mersenneforum.org > Math 2nd Hardy-Littlewood conjecture
 Register FAQ Search Today's Posts Mark Forums Read

2020-10-27, 06:44   #23
R2357

"Ruben"
Oct 2020
Nederland

2·19 Posts
It looks like "we" are certain that something is wrong...

Quote:
 Originally Posted by CRGreathouse the first counterexample may be large.
So, if I understand, "we" believe that there exist a counter-example for the 2nd Hardy-Littlewood 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 counter-example…

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.

2020-10-27, 14:08   #24
CRGreathouse

Aug 2006

2×2,969 Posts

Quote:
 Originally Posted by R2357 So, if I understand, "we" believe that there exist a counter-example for the 2nd Hardy-Littlewood conjecture, which is already surprising!
Actually, I don't think that's surprising at all.

Quote:
 Originally Posted by R2357 But even more surprising is that we think that it's by getting to "large" numbers that we'll get a counter-example…
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.

 2020-10-27, 14:42 #25 Dr Sardonicus     Feb 2017 Nowhere 3,779 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" 447-tuples 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 n-value given in the post. In the process, I found the following: The n-value n = 1566280308578217520031412816790827048467516641360946961779273951 in the above can be replaced by n = 1896824066411233857172894793338758314448971776013524011962105921, n = 427740698264494580988530453125730465642504510890959344482852721, n = 3930817549863319998911606428403486440368429511069909409490035981, n = 4261361307696336336053088404951417706349884645722486459672867951, or n = 2792277939549597059868724064738389857543417380599921792193614751, and the 447-tuples (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 Hardy-Littlewood conjecture, and will also consist entirely of numbers relatively prime to 163#. The given 447-tuple [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 n-values above, which force k*163# + n to be relatively prime to 163#. These give a total of 6 n-values, those given above. This is all contained in the information given at the k-tuples site referenced earlier. Last fiddled with by Dr Sardonicus on 2020-10-27 at 14:45 Reason: Completion of re-editing
2020-10-27, 18:17   #26
mart_r

Dec 2008
you know...around...

25·19 Posts

Quote:
 Originally Posted by R2357 (...) the proportion of primes found less than x is inversely proportionnal to the natural logarithm of x.
Correct - when the numbers have 1200 digits, on average about 1 in every 2762 numbers is a prime. This does not contradict the assumption that dense clusters of 447 primes of this size do exist. Maybe there will be a gap of one million numbers (consecutive integers, for the nitpickers) without a prime after such a cluster - improbable, but not impossible.
You're probably surprised to learn that you're even more likely to find prime-447-tuplets in numbers with 1201 digits than in numbers with 1200 digits. And even more so when searching all 1202-digit 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 2020-10-27 at 19:05

2020-10-27, 19:44   #27
R2357

"Ruben"
Oct 2020
Nederland

2·19 Posts

Quote:
 Originally Posted by mart_r Correct - when the numbers have 1200 digits, on average about 1 in every 2762 numbers is a prime. This does not contradict the assumption that dense clusters of 447 primes of this size do exist. Maybe there will be a gap of one million numbers (consecutive integers, for the nitpickers) without a prime after such a cluster - improbable, but not impossible.
Did you analyse the way Possible Prime Positions are taken away in patterns of p# (and I'm not even talking about the composites by primes higher than p)

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 2-53, 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 :)

2020-10-28, 02:05   #28
Dr Sardonicus

Feb 2017
Nowhere

1110110000112 Posts

Obviously, searching far enough to find a 447-tuple 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" k-tuples of the 447-tuple for small k might be within reach. I did a simpleminded numerical sweep using Pari-GP, 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 4-tuples" of (pseudo)primes.

I then tried t up to a million on the "initial 5-tuple"

p, p + 2, p + 6, p + 8, and p + 12.

Only 153040 -- the smallest value giving an initial 4-tuple -- filled the bill. The rest of the values giving "initial 4-tuples" failed to give an "initial 5-tuple." Talk about lucky...

I ran t up to ten million. The only additional values yielding "initial 5-tuples" were t = 4397928, 5193864, 5598490, 5942162, 9086440, and 9926981.

If you want an initial 6-tuple

(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:
 Originally Posted by CRGreathouse 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.
In order to try to get some idea of how large a value of x is required for a prime k-tuple, I cheerfully disregarded the constant multiplier, and looked at the equation

(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 k-tuple: 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 k-tuple grows faster than exponentially, even faster than factorially, with k.

 Similar Threads Thread Thread Starter Forum Replies Last Post didgogns Miscellaneous Math 1 2020-08-05 06:51 MathDoggy Miscellaneous Math 11 2019-04-17 07:05 reddwarf2956 Prime Gap Searches 2 2016-03-01 22:41 CRGreathouse Math 4 2010-07-06 17:47 GP2 Math 2 2003-09-14 04:36

All times are UTC. The time now is 13:07.

Tue Nov 24 13:07:36 UTC 2020 up 75 days, 10:18, 4 users, load averages: 1.67, 1.89, 2.00