OFFSET
1,2
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.
LINKS
Wikipedia, Fractal sequence
EXAMPLE
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,...
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 *)
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jun 16 2011
STATUS
approved