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 2005-09-17, 02:21 #1 Citrix     Jun 2003 62716 Posts Exponential prime search Exponential prime search This is not a distributed computing project but a top 10 tables of series that produce n consecutive primes. Rules:- 1.Sequences published in a journal are stored indefinitely else they are only stored as long they remain in the top 10 list 2.If 2 sequences have equal length, the series producing larger primes will have a higher rank 3.The general term of the series must be a*p^2+b*p+c, where a,b,c are integers and p must be a prime. 4.Note only the consecutive prime terms will be counted to calculate the length. 5.1 will not be counted as prime, but 2 will be. 6.I have the right to make rules, in case I have left out some details. :) Example p=2 and a=1,b=-1 and c=1 Produces 2,3,7,43 So length =4 Code: Rank # Starting p a b c Length Found by Date 1 3 1 -2 2 4 Fermat ? 2 2 1 -1 1 4 Euclid-Mullin ? Post your best sequences below. A software for sieving this type of primes is needed. Help with the above table is also needed. Any ideas on how to make tables in the forum. Last fiddled with by Citrix on 2005-09-17 at 02:28
 2005-09-17, 22:54 #2 grandpascorpion     Jan 2005 Transdniestr 50310 Posts Citrix, I also think it's quite interesting to find the first (even first 10) occurences of a polynomial / sequence length. As such, there should be an indication of how far sieving has occured so that effort isn't wasted in finding the minimum p for a given length. You might want to open this up to higher degree polynomials too.
 2005-09-17, 22:56 #3 grandpascorpion     Jan 2005 Transdniestr 503 Posts For f(n+1)= f(n)^2 - f(n) +1 (Euclid-Mullin), length=5 (results complete thru 10^9) rank len starting p a b c Found by Date 1 5 964146163 1 -1 1 Fred Schneider 09/16/2005 2 5 777378883 1 -1 1 Fred Schneider 09/16/2005 3 5 579155767 1 -1 1 Fred Schneider 09/16/2005 4 5 436352197 1 -1 1 Fred Schneider 09/16/2005 5 5 349058599 1 -1 1 Fred Schneider 09/16/2005 6 5 301526317 1 -1 1 Fred Schneider 09/16/2005 7 5 279861877 1 -1 1 Fred Schneider 09/16/2005 8 5 259231939 1 -1 1 Fred Schneider 09/16/2005 9 5 202100809 1 -1 1 Fred Schneider 09/16/2005 10 5 74448109 1 -1 1 Fred Schneider 09/15/2005 First occurence: 13115172 Fred Schneider 09/15/2005 5 is the max sequence value through p=1e9
 2005-09-17, 23:08 #4 grandpascorpion     Jan 2005 Transdniestr 7678 Posts For f(n+1)= f(n)^2 - f(n) -1 (Euclid-Mullin), length=5 (results complete thru 10^9) rank len starting p a b c Found by Date 1 5 912824551 1 -1 -1 Fred Schneider 20050917 2 5 893597069 1 -1 -1 Fred Schneider 20050917 3 5 805008689 1 -1 -1 Fred Schneider 20050917 4 5 800970271 1 -1 -1 Fred Schneider 20050917 5 5 799764341 1 -1 -1 Fred Schneider 20050917 6 5 793392217 1 -1 -1 Fred Schneider 20050917 7 5 793285429 1 -1 -1 Fred Schneider 20050917 8 5 775295392 1 -1 -1 Fred Schneider 20050917 9 5 756048229 1 -1 -1 Fred Schneider 20050917 10 5 736633275 1 -1 -1 Fred Schneider 20050917 First occurence: 2848669 Fred Schneider 20050917 5 is the max sequence value through p=1e9
 2005-09-18, 00:48 #5 grandpascorpion     Jan 2005 Transdniestr 503 Posts I'm looking at the following polynomials Folks, I'm using a simple script in PARI to find these numbers. If anyone wants, I can send it to them. I'm not looking at either "Euclid-Mullin" relations for the time being: f(n+1)=f(n)^2+f(n)+/- 1 any more. I think it's more fruitful to look at some of the polynomials at: http://mathworld.wolfram.com/Prime-G...olynomial.html From that page, I'm checking : f(n+1)=f(n)^2+f(n)+41 (a la Euler) through 10^11. So far, I have found 2 length 6 prime sequences. f(n+1)= 36*f(n)^2-810*f(n)+2753 (a la Fung and Ruby) through 10^9.
 2005-09-21, 12:58 #6 grandpascorpion     Jan 2005 Transdniestr 503 Posts f(n+1)= 36*f(n)^2-810*f(n)+2753 (a la Fung and Ruby) (results complete through 10^9) r l starting p a b c Found by Date 1 6 974646599 36 -810 2753 Fred Schneider 20050921 2 6 903232999 36 -810 2753 Fred Schneider 20050921 3 6 806881279 36 -810 2753 Fred Schneider 20050921 4 6 772617023 36 -810 2753 Fred Schneider 20050920 5 6 716182837 36 -810 2753 Fred Schneider 20050920 6 6 673119389 36 -810 2753 Fred Schneider 20050920 7 6 597237691 36 -810 2753 Fred Schneider 20050920 8 6 428162263 36 -810 2753 Fred Schneider 20050919 9 6 245011747 36 -810 2753 Fred Schneider 20050919 10 6 143888027 36 -810 2753 Fred Schneider 20050919 First occurence: 22144249 6 is the max sequence value through 10^9. There are 11 up to that limit.
 2005-09-21, 13:09 #7 grandpascorpion     Jan 2005 Transdniestr 503 Posts f(n+1)=f(n)^2+f(n)+41 (a la Euler) (results complete through 1.3e9) r l starting p a b c Found by Date 1 6 1184418737 1 1 41 Fred Schneider 20050920 2 6 1164167891 1 1 41 Fred Schneider 20050920 3 6 1103804483 1 1 41 Fred Schneider 20050920 4 6 644062919 1 1 41 Fred Schneider 20050920 5 6 610038257 1 1 41 Fred Schneider 20050920 6 6 428366489 1 1 41 Fred Schneider 20050920 7 6 320142863 1 1 41 Fred Schneider 20050920 8 6 196805633 1 1 41 Fred Schneider 20050919 9 6 169177649 1 1 41 Fred Schneider 20050919 10 6 167667607 1 1 41 Fred Schneider 20050919 First occurence: 16355881 6 is the max sequence value through 10^9. There are 13 up to that limit.
 2005-09-21, 15:23 #8 grandpascorpion     Jan 2005 Transdniestr 50310 Posts Sorry, that should read: 6 is the max sequence value through 1.3e9.
2005-09-21, 17:58   #9
grandpascorpion

Jan 2005
Transdniestr

1F716 Posts

Quote:
 Originally Posted by grandpascorpion rank len starting p a b c Found by Date 1 5 912824551 1 -1 -1 Fred Schneider 20050917 2 5 893597069 1 -1 -1 Fred Schneider 20050917 3 5 805008689 1 -1 -1 Fred Schneider 20050917 4 5 800970271 1 -1 -1 Fred Schneider 20050917 5 5 799764341 1 -1 -1 Fred Schneider 20050917 6 5 793392217 1 -1 -1 Fred Schneider 20050917 7 5 793285429 1 -1 -1 Fred Schneider 20050917 8 5 775295392 1 -1 -1 Fred Schneider 20050917 9 5 756048229 1 -1 -1 Fred Schneider 20050917 10 5 736633275 1 -1 -1 Fred Schneider 20050917 First occurence: 2848669 Fred Schneider 20050917 5 is the max sequence value through p=1e9
Ugh, please disregard this one. There are spurious results.
In addition, if you want a f(x+1)= -1 + product of terms 1 to x, the correct polynomial to use is: f(x+1)=f(x)^2+f(x)-1.

I'll search on that front.

Last fiddled with by grandpascorpion on 2005-09-21 at 17:59

 2005-09-23, 01:57 #10 Citrix     Jun 2003 32·52·7 Posts I have searched other series but have not had any luck. So I guess 6 is the record then. I estimate to find 7 primes in a row you will have to search a depth of x^2. where x is the average interval between seeds that produce 6 primes in a row.
 2005-09-23, 03:38 #11 grandpascorpion     Jan 2005 Transdniestr 1111101112 Posts I suspect 7 is possible. There's definitely an analogue between prime producing polynomials I linked to earlier and longer sequences. I found a few more good candidates at: http://www.shyamsundergupta.com/canyoufind.htm (question 13) I have been searching on: x^2 - 99999x + 1116427201 and found a sequence of 6 primes beginning at only 1851329***. There are 20 such primes through 350 million. *** this brings up a side question: what is the earliest prime combination at which a length 6 sequence occurs? On the sieving front, I find that the better you can sieve the results, the worse the hit ratio for longer sequences. For instance, using a sieve based on 510510 (product of primes 2 thru 17), for the "Euclid-Mullin+1" function, I found there were only 3200 values v where 510510x+v could lead to a sequence of more than 4. But, I could only find a single length 6 sequence through 17 billion (at just over a billion). On the other hand, the "prime-producing polynomials" search set can't be reduced further than phi (92160) for this set and perform much better. Inituitively, this makes sense, you have many more candidates to try with such "un-smooth" polynomials. =================================================== If you don't mind sharing, what technique did you use to pare down your search?

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