A082691 a(1)=1, a(2)=2, then if the first 3*2^k-1 terms are a(1), a(2), ..., a(3*2^k - 1), the first 3*2^(k+1)-1 terms are a(1), a(2), ..., a(3*2^k - 1), a(1), a(2), ..., a(3*2^k - 1), a(3*2^k-1) + 1.

1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 6, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 6, 7, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3

Offset: 1

Author

Benoit Cloitre, Apr 12 2003

Keywords

• nonn

Comments

Consider the subsequence b(k) such that a(b(k))=1. Then 3k - b(k) = A063787(k+1) and b(k) = 1 + A004134(k-1).

From Sam Alexander, Nov 27 2010: (Start)

A naive way to try and guess whether a sequence is periodic, based on its first k terms (n1, ..., nk), is to look at all sequences which have period less than k, and guess "periodic" if any of them extend (n1, ..., nk), "nonperiodic" otherwise.

a(1)=1, a(2)=2. Suppose a(1), ..., a(n) have been defined, n > 1.

1. If the above guessing method guesses that (a(1), ..., a(n)) is an initial segment of a periodic sequence, then let a(n+1) be the least nonzero number not appearing in (a(1), ..., a(n)).

2. Otherwise, let (a(n+1), ..., a(2n)) be a copy of (a(1), ..., a(n)).

This sequence thwarts the guessing attempt, tricking the guesser into changing his mind infinitely many times as n->infinity. (End)

As n increases, the average value of the first n terms approaches 7/3 = 2.333... - Maxim Skorohodov, Dec 15 2022

Examples

			To construct the sequence: start with (1, 2); concatenating those 2 terms gives (1,2,1,2). Appending 3 gives the first 5 terms: (1,2,1,2,3). Concatenating those 5 terms gives (1,2,1,2,3,1,2,1,2,3). Appending 4 gives the first 11 terms: (1,2,1,2,3,1,2,1,2,3,4), etc.
		

Links

• Maxim Skorohodov, Table of n, a(n) for n = 1..10000
• Samuel Alexander, On Guessing Whether A Sequence Has A Certain Property, arxiv:1011.6626 [math.LO], 2010-2012; J. Int. Seq. 14 (2011) # 11.4.4

Crossrefs

Cf. A082692 (partial sums), A182659, A182660 (other sequences engineered to spite naive guessers).