A330896 Lexicographically earliest sequence of positive integers such that for any m > 0, gaps between consecutive m's are all distinct.

1, 1, 2, 1, 2, 2, 1, 3, 2, 3, 1, 4, 2, 3, 3, 1, 4, 2, 4, 3, 5, 1, 3, 2, 4, 4, 5, 5, 1, 3, 2, 5, 4, 5, 6, 3, 1, 6, 2, 6, 4, 5, 7, 3, 4, 1, 5, 2, 6, 5, 7, 7, 3, 4, 6, 1, 4, 2, 5, 6, 6, 7, 3, 7, 6, 5, 1, 4, 2, 7, 8, 6, 7, 3, 8, 5, 7, 4, 1, 6, 2, 7, 8, 8, 9, 3, 5

Offset: 1

Author

Rémy Sigrist, May 01 2020

Keywords

• nonn
• look

Comments

Every positive integer appears infinitely many times in the sequence.

This sequence has similarities with A003602, where gaps between consecutive equal values are all distinct.

This sequence has similarities with A002260, where for any m > 0, gaps between consecutive m's are strictly increasing.

Apparently, for any m > 0:

- the k-th gap between consecutive m's equals k except for finitely many k's,

- the k-th occurrence of m appears at index A330897(m) + A000217(k-1) except for finitely many k's.

Examples

			The first terms, alongside the gaps for m = 1..4, are:
     n|  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23  ...
  a(n)|  1  1  2  1  2  2  1  3  2  3  1  4  2  3  3  1  4  2  4  3  5  1  3  ...
  ----+---------------------------------------------------------------------
   1's|     1,    2,       3,          4,             5,                6,    ...
   2's|              2, 1,       3,          4,             5,                ...
   3's|                             2,          4, 1,             5,       3, ...
   4's|                                                  5,    2,             ...
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, PARI program for A330896

Crossrefs

Cf. A000124, A000217, A002260, A003602, A330897.

Programs

PARI

\\ See Links section.

Python

from itertools import islice
def agen(): # generator of terms
    m, a = dict(), []
    while True:
        an, allnew = 0, False
        while not allnew:
            allnew, an, mn = True, an+1, set()
            for i in range(len(a)-1, -1, -1):
                if an == a[i]:
                    t = len(a)-i+1
                    if (an in m and t in m[an]) or t in mn: allnew = False; break
                    mn.add(t)
                    break
        yield an; a.append(an)
        if an not in m: m[an] = set()
        m[an] |= mn
print(list(islice(agen(), 87))) # Michael S. Branicky, Dec 06 2024

Formula

a(n) = 1 iff n belongs to A000124.