A191770 Lim f(f(...f(n)...)) where f(n) is the fractal sequence A022446.

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

Offset: 1

Author

Clark Kimberling, Jun 16 2011

Keywords

• nonn

Comments

Suppose that f(1), f(2), ... is a fractal sequence (such as 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, ..., which contains itself as a proper subsequence - if the first occurrence of each n is deleted, the remaining sequence is identical to the original; see the Wikipedia article for a rigorous definition). Then for each n>=1, the limit L(n) of composites f(f(f...f(n)...)) exists and is one of the numbers in the set {k : f(k)=k}. If f(2)>2, then L(n)=1 for all n; if f(2)=2 and f(3)>3, then L(n) equals 1 or 2 for all n. Examples: A020903, A191770, A191774.

Examples

			Write the counting numbers and A022446 like this:
1..2..3..4..5..6..7..8..9..10..11..12..13..14..15..
1..2..3..1..4..2..5..8..1..4...6...2...7...5...3...
It is then easy to check composites:
1->1, 2->2, 3->3, 4->1, 5->4->1, 6->2, 7->5->4->1,...
		

Links

• Wikipedia, Fractal sequence

Crossrefs

Cf. A020903, A191770, A191771, A191772, A191773, A191774.

Programs

Mathematica

g[n_] :=  Length[Select[Table[FixedPoint[i + PrimePi[#] + 1 &, i + PrimePi[i] + 1], {i, n}], # <= n &]];
f[n_] := PrimePi[NestWhile[g, n, ! PrimeQ[#] && # != 1 &]] + 1;
Array[f, 80]             (* A022446 *)
h[n_] := Nest[f, n, 40]; t = Table[h[n], {n, 1, 300}]  (* A191770 *)
Flatten[Position[t, 1]]  (* A191771 *)
Flatten[Position[t, 2]]  (* A191772 *)
Flatten[Position[t, 3]]  (* A191773 *)