%I #11 Apr 09 2014 10:16:58
%S 1,1,1,2,2,2,0,0,3,1,2,2,1,1,4,4,2,0,0,0,4,1,1,3,2,3,2,4,4,2,0,2,0,0,
%T 0,6,2,2,0,2,0,4,2,2,2,4,0,0,0,0,5,1,2,2,2,2,2,2,6,6,2,0,0,2,0,0,6,2,
%U 2,2,4,0,4,0,4,2,2,0,2,0,0,0,0,0,8,2,2,3,0,3,0,4,0,0,0,4,0,0,1,0,0,6,2,2,0
%N Array where a(1,1)=1 and m-th term of n-th row is number of terms of (n-1)th row which are coprime to the m-th divisor of n.
%C Number of terms in row n is A000005(n).
%e The irregular array's 5th row is [3,1]. The divisors of 6 are 1, 2, 3 and 6. In the 5th row there are 2 terms coprime to 1, 2 terms coprime to 2, 1 term coprime to 3 and 1 term coprime to 6. So the 6th row of the array is [2,2,1,1].
%e Array starts
%e 1;
%e 1,1;
%e 2,2;
%e 2,0,0;
%e 3,1;
%e 2,2,1,1;
%e 4,4;
%e 2,0,0,0;
%Y Cf. A000005, A112791.
%K nonn,tabf
%O 1,4
%A _Leroy Quet_, Dec 31 2005
%E More terms from _R. J. Mathar_, Aug 29 2007