5 ms·

Ah I had missed it too. Is there a bound on the gap between subsequent primes as well?

lmkg 13y ago
Just 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 ago
Nope :) 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 ago
  That is a neat proof :D
- atmosx 13y ago
  Weird, 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 ago
    I 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 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 ago
  That's a neat little proof.
- redthrowaway 13y ago
  That's a really damned cool proof. Why is it formulated as n+1 instead of merely n?
  - btilly 13y ago
    Because 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 ago
      Concrete 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 ago
        You 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.
        13y ago

lmkg 13y ago

Just 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 ago

Nope :) 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 ago
That is a neat proof :D
atmosx 13y ago
Weird, 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 ago
  I 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 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 ago
That's a neat little proof.
redthrowaway 13y ago
That's a really damned cool proof. Why is it formulated as n+1 instead of merely n?
- btilly 13y ago
  Because 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 ago
    Concrete 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 ago
      You 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.
      - 13y ago