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A094568 Triangle of binary products of Fibonacci numbers. 3
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 (list; table; graph; refs; listen; history; text; internal format)
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
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
Sequence in context: A226330 A226329 A325548 * A251607 A219276 A183871
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, May 12 2004
STATUS
approved

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Last modified July 15 21:59 EDT 2024. Contains 374334 sequences. (Running on oeis4.)