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

1,1

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.

Links

Table of n, a(n) for n=1..51.

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

Example

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

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);
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

Crossrefs

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

Sequence in context: A226330 A226329 A325548 * A251607 A219276 A183871

Adjacent sequences: A094565 A094566 A094567 * A094569 A094570 A094571

Keyword

nonn,tabl,changed

Author

Clark Kimberling, May 12 2004

Status

approved