

A020903


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


5



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
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OFFSET

1,2


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)<k for all 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)


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) 103117.
Wikipedia, Fractal sequence


EXAMPLE

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, ...


MAPLE

f:= proc(n) option remember; local t; t:= floor((sqrt(8*n+1)1)/2); procname(n+1t*(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 *)


CROSSREFS

Cf. A020904, A191777, A191770, A191774.
Sequence in context: A006337 A214848 A006338 * A133083 A083921 A105496
Adjacent sequences: A020900 A020901 A020902 * A020904 A020905 A020906


KEYWORD

nonn


AUTHOR

Clark Kimberling


STATUS

approved



