OFFSET
1,3
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.
LINKS
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.
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)
EXAMPLE
Rows 1 to 4:
1
1 2
3 4 5
8 9 10 13
PROG
(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
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, May 12 2004
STATUS
approved