A020903 Lim f(f(...f(n))) where f is the fractal sequence given by f(n)=A002260(n+1).

1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2

Offset: 1

Author

Clark Kimberling

Keywords

• nonn

Comments

Suppose that f(1), f(2), f(3),... is a fractal sequence (a sequence which contains itself as a proper subsequence, such as 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, ...; if the first occurrence of each n is deleted, the remaining sequence is identical to the original; see the Wikipedia article for details). 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) is 1 or 2 for all n, etc. Examples: A020903, A191770, A191774.

Conjecture: a(n) and a(n+1) are never both 2. - Robert Israel, Sep 03 2015

From Michel Dekking, Apr 09 2016: (Start)

Proof of the conjecture: Let f(n)=A002260(n+1)=1,2,1,2,3,1,2,3,4,... Then (f(n)) is a concatenation of ladders 1,2 followed by 1,2,3 followed by 1,2,3,4 etc. The proof is by induction. Note that the sequence (a(n)) can be seen as map from the positive integers to the positive integers. The induction starts from the observation that a(1) and a(2) are not both 2.

We use that f(k)2. Any pair (k,k+1) from a ladder has image (a(k),a(k+1)) = (a(f(k)),a(f(k+1))), which occurs either as image of two adjacent integers (j,j+1) earlier in the sequence, and so will not be equal to (2,2) by the induction hypothesis, or as image of a pair (j,1), whose image is also not equal to (2,2). The same holds for a pair consisting of the end of a ladder and the next entry. (End)

Examples

			f=(1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,...); write
n->n1->n2-> to mean n1=f(n), n2=f(n1),... Then
1->1, 2->2, 3->1, 4->2, 5->3->1, 6->1, 7->2, ...
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• C. Kimberling, Fractal sequences
• C. Kimberling, Numeration systems and fractal sequences, Acta Arithmetica 73 (1995) 103-117.
• Wikipedia, Fractal sequence

Crossrefs

Cf. A020904, A191777, A191770, A191774.

Programs

Maple

f:= proc(n) option remember; local t; t:= floor((sqrt(8*n+1)-1)/2); procname(n+1-t*(t+1)/2) end proc:
f(1):= 1: f(2):=2:
seq(f(i),i=1..1000); # Robert Israel, Sep 03 2015

Mathematica

m[n_] := Floor[(-1 + Sqrt[8 n - 7])/2];
b[n_] := n - m[n] (m[n] + 1)/2; f[n_] := b[n + 1];
Table[m[n], {n, 1, 100}]      (*A003056*)
Table[f[n], {n, 1, 100}]      (*A002260(n+1)*)
h[n_] := Nest[f, n, 40]
t = Table[h[n], {n, 1, 300}]  (* A020903 *)
Flatten[Position[t, 1]]       (* A191777 *)
Flatten[Position[t, 2]]       (* A020904 *)