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A216320 Irregular triangle: row n lists the Modd n order of the odd members of the reduced smallest nonnegative residue class modulo n.

%I A216320 #20 May 15 2025 05:28:00
%S A216320 1,1,1,1,2,1,2,1,2,1,3,3,1,4,4,2,1,3,3,1,4,4,2,1,5,5,5,5,1,2,2,2,1,3,
%T A216320 2,6,3,6,1,3,6,3,6,2,1,4,2,4,1,8,8,4,4,8,8,2,1,8,8,8,4,8,2,4,1,6,6,3,
%U A216320 3,2,1,9,9,3,9,3,9,9,9,1,4,4,2,2,4,4,2,1,3,6,2,3,6
%N A216320 Irregular triangle: row n lists the Modd n order of the odd members of the reduced smallest nonnegative residue class modulo n.
%C A216320 The length of row n is delta(n):=A055034(n).
%C A216320 For the multiplicative group Modd n see a comment on A203571, and also on A216319.
%C A216320 A216319(n,k)^a(n,k) == +1 (Modd n), n >= 1.
%C A216320 If the Modd n order of an (odd) element from row n of A216319 is delta(n) (the row length) then this element is a primitive root Modd n. There is no primitive root Modd n if no such element of order delta(n) exists. For example, n = 12, 20, ... (see A206552 for more of these n values). There are phi(delta(n)) = A216321(n) such primitive roots Modd n if there exists one, where phi=A000010 (Euler's totient). The multiplicative group Modd n is cyclic if and only if there exists a primitive root Modd n. The multiplicative group Modd n is isomorphic to the Galois group G(Q(rho(n))/Q) with the algebraic number rho(n) := 2*cos(Pi/n), n>=1.
%F A216320 a(n,k) = order of A216319(n,k) Modd n, n>=1, k=1, 2, ..., A055034(n). This means: A216319(n,k)^a(n,k) == +1 (Modd n), n>=1, and a(n,k) is the smallest positive integer exponent satisfying this congruence. For Modd n see a comment on A203571.
%e A216320 The table a(n,k) begins:
%e A216320   n\k 1  2  3  4  5  6  7  8  9 ...
%e A216320   1   1
%e A216320   2   1
%e A216320   3   1
%e A216320   4   1  2
%e A216320   5   1  2
%e A216320   6   1  2
%e A216320   7   1  3  3
%e A216320   8   1  4  4  2
%e A216320   9   1  3  3
%e A216320   10  1  4  4  2
%e A216320   11  1  5  5  5  5
%e A216320   12  1  2  2  2
%e A216320   13  1  3  2  6  3  6
%e A216320   14  1  3  6  3  6  2
%e A216320   15  1  4  2  4
%e A216320   16  1  8  8  4  4  8  8  2
%e A216320   17  1  8  8  8  4  8  2  4
%e A216320   18  1  6  6  3  3  2
%e A216320   19  1  9  9  3  9  3  9  9  9
%e A216320   20  1  4  4  2  2  4  4  2
%e A216320   ...
%e A216320 a(7,2) = 3 because A216319(7,2) = 3 and 3^1 == 3 (Modd 7);
%e A216320   3^2 = 9 == 5 (Modd 7) because floor(9/7)= 1 which is odd, therefore 9 (Modd 7) = -9 (mod 7) = 5; 3^3 == 5*3 (Modd n)
%e A216320   = +1 because floor(15/7)=2 which is even, therefore 15 (Modd 7) = 15 (modd 7) = +1.
%e A216320 Row n=12 is the first row without an order = delta(n) (row length), in this case 4. Therefore there is no primitive root Modd 12, and the multiplicative group Modd 12 is non-cyclic.
%e A216320   Its cycle structure is [[5,1],[7,1],[11,1]] which is the group Z_2 x Z_2 (the Klein 4-group).
%Y A216320 Cf. A203571, A216321.
%K A216320 nonn,easy,tabf
%O A216320 1,5
%A A216320 _Wolfdieter Lang_, Sep 21 2012