A272947 Number of factors Fibonacci(i) > 1 of A160009(n+1).

1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 3, 1, 2, 2, 2, 3, 3, 1, 2, 2, 2, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 3, 3, 3, 1, 4, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 1, 4, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 1, 4, 4, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 1

Author

Clark Kimberling, May 13 2016

Keywords

• nonn
• easy

Examples

			A160009(15) = 30 = 2*3*5, so that a(15) = 3.
		

Links

• MathOverflow, Distinctness of products of Fibonacci numbers

Crossrefs

Cf. A000045, A160009, A272948, A271354, A272949, A273950.

Programs

Mathematica

s = {1}; nn = 60; f = Fibonacci[2 + Range[nn]]; Do[s = Union[s, Select[s*f[[i]], # <= f[[nn]] &]], {i, nn}]; s =  Prepend[s, 0]; Take[s, 100]  (* A160009 *)
isFibonacciQ[n_] := Apply[Or, Map[IntegerQ, Sqrt[{# + 4, # - 4} &[5 n^2]]]];
ans = Join[{{0}}, {{1}}, Table[#[[Flatten[Position[Map[Apply[Times, #] &, #], s[[n]]]][[1]]]] &[Rest[Subsets[Rest[Map[#[[1]] &, Select[Map[{#, isFibonacciQ[#]} &, Divisors[s[[n]]]], #[[2]] &]]]]]], {n, 3, 500}]]
Map[Length, ans] (* A272947 *)
Flatten[Position[Map[Length, ans], 1]]  (* A272948 *)
Map[Apply[Times, #] &, Select[ans, Length[#] == 1 &]]  (* A000045 *)
Map[Apply[Times, #] &, Select[ans, Length[#] == 2 &]]  (* A271354 *)
Map[Apply[Times, #] &, Select[ans, Length[#] == 3 &]]  (* A272949 *)
Map[Apply[Times, #] &, Select[ans, Length[#] == 4 &]]  (* A272950 *)
(* Peter J. C. Moses, May 11 2016 *)