20220111, 16:38  #23  
Jun 2003
Oxford, UK
2,039 Posts 
Quote:
This is based on the mods of primes to 400 that never produce 0mod(the prime) for all values in the admissible sets. So 1mod2, 1mod3, 3mod5, 4mod7... this is approx. 5e160, compared to the prime at the start of the gap of 15900, which is 2e174, so it looks fine to play around with. Last fiddled with by robert44444uk on 20220111 at 17:08 

20220111, 22:52  #24 
Dec 2008
you know...around...
748_{10} Posts 
I get the same CRT offset with the pattern you linked to, so that's correct.
For my result, I didn't bother too much about sieving and just tested n*p#+c+x for primes, incrementing n when not enough primes were found above a customized threshold for x; something should be gained by applying an appropriate sieving technique. Up to 479#, there's only one open residue class for each prime, so it should merely be checked that not too many potential coprimes are cancelled out by the sieve. Just for fun, here are offsets for some larger p#: Code:
n*401#+ 39513451711353368972101707142676951932015103896038867201264946985487496815918734804213619530781605639321227211666671723990836697616026451458080515468952387537444673 n*409#+ 5300963833569209940057949054368721853533208296343484993741509551411018530966721606193800807153951251943913507869529328702625642328421257335999756718592000205492358083 n*419#+ 906110212932116653575732557665488316402090982314903912519073620662591471401157669028755216511381275347162712814920340507230752131956051717149856040611426373130703347023 
20220112, 09:04  #25 
Jun 2003
Oxford, UK
2,039 Posts 
I'm doing something wrong I think, although I am not sure (maybe mart_r could check)
Average number of primes in a range of x=15900 integers from a = 3483347771*409#/30 is = 15900/ln(a) or approx 39.65, given an average gap of 401. Average found number primes for n from 0 to 100 in n*p#+c+x, is 40.01 with a max prime count of 54 at n=93. c offset: 49268.... I am surprised to see such a small average pickup it is well within the bounds of statistics to be zero effect. if I am doing this right I am not sure the method pays off. Last fiddled with by robert44444uk on 20220112 at 09:11 
20220112, 12:27  #26  
Jun 2003
Oxford, UK
2,039 Posts 
Quote:
I think I was wrong to start at the deficient primorial 409#/30, I should have started with the primorial 409#, with a lower multiplier, in this case 3483347771/30 = 116111593 rounded up. The I don't multiply the offset c, I add one each time to n. My results for the first 100 n above 116111593 shows an average of 48.62 with a maximum of 63. That's more like it! 

20220112, 16:41  #27  
Jun 2003
Oxford, UK
2,039 Posts 
Quote:
Largest gap: 16690 Average merit: 1.229114171 First prime: 622973626447 Smallest gap: 10306 Average merit: 0.795722241 First prime: 177726413581 200 results: tested to approx. 1.39e12 Largest: 7338 Average merit: 1.414200628 First prime:185067242119 Smallest: 3646 Average merit: 0.694714106 First prime: 249072607711 100 results: tested to approx. 2.4e12 Largest: 4540 Average merit: 1.642701445 First prime: 1006401165853 Smallest: 1640 Average merit: 0.580014238 First prime: 1904361666929 Last fiddled with by robert44444uk on 20220112 at 16:46 

20220114, 09:44  #28  
Jun 2003
Oxford, UK
2,039 Posts 
Quote:
In n*p#+c+x, where p = 409, c = 492682..., x from 0 to 15900, then 87 primes are at n = 117575956 118482688 

20220114, 10:22  #29  
Jun 2003
Oxford, UK
2,039 Posts 
Quote:
As an aid to the factorials and offsets approach, it is relatively simple to show that there is no range of 100 primes p1..p100 at relatively small p1 where p1 is larger than any gap smaller than p1. In the exhaustive check of gaps between 101 primes ( a proxy for 100) highlighted with p1 <2.4e12, no range has an average merit of < 0.58, which equates to requiring a gap, where p1 is less than 2.4e12 whose merit is > 58. As the largest merit ever found is not even 42, there is the basis for the proof. 

20220114, 21:25  #30  
Dec 2008
you know...around...
2^{2}×11×17 Posts 
Quote:
It took me a couple of days, so I do think you have a chance to beat me at my own game I was also trying to squeeze even more primes into an interval of 8348, about two years ago, but a couple more days of searching and I didn't get past about 94 or 95 primes. 

20220115, 11:35  #31 
Jun 2003
Oxford, UK
11111110111_{2} Posts 
A slight improvement in the prime count for n in n*p#+c+x, where p = 409, c = 492682..., x from 0 to 15900
123733011 91 120673847 88 121848887 88 In terms of sieves, I found it was useful to break down the range x into four even parts and set cumulative targets for each range. I found at least 8 values at 50 or more primes at half way with a top value of 53. I do feel that it would be better to concentrate more on the 41 merit gap, maybe using a few new tweaks to get to the 101 level. Almost 42 merit is totally different to 39 in terms of space. Last fiddled with by robert44444uk on 20220115 at 11:41 
20220117, 08:56  #32 
Jun 2003
Oxford, UK
2,039 Posts 
A couple more, getting closer, but not that close
124977806 92 125316443 90 
20220117, 21:19  #33  
Dec 2008
you know...around...
2^{2}×11×17 Posts 
Quote:
Some possibly useful ideas, theorywise: Let \(p_n\) be large (well, say, > 10^{6}), let \(p_{n+k}\) be the kth prime after \(p_n\) (k < \(\sqrt{p_n}\), just to be safe). \(g = p_{n+k}  p_n\) \(m = G(p_{n+k})G(p_n)k+1\), G(x) being the formula for the blue line in the graph in http://www.primefan.ru/stuff/primes/table.html#theory \(CSG^* = \frac{m \cdot m}{g}\) \(m^* = CSG^* \cdot \log p_{n+k}\) One may be inclined to expect the distribution of \(m^*\) as behaving like the ones for the merits of usual prime gaps. This may even be halfway right, as experimental data suggests, for example these 1,751,000 samples of intervals of length 10^{5} with p in the vicinity of 34*10^{12} can be turned into this (similar results for other parameters): Code:
m*< #times r (#/total) log(1r) 1 1581721 0.903324386 2.336394091 1) 2 1695327 0.968205026 3.448447043 3 1731016 0.988587093 4.473010379 4 1743358 0.995635637 5.434282983 5 1747999 0.998286122 6.368996766 6 1749835 0.999334666 7.315221245 7 1750528 0.999730440 8.218718626 8 1750830 0.999902913 9.239899174 9 1750932 0.999961165 10.15618991 10 1750978 0.999987436 11.28465516 11 1750990 0.999994289 12.07311252 12 1750995 0.999997144 12.76625970 13 1750997 0.999998287 13.27708532 14 1750998 0.999998858 13.68255043 15 1750999 0.999999429 14.37569761 16 1751000 1 Would that lead to a way to conjecture that the gaps between nonconsecutive primes are bounded by a constant times \((\log(p)+k) \cdot \log(p)\) ? Or am I thinking way too complicated? 

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