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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.

%I A082691 #43 Feb 27 2025 12:45:17
%S A082691 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,
%T A082691 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,
%U A082691 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
%N 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.
%C A082691 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).
%C A082691 From _Sam Alexander_, Nov 27 2010: (Start)
%C A082691 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.
%C A082691 a(1)=1, a(2)=2. Suppose a(1), ..., a(n) have been defined, n > 1.
%C A082691 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)).
%C A082691 2. Otherwise, let (a(n+1), ..., a(2n)) be a copy of (a(1), ..., a(n)).
%C A082691 This sequence thwarts the guessing attempt, tricking the guesser into changing his mind infinitely many times as n->infinity. (End)
%C A082691 As n increases, the average value of the first n terms approaches 7/3 = 2.333... - _Maxim Skorohodov_, Dec 15 2022
%H A082691 Maxim Skorohodov, <a href="/A082691/b082691.txt">Table of n, a(n) for n = 1..10000</a>
%H A082691 Samuel Alexander, <a href="http://arxiv.org/abs/1011.6626">On Guessing Whether A Sequence Has A Certain Property</a>, arxiv:1011.6626 [math.LO], 2010-2012; <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Alexander/alex2.html">J. Int. Seq. 14 (2011) # 11.4.4</a>
%e A082691 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.
%Y A082691 Cf. A082692 (partial sums), A182659, A182660 (other sequences engineered to spite naive guessers).
%K A082691 nonn
%O A082691 1,2
%A A082691 _Benoit Cloitre_, Apr 12 2003