cp's OEIS Frontend

A323018 T(n,k) = A321479(n,k)/A321478(n,k), 0 <= k <= n - 1.

%I A323018 #17 Jan 23 2019 09:33:05
%S A323018 1,1,1,2,2,1,2,2,1,1,2,2,1,2,1,2,2,1,2,1,1,2,2,1,2,2,2,1,2,2,1,2,1,2,
%T A323018 1,1,2,2,1,2,2,1,2,2,1,2,2,1,2,1,2,1,1,1,1,2,2,1,1,2,2,2,1,2,2,1,2,2,
%U A323018 1,2,2,2,1,2,2,2,1,1
%N A323018 T(n,k) = A321479(n,k)/A321478(n,k), 0 <= k <= n - 1.
%C A323018 For Lucas sequences, say, rows in A316269, we are mainly concerned about the periods, ranks and the ratios of the periods to the ranks of them modulo a given integer n. The period of {A316269(k,m) modulo m} is given as A321479(n,k), and the rank, which is defined as the smallest l > 0 such that n divides A316269(k,l), is given as A321478(n,k). T(n,k) is their ratio, which is the multiplicative order of A316269(k, A321478(n,k)+1) modulo n.
%C A323018 T(n,k) has value 1 or 2. This is because A316269(k,m+1)^2 == 1 (mod A316269(k,m)). See A172236 for some further properties.
%C A323018 It seems that the n-th row contains more 2's than 1's unless n is a power of 2, in which case the numbers of 1's and 2's are always the same if n >= 4.
%e A323018 Table begins
%e A323018   1,
%e A323018   1, 1,
%e A323018   2, 2, 1,
%e A323018   2, 2, 1, 1,
%e A323018   2, 2, 1, 2, 1,
%e A323018   2, 2, 1, 2, 1, 1,
%e A323018   2, 2, 1, 2, 2, 2, 1,
%e A323018   2, 2, 1, 2, 1, 2, 1, 1,
%e A323018   2, 2, 1, 2, 2, 1, 2, 2, 1,
%e A323018   2, 2, 1, 2, 1, 2, 1, 1, 1, 1,
%e A323018   ...
%o A323018 (PARI) A316269(k, m) = ([k, -1; 1, 0]^m)[2, 1]
%o A323018 T(n, k) = my(i=1); while(A316269(k, i)%n!=0, i++); znorder(Mod(A316269(k, i+1), n))
%Y A323018 Cf. A316269, A321478, A321479.
%K A323018 nonn,tabl
%O A323018 1,4
%A A323018 _Jianing Song_, Jan 07 2019