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%I #20 Dec 31 2014 08:23:25
%S 1,1,1,1,1,1,1,1,0,2,1,1,1,1,2,0,0,2,1,3,1,1,2,0,0,2,1,1,0,2,1,1,0,0,
%T 4,0,0,0,0,4,1,3,1,1,2,2,0,2,0,0,0,0,0,4,1,1,2,0,0,2,1,3,0,4,1,3,0,4,
%U 1,1,0,2,0,0,0,4,0,0,0,0,0,0,0,8,1,1,2,0,0,2,1,1,2,0,0,2,0,0,6,0,0,0,0,0,0,0,0,6,1,3,0,4
%N Irregular triangular array read by rows: T(n,k) is the number of elements in the multiplicative group of integers modulo n that have order k, n>=1, 1<=k<=A002322(n).
%C Row sums are A000010.
%C Column 2 = A155828(n) = A060594(n) - 1.
%H Alois P. Heinz, <a href="/A252911/b252911.txt">Rows n = 1..250, flattened</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ModuloMultiplicationGroup.html">Modulo Multiplication Group.</a>
%e 1;
%e 1;
%e 1, 1;
%e 1, 1;
%e 1, 1, 0, 2;
%e 1, 1;
%e 1, 1, 2, 0, 0, 2;
%e 1, 3;
%e 1, 1, 2, 0, 0, 2;
%e 1, 1, 0, 2;
%e 1, 1, 0, 0, 4, 0, 0, 0, 0, 4;
%e 1, 3;
%e 1, 1, 2, 2, 0, 2, 0, 0, 0, 0, 0, 4;
%e 1, 1, 2, 0, 0, 2;
%e 1, 3, 0, 4;
%e T(15,2)=3 because the elements 4, 11, and 14 have order 2 in the modulo multiplication group (Z/15Z)*. We observe that 4^2, 11^2, and 14^2 are congruent to 1 mod 15.
%p with(numtheory):
%p T:= n-> `if`(n=1, 1, (p-> seq(coeff(p, x, j), j=1..degree(p)))(
%p add(`if`(igcd(n, i)>1, 0, x^order(i, n)), i=1..n-1))):
%p seq(T(n), n=1..30); # _Alois P. Heinz_, Dec 30 2014
%t Table[Table[
%t Count[Table[
%t MultiplicativeOrder[a, n], {a,
%t Select[Range[n], GCD[#, n] == 1 &]}], k], {k, 1,
%t CarmichaelLambda[n]}], {n, 1, 20}] // Grid
%Y Cf. A000010, A002322, A054522, A060594, A155828.
%K nonn,tabf
%O 1,10
%A _Geoffrey Critzer_, Dec 24 2014
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