A318916 - OEIS

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A318916

Minimum length of longest palindromic subsequence, where the minimum is taken over all binary circular words of length n.

1, 1, 3, 3, 5, 5, 6, 7, 7, 8, 9, 9, 11, 11, 12, 13, 13, 14, 15, 15, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

Here "subsequence" means "scattered subsequence" (not necessarily contiguous). A circular word "wraps around".

Slide 10 of Andrew Ryzhikov's talk contains the conjecture that a(n) >= 3n/4, but this conjecture fails for n = 2 and n = 62 (at least). The length-62 counterexample 11111111000000000001010001110110000011010011111001000111010111, where the longest palindromic subsequence is of length 45, was found by Antonio Molina Lovett.

LINKS

Table of n, a(n) for n=1..32.

Andrew Ryzhikov, Regular subsequences in finite words, talk for the Workshop on Words and Complexity, 19-23 Feb 2018, Villeurbanne, France.

EXAMPLE

Consider the circular word 0001011. A rotation of this circular word gives 0101100, and deleting the first 1 gives the subsequence 001100, which is a palindrome of length 6. This shows that a(7) >= 6.

CROSSREFS

Sequence in context: A054847 A335266 A067782 * A035299 A338215 A266251

Adjacent sequences: A318913 A318914 A318915 * A318917 A318918 A318919

KEYWORD

nonn,more

AUTHOR

Jeffrey Shallit, Sep 05 2018

EXTENSIONS

Terms a(19)-a(32) computed by Antonio Molina Lovett

STATUS

approved