Re^2: (OT) A different kind of 'combinatorics'

And the regex is:

m/(?=(....)).+\1/
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Comment on Re^2: (OT) A different kind of 'combinatorics' Download Code

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Re^3: (OT) A different kind of 'combinatorics' by tybalt89 (Monsignor) on Apr 20, 2017 at 14:11 UTC
That was me.	[reply]