A321479 -id:A321479 - cp's OEIS frontend

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

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, 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, 1, 2, 2, 2, 1, 2, 2, 2, 1, 1

Offset: 1

Jianing Song, Jan 07 2019

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.
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.
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.

			Table begins
  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, 1, 1,
  2, 2, 1, 2, 2, 1, 2, 2, 1,
  2, 2, 1, 2, 1, 2, 1, 1, 1, 1,
  ...

Cf. A316269, A321478, A321479.

A316269(k, m) = ([k, -1; 1, 0]^m)[2, 1]
T(n, k) = my(i=1); while(A316269(k, i)%n!=0, i++); znorder(Mod(A316269(k, i+1), n))

A321478 Regular triangle read by rows: T(n,k) is the rank of {A316269(k,m)} modulo n, 0 <= k <= n - 1.

Original entry on oeis.org

1, 2, 3, 2, 3, 3, 2, 3, 4, 3, 2, 3, 5, 5, 3, 2, 3, 6, 6, 6, 3, 2, 3, 7, 4, 4, 7, 3, 2, 3, 8, 3, 4, 3, 8, 3, 2, 3, 9, 6, 9, 9, 6, 9, 3, 2, 3, 10, 15, 6, 6, 6, 15, 10, 3, 2, 3, 11, 5, 5, 6, 6, 5, 5, 11, 3, 2, 3, 12, 6, 6, 3, 4, 3, 6, 6, 12, 3

Offset: 1

Jianing Song, Nov 11 2018

The rank of {A316269(k,m)} modulo n is the smallest l such that n divides A316269(k,l).
Though {A316269(0,m)} is not defined, it can be understood as the sequence 0, 1, 0, -1, 0, 1, 0, -1, ... So the first column of each row (apart from the first one) is always 2.
Though {A316269(1,m)} is not defined, it can be understood as the sequence 0, 1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, ... So the second column of each row is always 3.
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.

			Table begins
  1;
  2,  3;
  2,  3,  3;
  2,  3,  4,  3;
  2,  3,  5,  5,  3;
  2,  3,  6,  6,  6,  3;
  2,  3,  7,  4,  4,  7,  3;
  2,  3,  8,  3,  4,  3,  8,  3;
  2,  3,  9,  6,  9,  9,  6,  9,  3;
  2,  3, 10, 15,  6,  6,  6, 15, 10,  3;
  ...

Cf. A316269, A321479 (periods).

A316269(k, m) = ([k, -1; 1, 0]^m)[2, 1]
T(n, k) = my(i=1); while(A316269(k, i)%n!=0, i++); i

Let p be a prime >= 5. (i) If k^2 - 4 is not divisible by p, then T(p^e,k) is 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 >= 2 and 1 < k < 2^e - 1, T(2^e,k) = 3*2^(e-v(k^2-1,2)) for odd k and 2^(e-v(k,2)+1) for even k, where v(k,2) is the 2-adic valuation of k.
For e > 0 and k > 1, T(3^e,k) = 2*3^(e-v(k,3)) for k divisible by 3 and 3^(e-v(k^2-1,3)+1) otherwise.
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) <= (3/2)*n.

cp's OEIS Frontend

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

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