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Timeline for question in prime numbers

Current License: CC BY-SA 2.5

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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Sep 6, 2015 at 1:13 comment added Aaron Meyerowitz Do you know $l_n$ for the prime $229?$ If so, update the OEIS. There is this lower bound (if I understand correctly): $l_n \gt (2e^{\gamma + o(1)}) n (\log n)^2 \log \log \log n (\log \log n)^{-2}.$ Using the estimate $p_n \approx n\log n$ might help.
Sep 5, 2015 at 19:43 comment added Brad Graham Is there any reason to expect that the ratio $l_n/p_n$ from $[p_n, l_n]$ keeps increasing... I ask because I seem to have found a maximum limit for $l_n$; that is $l_n < 4p_n$
Dec 22, 2010 at 20:29 history edited Aaron Meyerowitz CC BY-SA 2.5
maximal gaps with no exceptions
Dec 22, 2010 at 16:50 vote accept Asterios Gkantzounis
Dec 22, 2010 at 16:43 vote accept Asterios Gkantzounis
Dec 22, 2010 at 16:44
Dec 22, 2010 at 16:43 vote accept Asterios Gkantzounis
Dec 22, 2010 at 16:43
Dec 17, 2010 at 8:03 comment added Gerhard Paseman Aaron, I will attempt to reach you by email. Gerhard "Ask Me About System Design" Paseman, 2010.12.17
Dec 16, 2010 at 22:25 comment added Aaron Meyerowitz Do you know if this is a recognized problem? I have a couple more ideas which are long shots, but who knows?
Dec 16, 2010 at 17:14 comment added Gerhard Paseman Indeed, Westzynthius does use CRT in his 1931 paper, not only for his $2p_{n-1}$ lower bound gap, but also for his larger lower bound gap which eventually grows mildly faster than Cp_n. I was hoping you would find p_n such that there would be a gap of 2p_{n+1} in the p# sequence. Gerhard "Ask Me About System Design" Paseman, 2010.12.16
Dec 16, 2010 at 11:07 comment added Aaron Meyerowitz Thanks for asking. See above for methods. A long prime free gap is no help if it has numbers like 257*103 or the like scattered inside it. After all, there is a gap 282 of following 436273009 which is a longer gap using smaller numbers. I suppose that by using the Chinese remainder theorem to make a good stretch better by plugging holes I thinned out the places where a prime could show up.
Dec 16, 2010 at 10:58 history edited Aaron Meyerowitz CC BY-SA 2.5
added 2605 characters in body
Dec 15, 2010 at 21:03 comment added Gerhard Paseman Aaron, of your example involving 89 consecutive numbers, did you search the primes looking for large gaps? If not, how did you find it? Gerhard "Ask Me About System Design" Paseman, 2010.12.15
Dec 15, 2010 at 20:58 comment added Gerhard Paseman if you are indexing by the interval length, Aaron's is an example for 19. If you are instead indexing by the highest small divisor (not as in the question) it is an example for 17. Gerhard "Ask Me About System Design" Paseman, 2010.12.15
Dec 15, 2010 at 9:03 comment added Asterios Gkantzounis for 19 you need 46 numbers with only 2 holes
Dec 15, 2010 at 9:00 comment added Asterios Gkantzounis thnx for the analytic answer
Dec 15, 2010 at 8:59 comment added S. Carnahan Aaron: How is this not a counterexample for $p=19$?
Dec 15, 2010 at 8:56 comment added Aaron Meyerowitz 87890 to 878928 inclusive is 39 numbers. The two holes are 87911 and 87917. Here are the smallest divisors of each number: $$[2, 3, 2, 13, 2, 5, 2, 3, 2, 7, 2, 11, 2, 3, 2, 5, 2, 17, 2, 3, 2, 87911, 2, 7, 2, 3, 2, 87917, 2, 13, 2, 3, 2, 11, 2, 5, 2, 3, 2]$$
Dec 15, 2010 at 7:47 comment added Asterios Gkantzounis if you have a counterexample for 17 you must give me a length of 38 numbers that you have only 2 "holes" according to the question so how is 87890 a counter example ?
Dec 15, 2010 at 7:16 history edited Aaron Meyerowitz CC BY-SA 2.5
correction
Dec 15, 2010 at 7:12 comment added Aaron Meyerowitz True enough. Actually 87931 is outside the range I mentioned so the claim is correct, just not the transcription. Thanks!
Dec 15, 2010 at 6:34 comment added Gerhard Paseman 87911 is also prime. Gerhard "Ask Me About System Design" Paseman, 2010.12.14
Dec 15, 2010 at 6:20 history edited Aaron Meyerowitz CC BY-SA 2.5
changed greatest to least
Dec 15, 2010 at 3:33 history edited Aaron Meyerowitz CC BY-SA 2.5
took out easy factorizations and added information for p=19
Dec 15, 2010 at 0:36 history edited Aaron Meyerowitz CC BY-SA 2.5
edited body
Dec 15, 2010 at 0:21 history edited Aaron Meyerowitz CC BY-SA 2.5
added factorizations
Dec 15, 2010 at 0:01 history answered Aaron Meyerowitz CC BY-SA 2.5