A366624 Lexicographically earliest sequence of positive integers such that each subsequence enclosed by two equal terms, not including the endpoints, is distinct.

1, 1, 2, 1, 2, 3, 1, 2, 4, 1, 2, 3, 2, 1, 3, 2, 1, 4, 1, 2, 3, 4, 1, 2, 3, 5, 1, 2, 3, 4, 2, 1, 3, 4, 2, 1, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 6, 1, 2, 3, 4, 5, 2, 1, 3, 4, 5, 2, 1, 4, 2, 1, 3, 5, 1, 2, 4, 3, 1, 2, 4, 3, 2, 1, 4, 3, 2, 1, 5, 2, 1, 3, 4, 6, 1, 2, 4

Offset: 1

Author

Samuel Harkness, Oct 14 2023

Keywords

• nonn

Comments

Every positive integer occurs infinitely many times in the sequence.

The subsequence between any two equal terms is unique. For example, consecutive values "A B A" prevents "C B C" because the subsequence "B" would be repeated between equal terms.

Two consecutive values create the empty subsequence, for this reason after {a(1), a(2)} = {1, 1}, no consecutive values will ever occur again.

A new value is always followed by 1.

Examples

			For a(9), we first try 1. If a(9) were 1, {a(8)} = {2} would be bounded by a(7) = a(9) = 1, but {2} is already bounded by a(2) = a(4) = 1.
Next, try 2. Note this would create consecutive values at {a(8), a(9)}, enclosing the empty subsequence, but this already occurred at {a(1), a(2)}.
Next, try 3. If a(9) were 3, {a(7), a(8)} = {1, 2} would be bounded by a(6) = a(9) = 3, but {1, 2} is already bounded by a(1) = a(4) = 1.
Next, try 4. 4 has yet to appear, so a(9) = 4.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..6804 (terms 1..1000 from Neal Gersh Tolunsky)
• Samuel Harkness, MATLAB program.
• Kevin Ryde, C Code.

Crossrefs

Cf. A366625 (with distinct multisets), A366631 (with distinct sets), A366493 (including endpoints), A330896, A366691.

Programs

C

/* See links */

MATLAB

See Links section.

Python

from itertools import islice
def agen(): # generator of terms
    m, a = set(), []
    while True:
        an, allnew = 0, False
        while not allnew:
            allnew, an, mn = True, an+1, set()
            for i in range(len(a)):
                if an == a[i]:
                    t = tuple(a[i+1:])
                    if t in m or t in mn: allnew = False; break
                    mn.add(t)
        yield an; a.append(an); m |= mn
print(list(islice(agen(), 87))) # Michael S. Branicky, Jan 15 2024