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Talk:Fibonacci word: Difference between revisions
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Them That Be. (I stopped after 20 because the algorithm I use for locating
repeated substrings is pretty naive and slow - hmmm, maybe <i>that</i> should be a separate
RosettaCode task?*).
In fact, thinking about it, the longest repeating substring has got to be <i>at least</i>
length(fword(n-2)) for all n > 2. Even expressing in hexidecimal isn't going to help much.
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