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A225331 A continuous "look-and-repeat" sequence (method 2).

%I A225331 #26 Apr 18 2021 02:15:40
%S A225331 1,1,1,1,3,3,1,2,2,3,1,1,1,2,2,2,1,1,3,3,3,1,3,3,2,2,2,1,3,3,3,1,1,1,
%T A225331 2,2,3,3,3,2,1,1,1,3,3,3,3,3,1,2,2,2,3,3,3,1,1,2,3,3,1,5,5,3,1,1,1,3,
%U A225331 3,2,3,3
%N A225331 A continuous "look-and-repeat" sequence (method 2).
%C A225331 A variant of the 'look-and-repeat' sequence A225329, without run cut-off. It describes at each step the preceding digits by repeating the frequency number.
%C A225331 The sequence is determined by triples of digits. The first two terms of a triple are the repeated frequency and the last term is the digit.
%C A225331 There are different optional rules to build such a sequence. This method 2 never considers twice the already said digits.
%C A225331 With this rule and seed, a(n) is always equal to 1, 2, 3 or 5, and the sequence is the simple concatenation of the look-and-repeat sequence by block A225329. This is because all blocks of A225329 begin with 2 or 3 and end with 2 and therefore, there is no possible interaction between blocks after concatenation.
%C A225331 It never contains runs of exactly four identical digits (except the first four ones), but it does contain runs of five identical digits. However, five 5's never appear. Proof: suppose '55555' appears for the first time in a(n)..a(n+4); because of 'five five 5' in 55555, it would imply that 55555 appears from a smaller n, which is a contradiction.
%e A225331 a(1) = 1, then a(2) = a(3) = a(4) = 1 (one one 1). Leaving out the first 1 already said, we now have three 1's, then a(5) = a(6) = 3, and a(7) = 1, etc.
%Y A225331 Cf. A225330 (a close variant with 4's), A225329 (look-and-repeat by block), A005150 (original look-and-say), A225224, A221646, A225212 (continuous look-and-say versions).
%K A225331 nonn,easy
%O A225331 1,5
%A A225331 _Jean-Christophe Hervé_, May 12 2013