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A193918 Mirror of the triangle A193917. 4

%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(n-2)*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,n-k).

%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

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