

A106291


Period of the Lucas sequence A000032 mod n.


17



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
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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^2x1.
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
Eric Weisstein's World of Mathematics, Fibonacci nStep


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, {n1}], {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

Cf. A106273 (discriminant of the polynomial x^nx^(n1)...x1).
Cf. A000032, A001175.
Sequence in context: A019604 A336079 A214726 * A137987 A212007 A187061
Adjacent sequences: A106288 A106289 A106290 * A106292 A106293 A106294


KEYWORD

nonn


AUTHOR

T. D. Noe, May 02 2005


STATUS

approved



