cp's OEIS Frontend

A193918 Mirror of the triangle A193917.

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