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Mirror of the triangle A193846.
4

%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