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Mirror of the triangle A194000.
3

%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^(n-1).

%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,n-k).

%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