A130405 Triangle where g.f. of row n = Product_{i=0..n} [F(i+1) + F(i)*x] for n>=0, where F(i) = A000045(i) is the i-th Fibonacci number.

1, 1, 1, 2, 3, 1, 6, 13, 9, 2, 30, 83, 84, 37, 6, 240, 814, 1087, 716, 233, 30, 3120, 12502, 20643, 18004, 8757, 2254, 240, 65520, 303102, 596029, 646443, 417949, 161175, 34342, 3120, 2227680, 11681388, 26630128, 34495671, 27785569, 14256879

Offset: 0

Author

Paul D. Hanna, May 24 2007

Keywords

• nonn
• tabl

Comments

First column equals A003266, the product of the first n nonzero Fibonacci numbers. Main diagonal and row sums are shifted forms of A003266.

Examples

			G.f. of row n = (1)(1+x)(2+x)(3+2x)(5+3x)*...*[F(n+1) + F(n)*x]:
row 3 g.f.: (1+x)(2+x)(3+2x) = 6 + 13x + 9x^2 + 2x^3;
row 4 g.f.: (1+x)(2+x)(3+2x)(5+3x) = 30 + 83x + 84x^2 + 37x^3 + 6x^4.
Triangle begins:
1;
1, 1;
2, 3, 1;
6, 13, 9, 2;
30, 83, 84, 37, 6;
240, 814, 1087, 716, 233, 30;
3120, 12502, 20643, 18004, 8757, 2254, 240;
65520, 303102, 596029, 646443, 417949, 161175, 34342, 3120;
...
Row vectors equal the product of flipped submatrices of Pascal's triangle;
for example, row vector 3 is equal to the matrix product:
[1 0 0 0] [1 1 0 0] [1 2 1 0] [1 3 3 1] = [6 13 9 2];
[0 0 0 0] [1 0 0 0] [1 1 0 0] [1 2 1 0]
[0 0 0 0] [0 0 0 0] [1 0 0 0] [1 1 0 0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [1 0 0 0]
likewise, row 4 may be obtained by the product:
[6 13 9 2 0] * [1 4 6 4 1] = [30 83 84 37 6] .
............. [1 3 3 1 0]
............. [1 2 1 0 0]
............. [1 1 0 0 0]
............. [1 0 0 0 0]
		

Crossrefs

Cf. A130406 (column 0), A130407 (diagonal); A003266, A000045.

Programs

PARI

{T(n,k)=polcoeff(prod(i=0,n,round((fibonacci(i+1)+x*fibonacci(i)))),k)}