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%I #15 Nov 18 2018 10:03:20
%S 1,2,3,2,4,4,2,6,4,6,2,5,3,3,5,2,12,4,6,4,12,2,8,6,8,8,6,8,2,6,8,6,4,
%T 6,8,6,2,12,12,6,4,4,6,12,12,2,15,6,3,10,6,10,3,6,15,2,10,12,4,10,12,
%U 12,10,4,12,10,2,12,4,6,4,12,4,12,4,6,4,12
%N Regular triangle read by rows: T(n,k) is the rank of {A172236(k,m)} modulo n, 0 <= k <= n - 1.
%C The rank of {A172236(k,m)} modulo n is the smallest l such that n divides A172236(k,l).
%C Though {A172236(0,m)} is not defined, it can be understood as the sequence 0, 1, 0, 1, ... So the first column of each row (apart from the first one) is always 2.
%C Every row excluding the first term is antisymmetric, that is, T(n,k) = T(n,n-k) for 1 <= k <= n - 1.
%C T(n,k) is the multiplicative order of -((k + sqrt(k^2 + 4))/2)^2 modulo n*sqrt(k^2 + 4), where the multiplicative order of u modulo z is the smallest positive integer l such that (u^l - 1)/z is an algebraic integer.
%F Let p be an odd prime. (i) If k^2 + 4 is not divisible by p: if p == 1 (mod 4), then T(p^e,k) is divisible by p^(e-1)*(p - ((k^2+4)/p))/2; if p == 3 (mod 4), then T(p^e,k) is divisible by p^(e-1)*(p - ((k^2+4)/p)) but not divisible by p^(e-1)*(p - ((k^2+4)/p))/2. Here (a/p) is the Legendre symbol. (ii) If k^2 + 4 is divisible by p, then T(p^e,k) = p^e.
%F For e >= 3 and k > 0, T(2^e,k) = 3*2^(e-2) for odd k and 2^(e-v(k,2)+1) for even k, where v(k,2) is the 2-adic valuation of k.
%F If gcd(n_1,n_2) = 1, then T(n_1*n_2,k) = lcm(T(n_1,k mod n_1),T(n_2, k mod n_2)).
%F T(n,k) <= 2*n.
%e Table begins
%e 1;
%e 2, 3;
%e 2, 4, 4;
%e 2, 6, 4, 6;
%e 2, 5, 3, 3, 5;
%e 2, 12, 4, 6, 4, 12;
%e 2, 8, 6, 8, 8, 6, 8;
%e 2, 6, 8, 6, 4, 6, 8, 6;
%e 2, 12, 12, 6, 4, 4, 6, 12, 12;
%e 2, 15, 6, 3, 10, 6, 10, 3, 6, 15;
%e ...
%o (PARI) A172236(k, m) = ([k, 1; 1, 0]^m)[2, 1]
%o T(n, k) = my(i=1); while(A172236(k, i)%n!=0, i++); i
%Y Cf. A172236, A321477 (periods).
%K nonn,tabl
%O 1,2
%A _Jianing Song_, Nov 11 2018
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