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Old 2021-09-25, 22:50   #23
Batalov
 
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Phi(4,2^7658614+1)/2

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Quote:
Originally Posted by Dr Sardonicus View Post
Just out of curiosity, I wrote a Pari-GP script to check this number for factors up to 2^28, and there were none.
I sieved to 2^37-ish iirc, so ..good! My sieve doesn't have a bug.

The middle part primes (array of Yi) are of course known, - and I do add a comment to them. That comment will make them live longer than they would given the rate the Top5000 grows.

Those "Arithmetic progression (2,d=...)" primes are:As for the "Arithmetic progression (1,d=...)" primes, they owe their digital life to me: they were not eligible for Top5000 when they were found (they are vanity primes and were found exclusively to pump up "PrimeGrid credits"), and now that I made them members of the AP-3 chain - they are. PrimeGrid is credited as a project that found them. ~Half of the seed array of Xj is actually in Top5000 in expired state; to those I will merely add a comment.
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Old 2021-09-26, 01:25   #24
Dr Sardonicus
 
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With a = 33*2^2939064 - 5606879602425*2^1290000 - 1 and d = 33*2^2939063 - 5606879602425*2^1290000, a == 2 (mod 5) and d == 4 (mod 5), so a + 2*d is divisible by 5. Thus a, a + d, a + 2d is not a 3-term AP of primes.

5606879602425*2^1290000 - 1 No prime factors < 2^28.
33*2^2939063 - 1 prime (table lookup)
33*2^2939064 - 5606879602425*2^1290000 - 1
99*2^2939063 - 5606879602425*2^1290001 - 1
33*2^2939065 - 16820638807275*2^1290000 - 1 Divisible by 5

Estimates of base-ten logs of a - d, a, a + d:

(log(33)+2939063*log(2))/log(10) 884747.64066010744143595512237352714687
(log(33)+2939064*log(2))/log(10) 884747.94169010310541715033611242187136 (Matches value given here.)
(log(99) + 2939063*log(2))/log(10) 884748.11778136216109839241740143040198

Thus a - d, a, a + d is (presumably) a 3-term AP of primes. Luckily, a - d is an 884748 decimal digit number as well as a, so assuming this is the first term of an AP-3 the digit count is still good.

If a - 2d = 5606879602425*2^1290000 - 1 happens to be prime, then a - 2d, a - d, a, a + d is a 4-term AP.

But a - 2d "only" has 388342 decimal digits (base-ten log is 388341.44312776625375912068666134298687, approximately)
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Old 2021-09-26, 01:33   #25
Batalov
 
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Phi(4,2^7658614+1)/2

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I did test for the 4th AP term.
  • for the lower AP-3, 5^2 | A(4)
  • for the higher AP-3, I have not found a trivial factor, but it is composite.
  • A1 = 5606879602425*2^1290000-1 prime
  • A2 = 33*2^2939063-1 prime
  • A3 = 33*2^2939064-5606879602425*2^1290000-1 prime
  • A4 = 99*2^2939063-5606879602425*2^1290001-1 composite
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Old 2021-09-26, 14:02   #26
Dr Sardonicus
 
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Quote:
Originally Posted by Batalov View Post
I did test for the 4th AP term.
  • for the lower AP-3, 5^2 | A(4)
  • for the higher AP-3, I have not found a trivial factor, but it is composite.
  • A1 = 5606879602425*2^1290000-1 prime
  • A2 = 33*2^2939063-1 prime
  • A3 = 33*2^2939064-5606879602425*2^1290000-1 prime
  • A4 = 99*2^2939063-5606879602425*2^1290001-1 composite
So, A3 is the top, as I originally guessed and you confirmed. Obviously, I don't know which end is up.

If the common difference is d, the AP-3 is A3 - 2d, A3 - d, A3.

I assume A4 was checked for small factors, then subjected to a compositeness test that proved it to be composite, but did not reveal any factors.
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Old 2021-09-29, 13:12   #27
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In musing about APs of primes, I came upon a tantalizingly simple formula which, on the one hand very likely gives infinitely many longest-possible AP's starting with a given odd prime, but the other hand, is likely to be computationally useless for producing either AP's with large last term, or APs with large numbers of terms. Using the asymptotic formula in the Bateman-Horn conjecture shows that in either case, the prospects of success are vanishingly small.

This may serve to illustrate how numerical data can fail to reflect what theory predicts. It may also serve to illustrate the level of artistry that goes into hunting for AP's of three or more terms which consist entirely of primes.

Let k be a positive integer, pk the kth prime number, N = pk#, the product of the first k prime numbers. Let

fj(x) = pk+1 + N*j*x, j = 1 to pk+1 - 1.

Then the pk+1 - 1 polynomials fj(x) satisfy the hypotheses of the Bateman-Horn conjecture. (Note that with f0 = pk+1 however, f0 would be identically 0 (mod pk+1))

Thus, it would seem that for each positive integer k, there are infinitely many AP-pk+1's beginning with pk+1 and having common difference pk#*x for positive integer x. The first few such (with the smallest x for which the AP's consist entirely of primes) are

3, 3 + 2x, 3 + 4x (x = 1)

5, 5 + 6x, 5 + 12x, 5 + 18x, 5 + 24x (x = 1)

7, 7 + 30x, 7 + 60x, 7 + 90x, 7 + 120x, 7 + 150x, 7 + 180x (x = 5)

11, 11 + 210x, 11 + 420x, 11 + 630x, 11 + 840x, 11 + 1050x, 11 + 1260x, 11 + 1470x, 11 + 1680x, 11 + 1890x, 11 + 2100x (x = 7315048)

The asymptotic formula in the Bateman-Horn conjecture indicates that for x in the vicinity of a large number X, the probability that the pk+1 - 1 degree-1 functions all yield prime values is

\frac{c_{k}}{(\log(X))^{p_{k+1}-1}}

where for each k ck is a positive constant.

The reader may verify that according to this estimate. in the vicinity of 10^884000, values of x for which 2x + 3 and 4x + 3 are both prime will be so thin on the ground, a simple-minded numerical sweep is very unlikely to find any. This is true a fortiori for the longer AP's.

Likewise, as k increases, the smallest value of x for which pk+1, pk+1 + x*pk#, ..., pk+1 + (pk+1 - 1)*x*pk#

are all prime, is likely to be too large for a simple-minded sweep to find it in a reasonable length of time.

Last fiddled with by Dr Sardonicus on 2021-09-29 at 15:57 Reason: fignix optsy
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Old 2021-10-15, 16:21   #28
Cybertronic
 
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Default AP Update

Site is updated.
http://www.pzktupel.de/JensKruseAndersen/aprecords.htm


Incorrect or news or..., let me know.

Last fiddled with by Cybertronic on 2021-10-15 at 16:22
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Old 2021-10-16, 10:43   #29
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I hope , I have caught all AP-updates in order of time.



http://www.pzktupel.de/JensKruseAnde...ds.htm#records


This link is now valid for later updates.


Norman

Last fiddled with by Cybertronic on 2021-10-16 at 10:43
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