%I #5 Mar 30 2012 18:57:38
%S 2,8,4,26,20,8,80,72,48,16,242,232,192,112,32,728,716,656,496,256,64,
%T 2186,2172,2088,1808,1248,576,128,6560,6544,6432,5984,4864,3072,1280,
%U 256,19682,19664,19520,18848,16832,12800,7424,2816,512,59048,59028
%N Mirror of the triangle A193850.
%C A193851 is obtained by reversing the rows of the triangle A193850.
%F Write w(n,k) for the triangle at A193850. The triangle at A193851 is then given by w(n,n-k).
%e First six rows:
%e 2
%e 8....4
%e 26...20....8
%e 80...72...40..16
%e 242...232...192...112...32
%e 728...716...656...496..256..64
%t z = 10;
%t p[n_, x_] := (x + 2)^n;
%t q[0, x_] := 1; q[n_, x_] := x*q[n - 1, x] + 1;
%t p1[n_, k_] := Coefficient[p[n, x], x^k];
%t p1[n_, 0] := p[n, x] /. x -> 0;
%t d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
%t h[n_] := CoefficientList[d[n, x], {x}]
%t TableForm[Table[Reverse[h[n]], {n, 0, z}]]
%t Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A193850 *)
%t TableForm[Table[h[n], {n, 0, z}]]
%t Flatten[Table[h[n], {n, -1, z}]] (* A193851 *)
%t TableForm[Table[Reverse[h[n]/2], {n, 0, z}]]
%t Flatten[Table[Reverse[h[n]]/2, {n, -1, z}]] (* A193852 *)
%t TableForm[Table[h[n]/2, {n, 0, z}]]
%t Flatten[Table[h[n]/2, {n, -1, z}]] (* A193853 *)
%Y Cf. A193850, A193853.
%K nonn,tabl
%O 0,1
%A _Clark Kimberling_, Aug 07 2011