)
[a(n) for n in (1..100)] # G. C. Greubel, Apr 27 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, May 02 2005
STATUS
approved