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2021-09-07, 10:43   #12
paulunderwood

Sep 2002
Database er0rr

23×491 Posts

Quote:
 Originally Posted by Puzzle-Peter I gave it a try and I found an AP-6 with 10593 decimal digits: N=(2,738,129,976+n*56,497,325)*24499#+1 for n=0..5
Congrats, Peter, for this AP-6. I am looking forward to your submissions to the UTM top20 and JK Andersen's list.

 2021-09-23, 22:14 #13 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23×1,201 Posts New record AP-3 with a 807,954-digit max element: 6325241166627*2^1290000-1+d*(1455*2^2683953-6325241166627*2^1290000), d=0,1,2
2021-09-24, 00:50   #14
paulunderwood

Sep 2002
Database er0rr

23·491 Posts

Quote:
 Originally Posted by Batalov New record AP-3 with a 807,954-digit max element: 6325241166627*2^1290000-1+d*(1455*2^2683953-6325241166627*2^1290000), d=0,1,2
This is huge! Surely it will top the second table for top20 APs, with almost unassailable points. Nice ingenuity!

 2021-09-24, 06:43 #15 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23·1,201 Posts Shoot! That didn't last long... 884748 digits is now the new top.
2021-09-24, 16:43   #16
Dr Sardonicus

Feb 2017
Nowhere

24×11×29 Posts

Quote:
 Originally Posted by Batalov Shoot! That didn't last long... 884748 digits is now the new top.
Congratulations, looks like you've found a huge 3-term AP of primes!

I'm a bit confused by your usage "new top." [Chekov voice]The top of what?[/Chekov voice]

I'm inferring the meaning "largest top term of any known 3-term AP of prime numbers." Is that correct?

 2021-09-24, 17:33 #17 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23·1,201 Posts Yes. I was too excited to speak in full sentences.
2021-09-24, 19:30   #18
Dr Sardonicus

Feb 2017
Nowhere

24·11·29 Posts

Quote:
 Originally Posted by Batalov Yes. I was too excited to speak in full sentences.
Quite understandable! Perhaps you are in need of sedation...

2021-09-25, 05:59   #19
Puzzle-Peter

Jun 2009

12648 Posts

Quote:
 Originally Posted by Batalov Shoot! That didn't last long... 884748 digits is now the new top.

WOW
Congratulations!

2021-09-25, 16:48   #20
Dr Sardonicus

Feb 2017
Nowhere

24×11×29 Posts

Actually, you may have done better than you realize. If

a = 33*2^2939064 - 5606879602425*2^1290000 - 1 and d = 33*2^2939063 - 5606879602425*2^1290000, then

a - d = 33*2^2939063 - 1. Just out of curiosity, I wrote a Pari-GP script to check this number for factors up to 2^28, and there were none.

So, I looked up a table, and what to my wondering eyes should appear...

List of primes k · 2^n − 1 for k < 300
Quote:
 k = 33 2, 3, 6, 8, 10, 22, 35, 42, 43, 46, 56, 91, 102, 106, 142, 190, 208, 266, 330, 360, 382, 462, 503, 815, 1038, 1651, 1855, 1858, 1992, 2232, 4462, 4726, 5475, 6702, 9710, 10931, 11503, 20552, 22291, 26575, 35931, 39271, 51326, 57695, 63115, 67182, 109848, 128635, 136971, 173110, 174182, 373446, 585400, 598248, 674050, 773030 [1000000] 1115902 L488, 1813526 L621, 2118570, 2212971, 2215291, 2499883, 2513872, 2939063, 3242126 L3345, 6894190 L4965
So, a + n*d, n = -1 to 3 is a four term AP of primes. Although a - d is only about half as big as a, it still might rank high as a first term of 4-term AP of primes.

I don't know whether a - 2*d = 5606879602425*2^1290000 - 1 is prime or composite.

Last fiddled with by Dr Sardonicus on 2021-09-25 at 16:49 Reason: Formatting

2021-09-25, 18:36   #21
paulunderwood

Sep 2002
Database er0rr

23×491 Posts

Quote:
 Originally Posted by Dr Sardonicus Actually, you may have done better than you realize. If ...... So, a + n*d, n = -1 to 3 is a four term AP of primes. Although a - d is only about half as big as a, it still might rank high as a first term of 4-term AP of primes.

Should that not be n=-1..2? And...

a=5606879602425*2^1290000-1 and d=33*2^2939063-5606879602425*2^1290000

So a-d is negative

a+d = 33*2^2939063 - 1

Now I can see DrS's mistake and Serge/Ryan's ingenuity.

Last fiddled with by paulunderwood on 2021-09-25 at 19:06

 2021-09-25, 22:32 #22 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23×1,201 Posts I usually wait for a year or two to pass, then use some old scripts and some new- for those AP-3 weightligting approaches. Clearly I must give kudos to David Broadhurst - because he invented this trickery and found three hits of this kind; I'd thought of how he did it and then re-engineered all of it once I saw his idea. If X and Y are some primes (and they have a big enough gcd(X+1,Y+1) ), then we can add 2Y-X, for which we want to have properties: a) can prove primality, b) can devise a relatively fast sieve. Because of a) and the particular choice of seeds, we cannot reach higher than 1.29M / 0.275-ish bits. I am ready to run CHG if we get a very high hit near the very end. Next technical problem is to sieve deep and reasonably fast. And to satisfy this criterion it really helps to have a large array of Xj that have a huge common part that I can remove (in this case k*21290000-1). I embrassingly-parallelize (simply separate processes) on Yi. There is an interesting restriction on eligible Yi, left as an thought experiment to the reader. The only input for that mini-puzzle is that in array of Xj is that _all_ of them by prior design are -1 (mod 3). By having a good array of X[5528] and Y[154], I have >850,000 "very" raw candidates. I would rather want > 10^6 (because of probability of a random odd ~1M digit number being prime is ~1/950,000, if I got it right, did I? *) but that's what we had at the moment. I was prepared for testing all sieved pool and getting no hits; the ME was ~1. Two hits was reasonably lucky and we are not finished yet... maybe we'll get a third. I call all my special-project sieves "EMsieve", before you ask. This proof-code simply means - there was a sieve and I wrote/tinkered it. After sieving, the value candidate list was ~51,000 to be run on Ryan's cluster. ___________ * footnote: 2Y-X as constructed and are not only odd (for that reason I already doubled the probability), they are also not divisible by 3 (because both X and Y are -1(mod 3) ), so I can triple the probability (these "very raw" candidates are natively pre-sieved for 2 and 3) and then perhaps it is not so lucky that we have found two hits and is not all that unlikely that we might find third hit.

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