A288936 Fixed point of the mapping 00->0010, 01->011, 10->011, starting with 00.

0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1

Offset

1

Comments

Conjecture: the number of letters (0's and 1's) in the n-th iterate of the mapping is given by A000124.

From Michel Dekking, Oct 16 2022: (Start)

Proof of the conjecture: let N0(n) be the number of 0's, and N1(n) be the number of 1's in the n-th iterate of the StringReplace mapping defining (a(n)).

CLAIM: N0(n) = n+2, N1(n) = n(n+1)/2 for all n=1,2,...

The proof is by induction. We have N0(1) =3, N1(1) = 1. The crucial observation is that 00 only occurs at position 1 and position 4, and that at position 4 it always is preceded by 1. Moreover, each 0 (except the first one), combines either as 01 or as 10, which both give 011 in the next iterate.

So N0(n+1) = N0(n) + 1 = n+3, and

N1(n+1) = N1(n) + N0(n) - 1 = n(n+1)/2 + n+1 = (n+1)(n+2)/2.

This proves the conjecture: N0(n)+N1(n) = (n^2 +3n+4)/2 = A000124(n+1) for n=1,2,.... (End)

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

Formula

a(n) = A023532(n-6), for n >= 7. - Alan Michael Gómez Calderón, Aug 01 2024

Example

The first six iterates of the mapping:
  00
  0010
  0010011
  00100110111
  0010011011011111
  0010011011011101111111

Mathematica

z = 11; (* number of iterates *)
s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"00" -> "0010", "01" -> "011", "10" -> "011"}]
TableForm[Table[w[n], {n, 0, 10}]]
st = ToCharacterCode[w[z]] - 48   (* A288936 *)
Flatten[Position[st, 0]]  (* A288937 *)
Flatten[Position[st, 1]]  (* A288938 *)
Table[StringLength[w[n]], {n, 0, 20}] (* A000124 conjectured *)

Crossrefs

Cf. A000124, A023532, A288937, A288938.

Sequence in context: A289057 A106138 A273129 * A064990 A284388 A289174

Adjacent sequences: A288933 A288934 A288935 * A288937 A288938 A288939

Keyword

nonn,easy

Author

Clark Kimberling, Jun 28 2017

Status

approved