Ten Squares Force an Overlap
classification
🧮 math.CO
cs.DMcs.FL
keywords
squaresconcatenationoverlapalphabetbestbinaryboundconsist
read the original abstract
We prove that every concatenation of $10$ or more binary squares contains an overlap. The bound $10$ is best possible. In contrast, over a ternary alphabet, there are infinitely long overlap-free words that consist of a concatenation of squares.
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