%I #23 Aug 11 2019 01:39:26
%S 1,1,1,1,4,1,2,2,1,4,2,1,4,2,2,2,2,1,2,2,2,2,1,2,4,4,1,2,4,2,2,2,2,2,
%T 2,1,4,2,2,2,4,2,1,2,2,1,2,2,2,4,2,2,1,1,2,2,2,4,2,2,1,2,2,2,4,2,2,2,
%U 1,2,2,2,4,4,2,2,2,2,1,2,1,4,2,2,2,1,2
%N The number of zeros in the fundamental Pisano period of the 3-Fibonacci numbers A006190 modulo n.
%C a(n) is the multiplicative order of A006190(A322907(n)+1) modulo n.
%C a(n) has value 1, 2 or 4. This is because A006190(k,m+1)^4 == 1 (mod A006190(k,m)).
%C Conjecture: For primes p == 1, 9, 17, 25, 49, 81 (mod 104), the probability of a(p^e) taking on the value 1, 2, 4 is 1/6, 2/3, 1/6, respectively; for primes p == 29, 53, 61, 69, 77, 101 (mod 104), the probability of a(p^e) taking on the value 1, 4 is 1/2, 1/2, respectively.
%H Jianing Song, <a href="/A322906/b322906.txt">Table of n, a(n) for n = 1..1000</a>
%F For n > 2, T(n,k) = 4 iff A322907(n) is odd; 1 iff A322907(n) is even but not divisible by 4; 2 iff A322907(n) is divisible by 4.
%F For primes p == 3, 23, 27, 35, 43, 51 (mod 52), a(p^e) = 1.
%F For primes p == 5, 21, 33, 37, 41, 45 (mod 52), a(p^e) = 4.
%F For primes p == 7, 11, 15, 19, 31, 47 (mod 52), a(p^e) = 2.
%F a(13^e) = 4. a(2^e) = 1 if e = 1, 2 and 2 if e >= 3.
%F a(n) = A175182(n)/A322907(n).
%o (PARI) A006190(m) = ([3, 1; 1, 0]^m)[2, 1]
%o a(n) = my(i=1); while(A006190(i)%n!=0, i++); znorder(Mod(A006190(i+1), n))
%Y Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = k*x(n+1) + x(n). Then the periods, ranks and the ratios of the periods to the ranks modulo a given integer n are given by:
%Y k = 1: A001175 (periods), A001177 (ranks), A001176 (ratios).
%Y k = 2: A175181 (periods), A214028 (ranks), A214027 (ratios).
%Y k = 3: A175182 (periods), A322907 (ranks), this sequence (ratios).
%K nonn
%O 1,5
%A _Jianing Song_, Jan 05 2019