5 ms·
To clarify, this result means that there are infinitely many pairs of primes separated by at most 600. This doesn't mean that the gap between subsequent primes
by davidjohnstone 13y ago
To clarify, this result means that there are infinitely many pairs of primes separated by at most 600. This doesn't mean that the gap between subsequent primes must always be less than 600.
- Lazare 13y agoThanks for the clarification. Reading back through the article I see where that's mentioned, but I totally missed it the first time through.
- raz32dust 13y agoAh I had missed it too. Is there a bound on the gap between subsequent primes as well?
- lmkg 13y agoJust the opposite. The frequency of primes is known to be logarithmic, and therefore asymptotically approaches zero. From this, there cannot be such a thing as the largest gap between primes, because that would put a lower bound on the frequency of primes.
- prezjordan 13y agoNope :) Consider the number 1000000! + 1 (one million factorial plus 1). 1000000! + 2 is divisible by 2. 1000000! + 3 is divisible by 3. ... 1000000! + 1000000 is divisible by 1000000. Here we have a gap of 999999 numbers, none of which are prime.
- undershirt 13y agoThat is a neat proof :D
- atmosx 13y agoWeird, I just read about this exact example yesterday in the "music of primes" by Markus De Sautoy. I'm not sure though if it was the German Siegel or the Norwegian guy (Sebel or something) who also end up to Princeton to figure it out, or maybe it was known already by Euler's time :-) Neat proof indeed.
- prezjordan 13y agoI first read it a few years back in Excursion's in Number Theory [0]. Super approachable (and short!), I read it in high school with only a little bit of calculus knowledge. [0]: http://www.amazon.com/Excursions-Number-Theory-Dover-Mathematics/dp/0486257789 http://www.amazon.com/Excursions-Number-Theory-Dover-Mathema...
- xyzzyz 13y agoFor 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.
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