A094568 Triangle of binary products of Fibonacci numbers.

2, 3, 5, 8, 10, 13, 21, 24, 26, 34, 55, 63, 65, 68, 89, 144, 165, 168, 170, 178, 233, 377, 432, 440, 442, 445, 466, 610, 987, 1131, 1152, 1155, 1157, 1165, 1220, 1597, 2584, 2961, 3016, 3024, 3026, 3029, 3050, 3194, 4181, 6765, 7752, 7896, 7917, 7920, 7922

Offset: 1

Author

Clark Kimberling, May 12 2004

Keywords

• nonn
• tabl

Comments

Start with the triangle in A094566: starting with row 2, expel from each row the term that is a square of a Fibonacci number (A007598). The remaining triangle is this sequence.

In each row, the difference between neighboring terms is a Fibonacci number. For n>1, row n consists of n numbers, first F(2n) and last F(2n+1).

Central numbers: (2,10,65,442,...), essentially A064170.

Alternating row sums: 2,2,11,11,78,78,...; the sequence b=(2,11,78,...) is A094569.

Examples

			First four rows:
2
3 5
8 10 13
21 24 26 34
		

Links

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

Crossrefs

Cf. A000045, A007598, A094565, A094566, A094569.

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);
isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8)); \\ from A010056
isfib2(x) = issquare(x) && isfib(sqrtint(x));
tabl(nn) = {for (n=2, nn, if (n % 2 == 0, for (k=1, n/2, if (! isfib2(x = pef(k,n)), print1(x, ", "));); forstep (k=n/2, 1, -1, if (! isfib2(x = pof(k,n)), print1(x, ", "));), for (k=1, n\2, if (! isfib2(x = pef(k,n)), print1(x, ", "));); forstep (k=n\2+1, 1, -1, if (! isfib2(x = pof(k,n)), print1(x, ", ")););); print(););} \\ Michel Marcus, May 04 2016