%I #5 Mar 30 2012 18:57:39
%S 1,3,2,9,5,3,24,15,8,5,64,39,24,13,8,168,104,63,39,21,13,441,272,168,
%T 102,63,34,21,1155,714,440,272,165,102,55,34,3025,1869,1155,712,440,
%U 267,165,89,55,7920,4895,3024,1869,1152,712,432,267,144,89,20736
%N Mirror of the triangle A194000.
%C A194001 is obtained by reversing the rows of the triangle A194000.
%C Here, we extend of the conjecture begun at A194000. Suppose n is an odd positive integer and r(n+1,x) is the polynomial matched to row n+1 of A194001 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+2)+x*F(n+3)
%C other: F(2)+F(4)*x^2+F(6)*x^4+...+F(n+1)*x^(n1).
%C Example, for n=5:
%C r(6,x)=168*x^5+104*x^4+63*x^3+39^x^2+21*x+13 factors as
%C 13+21x times 1+3x^2+8x^4.
%F Write w(n,k) for the triangle at A194000. The triangle at A194001 is then given by w(n,nk).
%e First six rows:
%e 1
%e 3....2
%e 9....5....3
%e 21...13...7....4
%e 41...28...17...9....5
%e 71...52...35...21...11...6
%t z = 11;
%t p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n  k), {k, 0, n}];
%t q[n_, x_] := p[n, x];
%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}]] (* A194000 *)
%t TableForm[Table[h[n], {n, 0, z}]]
%t Flatten[Table[h[n], {n, 1, z}]] (* A194001 *)
%Y Cf. A194000, A193918.
%K nonn,tabl
%O 0,2
%A _Clark Kimberling_, Aug 11 2011
