%I
%S 1,1,2,3,6,6,6,12,15,12,10,20,27,28,20,15,30,42,48,45,30,21,42,60,72,
%T 75,66,42,28,56,81,100,110,108,91,56,36,72,105,132,150,156,147,120,72,
%U 45,90,132,168,195,210,210,192,153,90,55,110,162,208,245,270,280
%N Mirror of the triangle A193895.
%C A193896 is obtained by reversing the rows of the triangle A193895.
%F Write w(n,k) for the triangle at A193895. The triangle at A193896 is then given by w(n,nk).
%e First six rows:
%e 1
%e 1....2
%e 3....6....6
%e 6....12...15...12
%e 10...20...27...28...20
%e 15...30...42...48...45...30
%t z = 9;
%t p[n_, x_] := x*p[n  1, x] + n + 1 (* #6 *) ; p[0, x_] := 1;
%t q[n_, x_] := (n + 1)*x^n + q[n  1, x] (* #7 *); q[0, x_] := 1;
%t t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x > 0;
%t w[n_, x_] := Sum[t[n, k]*q[n + 1  k, x], {k, 0, n}]; w[1, x_] := 1
%t g[n_] := CoefficientList[w[n, x], {x}]
%t TableForm[Table[Reverse[g[n]], {n, 1, z}]]
%t Flatten[Table[Reverse[g[n]], {n, 1, z}]] (* A193895 *)
%t TableForm[Table[g[n], {n, 1, z}]]
%t Flatten[Table[g[n], {n, 1, z}]] (* A193896 *)
%Y Cf. A193895.
%K nonn,tabl
%O 0,3
%A _Clark Kimberling_, Aug 08 2011
