20210907, 10:43  #12 
Sep 2002
Database er0rr
2×7×281 Posts 

20210924, 00:50  #14 
Sep 2002
Database er0rr
2·7·281 Posts 

20210924, 16:43  #16  
Feb 2017
Nowhere
5^{3}×41 Posts 
Quote:
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 3term AP of prime numbers." Is that correct? 

20210924, 17:33  #17 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×5×641 Posts 
Yes. I was too excited to speak in full sentences.

20210924, 19:30  #18 
Feb 2017
Nowhere
5^{3}·41 Posts 

20210925, 16:48  #20  
Feb 2017
Nowhere
5^{3}·41 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 PariGP 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:
I don't know whether a  2*d = 5606879602425*2^1290000  1 is prime or composite. Last fiddled with by Dr Sardonicus on 20210925 at 16:49 Reason: Formatting 

20210925, 18:36  #21  
Sep 2002
Database er0rr
2·7·281 Posts 
Quote:
Should that not be n=1..2? And... a=5606879602425*2^12900001 and d=33*2^29390635606879602425*2^1290000 So ad 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 20210925 at 19:06 

20210925, 22:32  #22 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×5×641 Posts 
I usually wait for a year or two to pass, then use some old scripts and some new for those AP3 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 reengineered 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 2YX, 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 / 0.275ish 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 X_{j} that have a huge common part that I can remove (in this case k*2^{1290000}1). I embrassinglyparallelize (simply separate processes) on Y_{i}. There is an interesting restriction on eligible Y_{i}, left as an thought experiment to the reader. The only input for that minipuzzle is that in array of X_{j} 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 specialproject sieves "EMsieve", before you ask. This proofcode 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: 2YX 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 presieved 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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