login
A321476
Regular triangle read by rows: T(n,k) is the rank of {A172236(k,m)} modulo n, 0 <= k <= n - 1.
2
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
1,2
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.
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.
EXAMPLE
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;
...
PROG
(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
CROSSREFS
Cf. A172236, A321477 (periods).
Sequence in context: A215182 A214906 A274228 * A214567 A259363 A087437
KEYWORD
nonn,tabl
AUTHOR
Jianing Song, Nov 11 2018
STATUS
approved