

A194001


Mirror of the triangle A194000.


3



1, 3, 2, 9, 5, 3, 24, 15, 8, 5, 64, 39, 24, 13, 8, 168, 104, 63, 39, 21, 13, 441, 272, 168, 102, 63, 34, 21, 1155, 714, 440, 272, 165, 102, 55, 34, 3025, 1869, 1155, 712, 440, 267, 165, 89, 55, 7920, 4895, 3024, 1869, 1152, 712, 432, 267, 144, 89, 20736
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OFFSET

0,2


COMMENTS

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.
Conjecture: r(n+1,x) is the product of the following two polynomials whose coefficients are Fibonacci numbers:
linear factor: F(n+2)+x*F(n+3)
other: F(2)+F(4)*x^2+F(6)*x^4+...+F(n+1)*x^(n1).
Example, for n=5:
r(6,x)=168*x^5+104*x^4+63*x^3+39^x^2+21*x+13 factors as
13+21x times 1+3x^2+8x^4.


LINKS



FORMULA

Write w(n,k) for the triangle at A194000. The triangle at A194001 is then given by w(n,nk).


EXAMPLE

First six rows:
1
3....2
9....5....3
21...13...7....4
41...28...17...9....5
71...52...35...21...11...6


MATHEMATICA

z = 11;
p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n  k), {k, 0, n}];
q[n_, x_] := p[n, x];
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x > 0;
d[n_, x_] := Sum[p1[n, k]*q[n  1  k, x], {k, 0, n  1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, 1, z}]] (* A194000 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, 1, z}]] (* A194001 *)


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



