A094566 Triangle of binary products of Fibonacci numbers.

1, 1, 2, 3, 4, 5, 8, 9, 10, 13, 21, 24, 25, 26, 34, 55, 63, 64, 65, 68, 89, 144, 165, 168, 169, 170, 178, 233, 377, 432, 440, 441, 442, 445, 466, 610, 987, 1131, 1152, 1155, 1156, 1157, 1165, 1220, 1597, 2584, 2961, 3016, 3024, 3025, 3026, 3029, 3050, 3194, 4181

Offset: 1

Author

Clark Kimberling, May 12 2004

Keywords

• nonn
• tabl

Comments

For n>1, row n consists of n numbers, first F(2n-2) and last F(2n-1).

Central numbers: (1,4,9,25,64,...), essentially A081016.

Row sums: A027991. Alternating row sums: 1,1,4,4,30,30,203,203; the sequence b=(1,4,30,203,1394,...) is A094567.

In each row, the difference between neighboring terms is a Fibonacci number.

Examples

			Rows 1 to 4:
1
1 2
3 4 5
8 9 10 13
		

Links

• Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

Crossrefs

Cf. A000045, A094565, A094567, A094568.

Programs

PARI

pef(k, n) = fibonacci(2*k)*fibonacci(2*n-2*k);
pof(k, n) = fibonacci(2*n-2*k+1)*fibonacci(2*k-1);
tabl(nn) = {for (n=1, nn, if (n==1, print1(1, ", "), if (n % 2 == 0, for (k=1, n/2, print1(pef(k,n), ", ");); forstep (k=n/2, 1, -1, print1(pof(k,n), ", "););, for (k=1, n\2, print1(pef(k,n), ", ");); forstep (k=n\2+1, 1, -1, print1(pof(k,n), ", ");););); print(););} \\ Michel Marcus, May 04 2016

Formula

Row 1 is the single number 1. For m>=1, Row 2m: F(2)F(4m-2), F(4)F(4m-4), ..., F(2m)F(2m), F(2m+1)F(2m-1), F(2m+3)F(2m-3), ..., F(4m-1)F(1) Row 2m+1: F(2)F(4m), F(4)F(4m-2), ..., F(2m+1)F(2m+1), F(2m+3)F(2m-1), F(2m+5)F(2m-3), ..., F(4m+1)F(1)