%I
%S 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,
%T 84,72,48,14,24,30,48,40,36,16,24,76,18,56,12,40,48,88,30,24,48,32,24,
%U 112,60,72,84,108,72,20,48,72,42,58,24,60,30,48,96,28,120,136,36,48,48
%N Period of the Lucas sequence A000032 mod n.
%C 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.
%C 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
%D 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
%H G. C. Greubel and D. Turner, <a href="/A106291/b106291.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonaccinStepNumber.html">Fibonacci nStep</a>
%F Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).
%e From _Klaus Purath_, Jul 10 2019: (Start)
%e 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.
%e 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.
%e 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)
%t 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}]
%o (Sage)
%o def a(n): return BinaryRecurrenceSequence(1, 1, 2, 1).period(n)
%o [a(n) for n in (1..100)] # _G. C. Greubel_, Apr 27 2019
%Y Cf. A106273 (discriminant of the polynomial x^nx^(n1)...x1).
%Y Cf. A000032, A001175.
%K nonn
%O 1,2
%A _T. D. Noe_, May 02 2005
