I found this problem quite interesting.

Sharky has a word of 2015 letters made of S and K letters only (e.g. SKSSS, KSKSK). A **palindrome** is a word which can be read the same as looking from left to right or right to left.

Sharky decides to cut up his word into sub-words such that each sub-word is a palindrome. Given that each sub-word must contain a natural number of letters (No cutting up a letter in half), find the minimum number of sub-words that can be cut out for any word of Sharky's.

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`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestGot an entry on OEIS. The formula there says 672.

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Damn, that looks neat. Anyone got a proof for that? The best I got was 674.

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