A321476 Regular triangle read by rows: T(n,k) is the rank of {A172236(k,m)} modulo n, 0 <= k <= n - 1.

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, 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, 12, 10, 4, 12, 10, 2, 12, 4, 6, 4, 12, 4, 12, 4, 6, 4, 12

Offset: 1

Author

Jianing Song, Nov 11 2018

Keywords

• nonn
• tabl

Comments

The rank of {A172236(k,m)} modulo n is the smallest l such that n divides A172236(k,l).

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.

Every row excluding the first term is antisymmetric, that is, T(n,k) = T(n,n-k) for 1 <= k <= n - 1.

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.

Examples

			Table begins
  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,  6,  8,  6;
  2, 12, 12,  6,  4,  4,  6, 12, 12;
  2, 15,  6,  3, 10,  6, 10,  3,  6, 15;
  ...
		

Crossrefs

Cf. A172236, A321477 (periods).

Programs

PARI

A172236(k, m) = ([k, 1; 1, 0]^m)[2, 1]
T(n, k) = my(i=1); while(A172236(k, i)%n!=0, i++); i

Formula

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.

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.

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)).

T(n,k) <= 2*n.