%I
%S 1,1,1,1,2,1,1,3,3,1,1,256,384,256,1,1,5,640,640,5,1,1,1146617856,
%T 2866544640,244611809280,2866544640,1146617856,1,1,7,4013162496,
%U 6688604160,6688604160,4013162496,7,1,1,35184372088832,123145302310912
%N An analog of Pascal's triangle based on A129454. T(n,k)= A129454(n)/(A129454(nk)*A129454(k)).
%C It appears that the T(n,k) are always integers. This would follow from the conjectured prime factorization given in A129454. Calculation suggests that the binomial coefficients C(n,k) divide T(n,k) and that T(n,k)/C(n,k) are perfect sixth powers.
%F T(n,k) = product_{h=1..n}product_{i=1..n}product_{j=1..n} gcd(h,i,j)/(product_{h=1..nk}product_{i=1..nk}product_{j=1..nk} gcd(h,i,j)*product_{h=1..k}product_{i=1..k}product_{j=1..k} gcd(h,i,j)).
%e Triangle starts:
%e 1
%e 1 1
%e 1 2 1
%e 1 3 3 1
%e 1 256 384 256 1
%Y Cf. A007318, A092287, A129453, A129454.
%K nonn,tabl
%O 0,5
%A _Peter Bala_, Apr 16 2007
