%I
%S 1,3,2,6,3,6,8,6,6,3,4,6,13,24,6,12,8,6,20,6,8,12,22,6,15,39,18,24,7,
%T 6,32,24,4,24,24,6,19,60,26,6,7,24,42,12,6,66,48,12,56,15,8,78,26,18,
%U 12,24,20,21,12,6,30,96,24,48,39,12,68,24,22,24,72,6
%N Entry points for the 3Fibonacci numbers A006190.
%C a(n) is the smallest k > 0 such that n divides A006190(k).
%C a(n) is also called the rank of A006190(n) modulo n.
%C For primes p == 1, 9, 17, 25, 29, 49 (mod 52), a(p) divides (p  1)/2.
%C For primes p == 3, 23, 27, 35, 43, 51 (mod 52), a(p) divides p  1, but a(p) does not divide (p  1)/2.
%C For primes p == 5, 21, 33, 37, 41, 45 (mod 52), a(p) divides (p + 1)/2.
%C For primes p == 7, 11, 15, 19, 31, 47 (mod 52), a(p) divides p + 1, but a(p) does not divide (p + 1)/2.
%C a(n) <= (12/7)*n for all n, with equality holds if and only if n = 2*7^e, e >= 1.
%H Jianing Song, <a href="/A322907/b322907.txt">Table of n, a(n) for n = 1..5000</a>
%F a(m*n) = a(m)*a(n) if gcd(m, n) = 1.
%F For odd primes p, a(p^e) = p^(e1)*a(p) if p^2 does not divide a(p). Any counterexample would be a 3WallSunSun prime.
%F a(2^e) = 3 if e = 1, 6 if e = 2 and 3*2^(e2) if e >= 3. a(13^e) = 13^e, e >= 1.
%o (PARI) A006190(m) = ([3, 1; 1, 0]^m)[2, 1]
%o a(n) = my(i=1); while(A006190(i)%n!=0, i++); i
%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), this sequence (ranks), A322906 (ratios).
%K nonn
%O 1,2
%A _Jianing Song_, Jan 05 2019
