20201027, 06:44  #23 
"Ruben"
Oct 2020
Nederland
46_{8} Posts 
It looks like "we" are certain that something is wrong...
So, if I understand, "we" believe that there exist a counterexample for the 2nd HardyLittlewood conjecture, which is already surprising!
But even more surprising is that we think that it's by getting to "large" numbers that we'll get a counterexample… I really don't see how, when we know that the proportion of primes found less than x is inversely proportionnal to the natural logarithm of x. 
20201027, 14:08  #24  
Aug 2006
2·2,969 Posts 
Quote:
I don't think that's surprising either. Both of these are totally ordinary and expected. Estimating exactly how large isn't particularly easy, but it's not hard to see that it should be more like hundreds of digits than dozens of digits. If the first counterexample happened near 10^22 I would be very surprised. 

20201027, 14:42  #25 
Feb 2017
Nowhere
EC3_{16} Posts 
Admissible prime constellations is IMO a good overview. The second graph, of p(k)  l(k)  1, tells the story.
For small constellation size k, the second conjecture looks good. It is only as k increases that admissible constellations of size k in an interval of length x, with k > pi(x), begin to show up. Little wonder, then, that the inconsistency of the two conjectures went undetected for over 50 years. The asymptotic formula in the first conjecture indicates why the "interesting" 447tuples of primes are unlikely to be found any time soon. I have already posted a link to the page with the computations. BTW I started with the posted 447 values, and "reverse engineered" then nvalue given in the post. In the process, I found the following: The nvalue n = 1566280308578217520031412816790827048467516641360946961779273951 in the above can be replaced by n = 1896824066411233857172894793338758314448971776013524011962105921, n = 427740698264494580988530453125730465642504510890959344482852721, n = 3930817549863319998911606428403486440368429511069909409490035981, n = 4261361307696336336053088404951417706349884645722486459672867951, or n = 2792277939549597059868724064738389857543417380599921792193614751, and the 447tuples (k*163# + n + 0, k*163# + n + 2, k*163# + n + 6, k*163# + n + 8, k*163# + n + 12, ... , k*163# + n + 3158) k = integer will be admissible for the first HardyLittlewood conjecture, and will also consist entirely of numbers relatively prime to 163#. The given 447tuple [0, 2, 6, 8, 12, ... 3158] determines a unique class modulo M = 137# * 149 * 151 * 163, namely that of r = 226869188698870760778384337119124029399468630472000601300261 (mod M), for which (k*M + r + 0, k*M + r + 2, k*M + r + 6, k*M + r + 8, k*M + r + 12, ... , k*M + r + 3158) k = integer are all relatively prime to M. There are, in addition, two choices of residue modulo 139, and three choices modulo 157, that determine the nvalues above, which force k*163# + n to be relatively prime to 163#. These give a total of 6 nvalues, those given above. This is all contained in the information given at the ktuples site referenced earlier. Last fiddled with by Dr Sardonicus on 20201027 at 14:45 Reason: Completion of reediting 
20201027, 18:17  #26  
Dec 2008
you know...around...
2^{5}×19 Posts 
Quote:
You're probably surprised to learn that you're even more likely to find prime447tuplets in numbers with 1201 digits than in numbers with 1200 digits. And even more so when searching all 1202digit numbers, and so on... But I think I'm writing too much... P.S.: @ Dr. Sardonicus: Too much info Last fiddled with by mart_r on 20201027 at 19:05 

20201027, 19:44  #27  
"Ruben"
Oct 2020
Nederland
2×19 Posts 
Quote:
2 takes away each of its multiples (as there is none before) 3 takes away each of those none taken away by 2, so 1 over 2, thus 1 over 6 in the numbers. 5 takes away 2/30 or 1/15 numbers Now if we look at the way they are taken, for example p4 7#+{7, 49, 77, 91, 119, 133, 161, 203} We notice that they are evenly distributed across the sequence, which doesn't contradict the fact that the sequence of primes is chaotic (we're only looking at the possible prime positions left after the divisors of the primorial, not the actual primes), clusters of 447 in 3159 would mean a sequence length 3159 in which there would be 447 numbers not only relatively prime to 253, but to all numbers up to the last prime before the square root of the numbers reached by the sequence, so you are right, I am surprised when you announce that a 447 cluster in 3159 is more probable between 10^1202 than 10^1201... But I would be even more surprised if you could prove it :) 

20201028, 02:05  #28  
Feb 2017
Nowhere
111011000011_{2} Posts 
Obviously, searching far enough to find a 447tuple of primes
p, p + 2, p + 6, p + 8, p + 12, ..., p + 3158 is a bit much to ask for. However, it occurred to me that finding "initial" ktuples of the 447tuple for small k might be within reach. I did a simpleminded numerical sweep using PariGP, looking for positive integer t such that p = 1566280308578217520031412816790827048467516641360946961779273951 + t*163#, p + 2, p + 6, and p + 8 were all (pseudo)prime. The smallest t is 153040. The values 174049, 194993, 211745, 523061, 634968, 640804, 646566, 681614, 693501, 831389, 846672, 969176, and 999701 also yielded "initial 4tuples" of (pseudo)primes. I then tried t up to a million on the "initial 5tuple" p, p + 2, p + 6, p + 8, and p + 12. Only 153040  the smallest value giving an initial 4tuple  filled the bill. The rest of the values giving "initial 4tuples" failed to give an "initial 5tuple." Talk about lucky... I ran t up to ten million. The only additional values yielding "initial 5tuples" were t = 4397928, 5193864, 5598490, 5942162, 9086440, and 9926981. If you want an initial 6tuple (p, p + 2, p + 6, p + 8, p + 12, p + 20) of primes, with p = 1566280308578217520031412816790827048467516641360946961779273951 + t*163#, you'll have to try t > 10000000. Quote:
(log(x))^{k} = x defining x as a function of k. Taking logs, k*log(log(x)) = log(x), or k = log(x)/log(log(x)). As a first whack, I tried log(x) = k*log(k). It's a bit too small, but not ridiculously far off. It does give some idea of how large x has to be (in terms of k), in order to reasonably expect to find a ktuple: log(x) grows something like k*log(k). For k = 447, 447*log(447)/log(10) = 1184.687, approximately. That's not even considering the multiplicative constant in the formula, and it's still in the right ball park. At least as an estimate of log(x)... So it seems that if the asymptotic formula is right, the likely size of the smallest prime ktuple grows faster than exponentially, even faster than factorially, with k. 

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