%I #12 May 11 2014 22:38:56
%S 1,1,2,0,2,0,3,5,7,5,10,4,12,9,6,5,16,3,18,2,17,5,22,8,22,12,13,8,28,
%T 5,30,5,30,15,22,9,36,23,18,16,40,2,42,15,20,9,46,20,46,12,23,21,52,
%U 10,33,26,40,27,58,21,60,23,41,31,55,11,66,32,44,21,70,19,72,23,40,28,63,17
%N Rightmost term of each row of triangle A112599. (For n >= 2, a(n) = number of terms in the (n-1)th row of triangle A112599 which are coprime to n.)
%C It appears that there are no 0's in A112599 after the 6th row. If so, a(p) = p-1 for every prime p except 5 and 7. - _Franklin T. Adams-Watters_, Nov 15 2006
%e The 6th row of triangle A112599 is [5,0,4,0,4,0]. So a(7) is the number of these terms which are coprime to 7. Now 5, 4 and 4 are coprime to 7, but the 0's are not; so a(7) = 3.
%Y Cf. A112599, A112635.
%K nonn
%O 1,3
%A _Leroy Quet_, Dec 22 2005
%E More terms from _Franklin T. Adams-Watters_, Nov 15 2006