This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A321479 #9 Nov 18 2018 10:03:46
%S A321479 1,2,3,4,6,3,4,6,4,3,4,6,5,10,3,4,6,6,12,6,3,4,6,7,8,8,14,3,4,6,8,6,4,
%T A321479 6,8,3,4,6,9,12,18,9,12,18,3,4,6,10,30,6,12,6,15,10,3,4,6,11,5,10,12,
%U A321479 12,5,10,22,3,4,6,12,12,12,6,4,6,12,12,12,3
%N A321479 Regular triangle read by rows: T(n,k) is the period of {A316269(k,m)} modulo n, 0 <= k <= n - 1.
%C A321479 The period of {A316269(k,m)} modulo n is the smallest l such that A316269(k,m) == A316269(k,m+l) (mod n) for every m >= 0. Clearly, T(n,k) is divisible by A321478(n,k). Actually, the ratio is always 1 or 2.
%C A321479 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 and second ones) is always 4.
%C A321479 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 (apart from the second one) is always 6.
%C A321479 T(n,k) is the LCM of A321478(n,k) and the multiplicative order of (k + sqrt(k^2 - 4))/2 modulo n, 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 A321479 Let p be a prime >= 5. (i) If ((k+2)/p) = 1, then T(p^e,k) is divisible by p^(e-1)*(p - ((k-2)/p))/2. Here (a/p) is the Legendre symbol (ii) If ((k+2)/p) = 1, then T(p^e,k) is divisible by p^(e-1)*(p + ((k-2)/p)), but not divisible by p^(e-1)*(p + ((k-2)/p))/2. (iii) If p divides k - 2, then T(p^e,k) = p^e. (iv) If p divides k + 2, then T(p^e,k) = 2*p^e.
%F A321479 if p == 1 (mod 4), then T(p^e,k) is divisible by p^(e-1)*(p - 1), and T(p^e,k) is even; if p == 3 (mod 4), then T(p^e,k) is divisible by p^(e-1)*(p - 1) but not divisible by p^(e-1)*(p - 1)/2. Here (a/p) is the Legendre symbol. (ii) If ((k^2+4)/p) = -1, then T(p^e,k) is divisible by 2*p^(e-1)*(p + 1) but not divisible by p^(e-1)*(p + 1). (iii) If k^2 + 4 is divisible by p, then T(p^e,k) = 4*p^e.
%F A321479 For e >= 2 and k > 1, T(2^e,k) = 3*2^(e-v(k^2-1,2)+1) 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 A321479 For e > 0 and k > 1, T(3^e,k) = 4*3^(e-v(k,3)) for k divisible by 3, 2*3^(e-v(k-1,3)+1) for k == 1 (mod 3) and 3^(e-v(k+1,3)+1) for k == 2 (mod 3).
%F A321479 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 A321479 If p is an odd prime such that ((k+2)/p) = -1, then T(p^e,k)/A321478(p^e,k) = 2.
%F A321479 T(n,k) <= 3*n.
%e A321479 Table begins
%e A321479 1;
%e A321479 2, 3;
%e A321479 4, 6, 3;
%e A321479 4, 6, 4, 3;
%e A321479 4, 6, 5, 10, 3;
%e A321479 4, 6, 6, 12, 6, 3;
%e A321479 4, 6, 7, 8, 8, 14, 3;
%e A321479 4, 6, 8, 6, 4, 6, 8, 3;
%e A321479 4, 6, 9, 12, 18, 9, 12, 18, 3;
%e A321479 4, 6, 10, 30, 6, 12, 6, 15, 10, 3;
%e A321479 ...
%o A321479 (PARI) A316269(k, m) = ([k, -1; 1, 0]^m)[2, 1]
%o A321479 T(n, k) = my(i=1); while(A316269(k, i)%n!=0||(A316269(k, i+1)-1)%n!=0, i++); i
%Y A321479 Cf. A316269, A321478 (ranks).
%K A321479 nonn,tabl
%O A321479 1,2
%A A321479 _Jianing Song_, Nov 11 2018