A281900 -id:A281900 - cp's OEIS frontend

A281579 Lexicographically earliest sequence such that, for any n>0, a(n)=length of the n-th run of consecutive terms in arithmetic progression in this sequence.

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

Offset: 1

Rémy Sigrist, Jan 29 2017

nonn

Runs of consecutive terms in arithmetic progression overlap: the last term of the n-th run corresponds to the first term of the (n+1)-st run.
See A281772 for the common difference of the n-th run of consecutive terms in arithmetic progression.
See A281783 for the index of the first term of the n-th run of consecutive terms in arithmetic progression.
See A281900 for the index of the first occurrence of n in the sequence.
We can show that:
1) a(n)>=2 for any n>0,
2) a(n+1)<=a(n)+1 for any n>0,
3) runs of consecutive 2's have at least length 2.
Conjectures:
4) there are infinitely many runs of consecutive 2's,
5) the sequence is unbounded.
This sequence has connections with the Kolakoski sequence (A000002) and Golomb's sequence (A001462) in the sense that they all establish a link between their terms and the lengths of inner runs.
This sequence has similarities with A113138. - Rémy Sigrist, Feb 08 2017
A380317 is an essentially identical sequence. - N. J. A. Sloane, Feb 17 2025

			a(1)=2 fits the definition (and a(1)=1 would not, because whatever a(2) is, (a(1),a(2)) is an arithmetic progression of length 2).
a(2)=2 also fits the definition.
(a(1), a(2)) constitutes the first run, and has length a(1)=2.
a(3) cannot equal 2 (as it would extend the previous run).
a(3)=3 fits the definition.
(a(2),a(3)) constitutes the second run, and has length a(2)=2.
a(4) cannot equal 2 (as a(5) would be equal to 1, which is impossible).
a(4)=3 fits the definition.
We complete the 3rd run with a(5)=3.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program for A281579

Cf. A000002, A001462, A113138, A281772, A281783, A281900, A380317.

A380317 The lexicographically earliest sequence of positive numbers which is identical to the run lengths of its first differences.

Original entry on oeis.org

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

Offset: 1

Dominic McCarty, Feb 13 2025

34 is the smallest value that does not appear in the first 10000 terms.
Conjecture: Every positive integer eventually appears.
Shortly after submitting this sequence the author, Dominic McCarty, discovered that it is almost identical to A281579: in fact, a(n) = A281579(n) - 1 for all n. However, since A281579 has seniority and the present sequence has a simpler definition and is more likely to be searched for, it has been decided to retain both entries. - N. J. A. Sloane, Feb 17 2025
The index of n in the present sequence is given by A281900(n+1).

			The sequence of first differences (where the n-th term is a(n+1)-a(n)) is:
0, 1, 0, 0, 1, 1, -1, -1, 0, 0, 0, 1, 1, 1, 1, -1, -1, -1, 0, 0, ...
The run lengths of consecutive values are:
1, 1, 2, 2, 2, 3, 4, 3, 2, ...
Which is the original sequence.

Dominic McCarty, Table of n, a(n) for n = 1..10000

Cf. A281579, A281900.

from itertools import groupby
def runs(l): return [len(list(group)) for i, group in groupby(l)]
def firstDifs(l): return [l[i]-l[i-1] for i in range(1,len(l))]
a = [1,1]
while len(runs(firstDifs(a))) <= 100:
    a.append(1)
    b, m = runs(firstDifs(a)), max(firstDifs(a))
    while not (all(b[n] == a[n] for n in range(len(b)-1)) and b[-1] <= a[len(b)-1]):
        a[-1] += 1
        if a[-1] > m+a[-2]+1: a.pop(); a[-1] += 1 #Backtracking needed
        b = runs(firstDifs(a))
print(a[:len(runs(firstDifs(a)))])

cp's OEIS Frontend

A281579 Lexicographically earliest sequence such that, for any n>0, a(n)=length of the n-th run of consecutive terms in arithmetic progression in this sequence.

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