%I #26 Feb 07 2021 19:05:47
%S 1,21,12,112,121,11212,1112,12121,11,2121,1121,21211,121112,1212,
%T 111212,121211,1212121,11211,11121212,11121112,12121112,121212,
%U 12121211,121212121,112121,211,1211,111,212121,112111,2121211,12121212121,212111,21212121,11212121,1121111,121212111,2111,21212111
%N Underline all even terms; concatenations of underlined terms and of non-underlined terms both equal the concatenation of the entire sequence. This is the lexicographically earliest such sequence without duplicate terms and with an even digit among the first two terms.
%C This can be seen as a fractal sequence (from the digit's point of view).
%C If we don't require the sequence to have an even digit among the first two terms, there are infinitely many lexicographically earlier solutions with the first even digit appearing almost arbitrarily late, e.g., T = 1, 3, 21, 132, 13, 213, 1132, 2131, 13221, ..., or U = 1, 3, 5, 21, 13, 135, 1352, 1131351352, 13521, ... or V = 1, 3, 5, 7, 21,... or W = 1, 3, 5, 7, 9, 21, ... etc.
%C The lexicographically earliest sequence with no such constraint on a(2) would be A = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, ... (2k+1) ..., 1357911131517189, 1357911131517191, 1357911131517192, 12, 32, 52, 72, 931333537394, 14, 34, 54, 74, 951535557596, 16, 36, 56, ... So A(678,955,565,758,597) = 1,357,911,131,517,192 would be the earliest term different from 2n+1.
%C We see here how the three sets of A's digits are the same in their succession: the underlined digits, the non-underlined digits, and the digits of the sequence itself.
%H Jean-Marc Falcoz, <a href="/A307312/b307312.txt">Table of n, a(n) for n = 1..1003</a>
%e The sequence cannot start with 0 (which cannot be reproduced by the concatenation of odd terms), so it must start with 1. This cannot be followed by 0, 2, 4, ..., 8, 10, ..., 18, 20, so the smallest possible a(2) = 21. - _M. F. Hasler_, Apr 04 2019
%e The sequence starts with 1,21,12,112,121,11212,1112,12121,11,2121,...
%e Let's parenthesize the even terms:
%e 1,21,(12),(112),121,(11212),(1112),12121,11,2121,...
%e We see that the parenthesized digits appear in the same order as the digits of the sequence.
%e Let's do the same with the odd terms:
%e (1),(21),12,112,(121),11212,1112,(12121),(11),(2121),...
%e We see again that the parenthesized digits appear in the same order as the digits of the sequence.
%K base,nonn
%O 1,2
%A _Eric Angelini_ and _Jean-Marc Falcoz_, Apr 02 2019
%E Edited by _M. F. Hasler_, Apr 04 2019
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