

A193918


Mirror of the triangle A193917.


4



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, 8, 441, 273, 168, 104, 63, 39, 21, 13, 1155, 714, 441, 272, 168, 102, 63, 34, 21, 3025, 1870, 1155, 714, 440, 272, 165, 102, 55, 34, 7920, 4895, 3025, 1869, 1155
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OFFSET

0,4


COMMENTS

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.
Conjecture: r(n+1,x) is the product of the following two polynomials whose coefficients are Fibonacci numbers:
linear factor: F(n+1)+x*F(n+2)
other factor: F(n+2)+F(n)*x^2+F(n2)*x^4+...+F(2)*x^n.
Example, for n=4:
r(5,x)=64*x^5+40*x^4+24*x^3+15^x^2+8*x+5 factors as
8x+5 times 8x^4+3x^2+1.


LINKS



FORMULA

Write w(n,k) for the triangle at A193917. The triangle at A193918 is then given by w(n,nk).


EXAMPLE

First six rows:
1
1....1
3....2....1
9....6....3....2
24...15...9....5....3
64...40...24...15...8...5


MATHEMATICA

z = 12;
p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n  k), {k, 0, n}];
q[n_, x_] := p[n, x];
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x > 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1  k, x], {k, 0, n}]; w[1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, 1, z}]]
Flatten[Table[Reverse[g[n]], {n, 1, z}]] (* A193917 *)
TableForm[Table[g[n], {n, 1, z}]]
Flatten[Table[g[n], {n, 1, z}]] (* A193918 *)


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



