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%I #5 Mar 30 2012 18:57:38
%S 2,8,4,26,28,8,80,136,80,16,242,568,512,208,32,728,2188,2672,1648,512,
%T 64,2186,8020,12392,10288,4832,1216,128,6560,28432,53216,55648,35072,
%U 13312,2816,256,19682,98416,216512,273376,216512,110080,35072
%N Mirror of the triangle A193846.
%C A193847 is obtained by reversing the rows of the triangle A193846.
%F Write w(n,k) for the triangle at A193846. The triangle at A193847 is then given by w(n,n-k).
%e First six rows:
%e 2
%e 8.....4
%e 26....28....8
%e 80....136...80....16
%e 242...568...512...208...32
%e 728...2188..2672..1648..512..64
%t p[n_, x_] := (x + 2)^n;
%t q[n_, x_] := (x + 1)^n
%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}]] (* A193846 *)
%t TableForm[Table[h[n], {n, 0, z}]]
%t Flatten[Table[h[n], {n, -1, z}]] (* A193847 *)
%t TableForm[Table[Reverse[h[n]/2], {n, 0, z}]]
%t Flatten[Table[Reverse[h[n]]/2, {n, -1, z}]] (* A193848 *)
%t TableForm[Table[h[n]/2, {n, 0, z}]]
%t Flatten[Table[h[n]/2, {n, -1, z}]] (* A193849 *)
%Y Cf. A193846, A193849.
%K nonn,tabl
%O 0,1
%A _Clark Kimberling_, Aug 07 2011
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