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Unfortunately, if you spot that pattern then you will not guess the correct generalization. It seems that this may be a puzzle that's easier to solve if you are
by gjm11 7y ago
Unfortunately, if you spot that pattern then you will not guess the correct generalization. It seems that this may be a puzzle that's easier to solve if you aren't good at spotting easy things.
- mankeysee 7y agoWell tbh, it seems 2^k-1 does actually match the sequence as far as it is shown. Perhaps there is a proof that n=2^k-1 for the numbers where 2^n is divisible by (n+1). I'll have to see.