%I #29 Feb 26 2022 21:20:39
%S 1,2,3,11,12,13,21,22,23,31,32,33,111,112,113,121,122,123,131,132,133,
%T 211,212,213,221,222,223,231,232,311,312,321,322,331,332,1112,1113,
%U 1121,1122,1123,1131,1132,1133,1211,1213,1221,1222,1223,1231,1232,1311,1312,1321,1322,1331,1332,2111
%N Lexicographically earliest numbers containing only digits 1,2,3 that appear again in the iterative cycle of their own 'Look and Say' description (Cf. A005150).
%C Take a number that contains only digits 1,2,3 and then describe the given number with its 'Look and Say' string, see A005150. Repeat this process until the starting number appears in the resulting string. The sequence lists the numbers that eventually reappear.
%C A334055 gives the number of iterations for the starting number to reappear and details of the final string for all recurrent numbers.
%C The number 233 is the first number that does not appear in its iterative Look and Say cycle. See examples below.
%e 1 is a term as the Look and Say description of 1 is '11', which contains '1' as a substring.
%e 111 is a term as the Look and Say description of 111 is '31'. Repeating this process leads to '1311' and then '111321', which contains '111' as a substring.
%e 1112 is a term as the Look and Say description of 1112 is '3112'. Repeating this process leads to '132112', '1113122112', '311311222112', '13211321322112', '1113122113121113222112', and then '31131122211311123113322112', which contains '1112' as a substring.
%e 233 is not a term. If it were, it would have to have a parent string that could produce 233 from its Look and Say description, and one can show that that is not possible. The '33' part cannot be interpreted as 'three 3's' as '333' cannot be inside the parent string. To prove that, consider that the string '333' can be interpreted in two ways. 1. "x number of 3's followed by three 3's" which is not possible as that would be written as "x+three 3's". 2. "three 3's followed by three x's", where x cannot be a 3, else it would be written as "six 3's". So it has to be either '333111' or '333222'. But the first case has to be interpreted as either "x 3's followed by three 3's followed by one 1 ..." which is not possible as that would be written as "x+three 3's", else it has to be interpreted as "three 3's followed by three 1's followed by one 1", which is not possible as that would be written as "... four 1's". Likewise for '333222', hence '333' has no valid parent string. Therefore '233' must be interpreted as "two 3's followed by 3 x's", where the x's cannot be 3's. Thus the two possible strings are '33111' or '33222'. But the '33' part of these strings cannot be "three 3's" as shown, so the first string has to be interpreted as "x 3's followed by three 1's followed by one 1", but that is not possible as the 1's would be combined. Similarly for the second string. Hence '233' has no parent string, thus it cannot reappear in the iterative process of this sequence.
%e A similar argument can prove that numbers like 313, 323, 1111, 2232, 2313, and with more effort, 22213, 112321, 213321, do not have any parent string and thus do not appear in this sequence.
%Y Cf. A334055, A005150, A119566, A007932, A045918.
%K nonn
%O 1,2
%A _Scott R. Shannon_, Sep 07 2020
