%I #5 Mar 30 2012 18:57:39
%S 1,5,2,14,8,3,34,23,13,5,74,55,37,21,8,152,120,89,60,34,13,299,246,
%T 194,144,97,55,21,571,484,398,314,233,157,89,34,1066,924,783,644,508,
%U 377,254,144,55,1956,1725,1495,1267,1042,822,610,411,233,89,3540
%N Mirror of the triangle A194007.
%C A194008 is obtained by reversing the rows of the triangle A194007.
%F Write w(n,k) for the triangle at A194007. The triangle at A194008 is then given by w(n,n-k).
%e First six rows:
%e 1
%e 5....2
%e 14...8....3
%e 34...23...13...5
%e 74...55...37...21...8
%e 152..120..89...60...34...13
%t z = 11;
%t p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n - k), {k, 0, n}];
%t q[n_, x_] := x*q[n - 1, x] + n + 1; q[0, n_] := 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}]] (* A194007 *)
%t TableForm[Table[h[n], {n, 0, z}]]
%t Flatten[Table[h[n], {n, -1, z}]] (* A194008 *)
%Y Cf. A194007.
%K nonn,tabl
%O 0,2
%A _Clark Kimberling_, Aug 11 2011