%I #5 Mar 30 2012 18:57:39
%S 1,3,2,6,8,3,11,23,18,5,19,55,66,37,8,32,120,196,167,73,13,53,246,511,
%T 587,388,139,21,87,484,1225,1777,1578,853,259,34,142,924,2765,4857,
%U 5428,3933,1799,474,55,231,1725,5969,12333,16642,15147,9275,3678
%N Mirror of the triangle A193997.
%C A193998 is obtained by reversing the rows of the triangle A193997.
%F Write w(n,k) for the triangle at A193997. The triangle at A193998 is then given by w(n,n-k).
%e First six rows:
%e 1
%e 3....2
%e 6....8.....3
%e 11...23....18....5
%e 19...55....66....37....8
%e 32...120...196...167...73...13
%t z = 11;
%t p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n - k), {k, 0, 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}]] (* A193997 *)
%t TableForm[Table[h[n], {n, 0, z}]]
%t Flatten[Table[h[n], {n, -1, z}]] (* A193998 *)
%Y Cf. A193997.
%K nonn,tabl
%O 0,2
%A _Clark Kimberling_, Aug 11 2011