4 ms·
For any natural n, the n consecutive numbers (n+1)! + 2, (n+1)! + 3, ... (n+1)! + (n+1) are all composite -- the first one is divisible by 2, the second one by
by xyzzyz 13y ago
For any natural n, the n consecutive numbers (n+1)! + 2, (n+1)! + 3, ... (n+1)! + (n+1) are all composite -- the first one is divisible by 2, the second one by three and so on.
Thus there are arbitrarily long sequences of consecutive numbers that have no primes in them.
- Pitarou 13y agoThat's a neat little proof.
- redthrowaway 13y agoThat's a really damned cool proof. Why is it formulated as n+1 instead of merely n?
- btilly 13y agoBecause n! + 1 can be prime, so from n! + 2 to n! + n you only get n - 1 numbers. That's why you need to start the sequence at (n+1)! + 2.
- wolfgke 13y agoConcrete examples: 3!+1 = 7 is prime 11!+1 = 39916801 is prime (according to http://www.wolframalpha.com/input/?i=is+11!%2B1+prime%3F http://www.wolframalpha.com/input/?i=is+11!%2B1+prime%3F) 27!+1 = 10888869450418352160768000001 is prime (according to http://www.wolframalpha.com/input/?i=is+27!%2B1+prime%3F http://www.wolframalpha.com/input/?i=is+27!%2B1+prime%3F)
- btilly 13y agoYou forgot a couple. 1!+1 = 2 is prime. 2!+1 = 3 is prime. Based on a back of the envelope estimate, the number of values of n up to N for which n!+1 is prime should be O(log(log(N))) so it should happen infinitely often. But proving that is likely to be difficult.
- xyzzyz 13y agoDid you get that estimate by some other way than using prime number theorem and Stirling approximation for n! ?
- btilly 13y agoThat is exactly how I estimated it. But when I think about it, it is the wrong way to estimate it. Because you've eliminated all of the most common causes of those numbers not being prime, so the density of primes in that set should be expected to be massively higher than normal. If I thought about it, I could come up with a much better estimate. But it would take some thought.
- redthrowaway 13y agoPresumably there's no semantic difference, though? Saying that, for any counting number n, there are at least n-1 consecutive composites is semantically no different than saying there are n consecutive composites. It just makes for a cleaner proof, no?
- eru 13y agoI seem to remember that there's always a prime between n and 2*n? (Still an arbitrarily large absolute gap, but a fixed relative gap.) Can anyone deny or confirm?
- robinhouston 13y agoYes, you are right. http://mathworld.wolfram.com/BertrandsPostulate.html http://mathworld.wolfram.com/BertrandsPostulate.html
- DerekL 13y agoYes, that's called Bertrand's postulate, first proven in 1850 by Chebyshev. For any integer n > 3, there's at least one prime strictly between n and 2n-2. http://en.wikipedia.org/wiki/Bertrand's_postulate http://en.wikipedia.org/wiki/Bertrand's_postulate