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Mirror of the triangle A193850.
4

%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