A362815 -id:A362815 - cp's OEIS frontend

A362816 Lexicographically earliest sequence such that nowhere is a term a(n) contained in an arithmetic progression of length greater than a(n).

2, 2, 3, 2, 2, 3, 3, 3, 5, 2, 2, 3, 2, 2, 3, 3, 3, 5, 3, 5, 5, 5, 3, 3, 3, 5, 5, 2, 2, 3, 2, 2, 5, 5, 3, 3, 2, 2, 3, 2, 2, 5, 3, 3, 5, 3, 5, 5, 3, 3, 5, 5, 3, 5, 5, 5, 6, 5, 3, 5, 5, 6, 5, 3, 3, 3, 5, 3, 5, 5, 5, 3, 3, 3, 5, 5, 5, 6, 5, 5, 3, 2, 2, 5, 2, 2, 6

Offset: 1

Samuel Harkness, May 04 2023

nonn

Progressions are terms at indices in arithmetic progression and with values which are some arithmetic progression too.
1 is never in the sequence, because if a(n) = 1, then {a(n),a(n+1)} would form an arithmetic progression greater than 1 in length.
Conjecture: only terms in A362815 appear in this sequence. This is true through the first 10^5 terms.
If this is true, then a(A003278) = 2, because the only way to constrain 2 would be {2,2,2}, and A003278 is defined by adding the smallest term which avoids any 3 term arithmetic progressions. If the conjecture is false, arithmetic progressions {4,3,2}, {8,5,2}, etc. may further constrain 2s.

			For n=9 first we check 1 (never in the sequence). If a(9) were 2, {a(1),a(5),a(9)} = {2,2,2} would form an arithmetic progression of length 3 with a minimum value of 2; this is not allowed. Next, if a(9) were 3, {a(6),a(7),a(8),a(9)} = {3,3,3,3} would form an arithmetic progression of length 4 with a minimum value of 3; this is not allowed. Next, if a(9) were 4, {a(5),a(7),a(9)} = {2,3,4} would form an arithmetic progression of length 3 with a minimum value of 2; this is not allowed. Last, a(9) = 5 fits the definition, as no arithmetic progressions p can be made such that length(p) > min (p) and 5 is the least positive integer where this is satisfied, so a(9) = 5.

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Samuel Harkness, MATLAB program

Cf. A362815, A363011 (indices of record highs), A003278, A090822, A281579.

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A363011 Indices of record highs in A362816.

Original entry on oeis.org

1, 3, 9, 57, 504, 3847, 41170

Offset: 1

Samuel Harkness, May 12 2023

It is conjectured that only terms in A362815 appear in A362816, and novel terms in A362816 seen to be in ascending order. Does every term in A362815 appear in A362816?

If the conjecture is true and novel terms are in ascending order, then these are the smallest k such that A362816(k) = A362815(n).

			Cf. A362815, A362816.

Samuel Harkness, MATLAB program

Cf. A362815, A362816.

MATLAB
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A381537 Lexicographically least sequence of natural numbers such that for all arithmetic progressions p, length(p) <= sqrt(max(p)).

Original entry on oeis.org

1, 4, 5, 8, 9, 10, 12, 15, 16, 17, 18, 22, 23, 24, 25, 26, 28, 29, 30, 31, 33, 35, 36, 37, 38, 39, 40, 42, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 61, 62, 63, 64, 65, 66, 67, 68, 70, 71, 72, 73, 74, 75, 76, 81, 82, 83, 84, 85, 86, 87, 88, 90, 91, 92

Offset: 1

Samuel Harkness, Feb 26 2025

nonn

Up to a(n) the longest possible arithmetic progression is sqrt(a(n)).

Does the density of this sequence approach 1?

			1 is in the sequence, as 1 creates the arithmetic progression p = {1}, where length(p) = 1 and sqrt(max(p)) = 1.
For 2: the arithmetic progression p = {1,2} would be created. Here, length(p) = 2, and sqrt(max(p)) = sqrt(2), so length(p) > sqrt(max(p)), thus 2 is not in the sequence. Similarly, 3 is not in the sequence.
For 4: p = {1,4} is the only new arithmetic progression. Here, length(p) = 2, and sqrt(max(p)) = 2, so 4 is in the sequence. Similarly, 5 is in the sequence.
For 6: the arithmetic progression p = {4,5,6} would be created. Here, length(p) = 3, and sqrt(max(p)) = sqrt(6), so length(p) > sqrt(max(p)), thus 6 is not in the sequence.

Samuel Harkness, MATLAB program

Cf. A281579, A362815.

MATLAB
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cp's OEIS Frontend

A362816 Lexicographically earliest sequence such that nowhere is a term a(n) contained in an arithmetic progression of length greater than a(n).

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A363011 Indices of record highs in A362816.

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A381537 Lexicographically least sequence of natural numbers such that for all arithmetic progressions p, length(p) <= sqrt(max(p)).

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Examples

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Crossrefs

Programs

MATLAB