OFFSET
1,1
COMMENTS
For n > 5, the numbers F(i)*F(j) satisfying F(n-1) <= F(i)*F(j) <= F(n) also satisfy F(n-1) < F(i)*F(j) < F(n). They are the numbers for which i + j = n + 1, where 2 < i < j, so that the number of such F(i)*F(j) is floor(n/2) - 2. The least is 3*F(n-3) and the greatest is 2*F(n-2).
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.
FORMULA
EXAMPLE
2*3 = 6, 2*5 = 10, 3*5 = 15, 2*8 = 16.
MATHEMATICA
z = 200; f[n_] := Fibonacci[n];
Take[Sort[Flatten[Table[f[m] f[n], {n, 3, z}, {m, 3, n - 1}]]], 100]
PROG
(PARI) list(lim)=my(v=List, F=vector(A130233(lim\2), k, fibonacci(k)), t); for(i=2, #F, for(j=1, i-1, t=F[i]*F[j]; if(t>lim, break); listput(v, t))); Set(v) \\ Charles R Greathouse IV, Oct 07 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 02 2016
STATUS
approved