A362881 -id:A362881 - cp's OEIS frontend

A362909 a(n) is the smallest number k such that A362881(k) = n.

1, 3, 6, 10, 22, 31, 34, 37, 40, 121, 1035, 1078, 1121, 7607, 9453, 13667, 13729, 50040, 50123, 50206, 615964, 616448, 616932, 617416, 2828280, 2851232, 2874184, 2897136, 5350614, 5583250, 5594326, 17310256, 17312210, 17994974, 18008820, 19051432, 19069162

Offset: 1

Samuel Harkness, May 14 2023

nonn

All positive integers will appear in A362881. Rémy Sigrist showed that as a consequence of Van der Waerden's theorem, A362881 is unbounded.

The present sequence is monotonically increasing since a progression of length n begins with a progression of length n-1 so a(n-1) < a(n).

Kevin Ryde, Table of n, a(n) for n = 1..62
Samuel Harkness, MATLAB program for A362909
Kevin Ryde, C Code

Cf. A362881.

C
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a(25)-a(31) from Yang Haoran, May 17 2023

a(32)-a(37) from Kevin Ryde, May 22 2023

A368290 a(n) is the length of the longest palindromic subsequence at symmetrically-spaced indices ending at a(n-1); a(1)=1.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 3, 3, 2, 5, 1, 7, 1, 6, 1, 6, 3, 5, 5, 2, 4, 1, 7, 4, 2, 9, 1, 11, 1, 6, 5, 4, 5, 4, 3, 7, 3, 9, 5, 6, 7, 7, 7, 5, 9, 6, 5, 7, 5, 5, 8, 1, 11, 6, 7, 7, 9, 10, 1, 9, 9, 6, 9, 6, 11, 7, 13, 1, 12, 1, 14, 1, 16, 1, 17, 1, 19, 1, 14, 7, 9, 7, 11

Offset: 1

Neal Gersh Tolunsky, Dec 19 2023

nonn

A set of indices is symmetric if, listed in increasing or decreasing order, its first differences are a palindromic sequence.
A new value is always followed by 1.
An alternate definition: a(n) is the largest number of coincidences between the subsequence enclosed by m..n-1 and its reverse, where a(n-1)=a(m), maximized over m.

			a(10)=5 because we find the following length-5 palindromic subsequence at symmetric indices ending at i=a(n-1)=a(9)=2:
  S: 1,1,2,1,3,1,3,3,2
  P:     2,  3,1,3,  2
a(14)=6 because of the following length-6 palindromic subsequence:
  S: 1,1,2,1,3,1,3,3,2,5,1,7,1
  P:   1,  1,    3,3,    1,  1

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Neal Gersh Tolunsky, Ordinal transform of first 30000 terms.
Neal Gersh Tolunsky, Graph of first 100000 terms.

Cf. A308659, A362881.

{ for (n = 1, #a = vector(83, n, 1), for (k = 1, n-1, if (a[k] == a[n-1], a[n] = max(a[n], sum (i = k, n-1, a[i] == a[n-1+k-i]);););); print1 (a[n]", ");); } \\ Rémy Sigrist, Dec 20 2023

More terms from Rémy Sigrist, Dec 20 2023

A382559 a(n) is the length of the longest subsequence at indices in arithmetic progression ending at a(n-1) whose terms form an arithmetic progression in some order; a(1)=1.

Original entry on oeis.org

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

Offset: 1

Neal Gersh Tolunsky, Apr 01 2025

nonn

First differs from A362881 at a(15).

This is a variant of A362881 in which the terms of an arithmetic progression can occur in any order.

			a(21) = 4: The subsequence at indices i = 2,8,14,20 (common difference 6) is {1,3,2,4} which can be rearranged to form the arithmetic progression {1,2,3,4}. We find that the longest such subsequence ending at a(20) has length 4, so a(21) = 4.

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Neal Gersh Tolunsky, Graph of the ordinal transform of the first 10000 terms, with lines labeled by corresponding values of this sequence.

Cf. A362881, A381629, A361933.

A384450 a(1) = 0; thereafter, a(n) is the number of arithmetic progressions of length 3 or greater at indices in an arithmetic progression ending at a(n-1).

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 2, 0, 4, 0, 5, 0, 8, 0, 9, 0, 12, 1, 0, 1, 0, 0, 5, 0, 5, 0, 5, 1, 0, 3, 0, 3, 0, 4, 0, 2, 2, 2, 2, 3, 0, 3, 0, 2, 1, 0, 7, 0, 5, 0, 5, 0, 7, 0, 10, 1, 1, 2, 1, 1, 0, 9, 0, 6, 3, 0, 6, 1, 0, 6, 3, 3, 1, 2, 2, 3, 0, 7, 0, 6, 3, 1, 0, 4, 4

Offset: 1

Neal Gersh Tolunsky, May 27 2025

nonn

In other words, take the longest arithmetic progression at indices with common difference k ending at a(n-1) and call that length j. a(n) is the sum of each j-2 that corresponds to a distinct common difference k. This means that an arithmetic progression of length 3 is worth 1 point, length 4 is worth 2 points, and so on.

			To find a(10), we see that there are 4 arithmetic progressions ending in a(9) = 0. These occur at indices i = 5,7,9; i = 3,5,7,9; i = 1,3,5,7,9; and i = 1,5,9. So a(10) = 4.

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000

Cf. A308638, A362881.

a(32)-a(86) from Pontus von Brömssen, May 30 2025

cp's OEIS Frontend

A362909 a(n) is the smallest number k such that A362881(k) = n.

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A368290 a(n) is the length of the longest palindromic subsequence at symmetrically-spaced indices ending at a(n-1); a(1)=1.

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A382559 a(n) is the length of the longest subsequence at indices in arithmetic progression ending at a(n-1) whose terms form an arithmetic progression in some order; a(1)=1.

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Author

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Examples

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A384450 a(1) = 0; thereafter, a(n) is the number of arithmetic progressions of length 3 or greater at indices in an arithmetic progression ending at a(n-1).

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Author

Keywords

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Examples

Links

Crossrefs

Extensions