%I #6 Mar 30 2012 18:57:39
%S 1,1,1,3,2,1,9,6,3,2,24,15,9,5,3,64,40,24,15,8,5,168,104,64,39,24,13,
%T 8,441,273,168,104,63,39,21,13,1155,714,441,272,168,102,63,34,21,3025,
%U 1870,1155,714,440,272,165,102,55,34,7920,4895,3025,1869,1155
%N Mirror of the triangle A193917.
%C A193918 is obtained by reversing the rows of the triangle A193917.
%C Here, we extend of the conjecture begun at A193917. Suppose n is an even positive integer and r(n+1,x) is the polynomial matched to row n+1 of A193918 as in the Mathematica program, where the first row is counted as row 0.
%C Conjecture: r(n+1,x) is the product of the following two polynomials whose coefficients are Fibonacci numbers:
%C linear factor: F(n+1)+x*F(n+2)
%C other factor: F(n+2)+F(n)*x^2+F(n2)*x^4+...+F(2)*x^n.
%C Example, for n=4:
%C r(5,x)=64*x^5+40*x^4+24*x^3+15^x^2+8*x+5 factors as
%C 8x+5 times 8x^4+3x^2+1.
%F Write w(n,k) for the triangle at A193917. The triangle at A193918 is then given by w(n,nk).
%e First six rows:
%e 1
%e 1....1
%e 3....2....1
%e 9....6....3....2
%e 24...15...9....5....3
%e 64...40...24...15...8...5
%t z = 12;
%t p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n  k), {k, 0, n}];
%t q[n_, x_] := p[n, x];
%t t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x > 0;
%t w[n_, x_] := Sum[t[n, k]*q[n + 1  k, x], {k, 0, n}]; w[1, x_] := 1
%t g[n_] := CoefficientList[w[n, x], {x}]
%t TableForm[Table[Reverse[g[n]], {n, 1, z}]]
%t Flatten[Table[Reverse[g[n]], {n, 1, z}]] (* A193917 *)
%t TableForm[Table[g[n], {n, 1, z}]]
%t Flatten[Table[g[n], {n, 1, z}]] (* A193918 *)
%Y Cf. A193917.
%K nonn,tabl
%O 0,4
%A _Clark Kimberling_, Aug 09 2011
