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A366625 Lexicographically earliest sequence of positive integers such that each multiset enclosed by two equal terms, excluding the endpoints, is distinct.

%I A366625 #49 Jan 15 2024 18:45:04
%S A366625 1,1,2,1,2,3,1,2,4,1,2,3,2,1,5,1,2,3,4,1,2,3,4,2,1,5,3,1,2,4,2,1,5,3,
%T A366625 4,1,2,6,1,2,3,4,5,1,2,4,3,7,1,2,3,4,5,3,1,2,6,2,1,3,4,5,7,1,2,3,4,5,
%U A366625 6,1,2,3,4,7,2,1,3,4,5,6,1,2,4,3,5,8,1
%N A366625 Lexicographically earliest sequence of positive integers such that each multiset enclosed by two equal terms, excluding the endpoints, is distinct.
%C A366625 Every positive integer occurs infinitely many times in the sequence.
%C A366625 The multiset between any two equal terms is unique. For example: once consecutive values "A B C A" occur, both "D B C D" and "D C B D" can never occur, because the multiset "B C" would be repeated between equal terms.
%C A366625 Two consecutive values enclose the empty multiset. For this reason, after [a(1), a(2)] = [1, 1], no consecutive equal values will occur again.
%C A366625 A new value is always followed by 1.
%C A366625 The sequence first differs from A366624 at a(15).
%H A366625 Michael S. Branicky, <a href="/A366625/b366625.txt">Table of n, a(n) for n = 1..7522</a>
%H A366625 Samuel Harkness, <a href="/A366625/a366625.m.txt">MATLAB program</a>.
%H A366625 Neal Gersh Tolunsky, <a href="/A366625/a366625.png">Ordinal Transform of 5000 terms</a>.
%e A366625 a(15) = 5: a(15) cannot be 1 since this would form the empty multiset with a(14) = 1. a(15) cannot be 2 because this would form the multiset [2 1 2] = {1}, which already occurred at [2 1 2] = {1}. a(15) cannot be 3 because this would form the multiset [3 2 1 3] = {1, 2}, which already occurred at [1 1 2 1] = {1, 2}. a(15) cannot be 4 because this would form the multiset [4 1 2 3 2 1 4] = {1, 1, 2, 2, 3}, which already occurred at [1 1 2 1 2 3 1] = {1, 1, 2, 2, 3}. a(15) = 5 because 5 is a first occurrence and thus creates no new multisets.
%e A366625 For this sequence, the multisets between k and all other occurrences of k must be checked. The first instance such that this is the sole reason for restricting a possible value is when considering 2 for a(27).
%e A366625 a(27) != 2 since 2 there would cause two enclosed multisets with the same 5 terms (in different order, which doesn't matter for a multiset),
%e A366625   n    =   15      19    22      26 27
%e A366625   a(n) =  1 5 1 2 3 4 1 2 3 4 2 1 5 [2]
%e A366625             |-------|     |-------|
%e A366625 There are also instances where overlapping conflicting regions are the sole reason for restricting a possible value.
%e A366625 a(74) != 5 since 5 there would cause two enclosed multisets with the same 20 terms,
%e A366625   n    =  38                              54
%e A366625   a(n) = 2 6 1 2 3 4 5 1 2 4 3 7 1 2 3 4 5 3 1
%e A366625            |----------------------------------
%e A366625                                            |--
%e A366625   n    =  57                              73
%e A366625   a(n) = 2 6 2 1 3 4 5 7 1 2 3 4 5 6 1 2 3 4[5]
%e A366625          --|
%e A366625          ----------------------------------|
%o A366625 (MATLAB) See Links section.
%o A366625 (Python)
%o A366625 from itertools import islice
%o A366625 def agen(): # generator of terms
%o A366625     m, a = set(), []
%o A366625     while True:
%o A366625         an, allnew = 0, False
%o A366625         while not allnew:
%o A366625             allnew, an, mn = True, an+1, set()
%o A366625             for i in range(len(a)):
%o A366625                 if an == a[i]:
%o A366625                     t = tuple(sorted(a[i+1:]))
%o A366625                     if t in m or t in mn: allnew = False; break
%o A366625                     mn.add(t)
%o A366625         yield an; a.append(an); m |= mn
%o A366625 print(list(islice(agen(), 87))) # _Michael S. Branicky_, Oct 25 2023
%Y A366625 Cf. A366624, A366631, A366493, A330896.
%K A366625 nonn
%O A366625 1,3
%A A366625 _Samuel Harkness_, Oct 14 2023