|
|
A106291
|
|
Period of the Lucas sequence A000032 mod n.
|
|
18
|
|
|
1, 3, 8, 6, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, 16, 30, 48, 24, 20, 84, 72, 48, 14, 24, 30, 48, 40, 36, 16, 24, 76, 18, 56, 12, 40, 48, 88, 30, 24, 48, 32, 24, 112, 60, 72, 84, 108, 72, 20, 48, 72, 42, 58, 24, 60, 30, 48, 96, 28, 120, 136, 36, 48, 48
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
FORMULA
|
Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).
|
|
EXAMPLE
|
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)
|
|
CROSSREFS
|
Cf. A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|