20080108, 06:27  #1  
May 2007
Kansas; USA
2^{6}·3·53 Posts 
Riesel/Sierp base 2 evenk/evenn/oddn testing
Quote:
Robert, Citrix, Axn, Geoff, Masser, or other people with intimate knowledge of the math's behind the conjectures, here is what I would propose (if it has not been proposed already) that Jean agrees with: In order to prove the Sierpinski conjecture for any base, all Generalized Fermat #'s as well as very Fermat #'s, i.e. any form that reduces to 2^n+1, should be excluded from those conjectures. In a nutshell, here is what I'm working towards as determining the Riesel/Sierpinski conjecture proofs on the "Conjectures 'R Us" web pages: 1. All generalized Fermat #'s (18*18^n+1) and very Fermat #'s (65536*4^n+1 or 65536*16^n+1) will be excluded from the conjectures. 2. Any k that obtains a full covering set in any manner from ALGEBRAIC factors will be excluded. In many instances, this includes k's where there is a partial covering set of numeric factors (or a single numeric factor) and a partial covering set of algebraic factors that combine to make a full covering set. 3. All k below the lowest k found to have a NUMERIC covering set must have a prime including multiples of the base (MOB) but excluding the conditions in #1 and #2 above. I will add MOB and exclude GFn's in the near future on the pages. There are very few MOB if GFNs are excluded. 4. All n must be >= 1. All input and opinions are welcome. Citrix, if you think this is misplaced, feel free to move it around somewhere. Thanks, Gary Last fiddled with by gd_barnes on 20080108 at 06:32 

20080108, 20:33  #2 
May 2004
FRANCE
2^{4}×5×7 Posts 
LiskovetsGallot numbers are beautiful for us!
Hi,
On the 23 May 2006, Citrix warned us, in the Sierpinski base 4 thread, about this problem : http://www.primepuzzles.net/problems/prob_036.htm To be short, the Liskovets assertion is : There are some k values such that k*2^n+1 is composite for all n values of certain fixed parity, and some k values such that k*2^n1 is composite for all n values of certain fixed parity. It is almost evident that these k values must be searched only amongst the multiples of 3 (the assertion is trivial if 3 does not divide k) : If k == 1 mod 3, then 3  k*2^n1 if n is even, and 3  k*2^n+1 if n is odd. If k == 2 mod 3, then 3  k*2^n+1 if n is even, and 3  k*2^n1 if n is odd. Almost immediately after, Yves Gallot discovered the first four LiskovetsGallot numbers ever produced : k*2^n+1=composite for all n=even: k=66741 k*2^n+1=composite for all n=odd: k=95283 k*2^n1=composite for all n=even: k=39939 k*2^n1=composite for all n=odd: k=172677 And Yves said that "I conjecture that 66741, 95283, 39939, ... and 172677 are the smallest solutions for the forms  having no algebraic factorization (such as 4*2^n1 or 9*2^1)  but I can't prove it." For several reasons, I think it would be interesting for us to coordinate the search in order to prove these four conjectures : 1) They involve only k values that are multiples of 3, so the success will no more be depending of the SoB or Rieselsieve one. 2) For the n even Sierpinski case, only k = 23451 and k = 60849 are remaining, with n up to more than 1,900,000 that is to say there are only two big primes to found, then the conjecture is proven! 3) For the n even Riesel (third line above) there are only four k values remaining : 9519, 14361, 19401 and 20049, although the search is only at the beginning! 4) For the two remaining n odd Sierpinski / Riesel (which can be tranlated as base 4, k even, and doubling the Gallot values : 190566 for k*4^n+1, 345354 for k*4^n1) I began to explore the problem, by eliminating all k's yielding a prime for n < 4096, eliminating the perfect square k values for Riesel, eliminating the MOB that are redondant, etc... Finally, there were 42 k values remaining for +1, 114 for 1, and after sieving rapidly with NewPGen, and LLRing up to ~32K, I have now 21 values remaining for +1 and 37 values for 1. I would be happy to know your opinion about all that... Regards, Jean Last fiddled with by LaurV on 20190614 at 16:51 Reason: fixed few typos 
20080110, 19:15  #3  
May 2007
Kansas; USA
2^{6}·3·53 Posts 
Quote:
I responded in the "Conjecture 'R Us" thread to this. Gary 

20080110, 21:44  #4 
"Erling B."
Dec 2005
3^{4} Posts 
I will probably reserve more of 2036 base 9 number (if no prime will be in my range) but can then slow down on LLR testing to free upp
one older AMD PC to help on LiskovetsGallot numbers. Maybe not wery fast PC for LLR but will do its job 24/7. Back to this later. 
20080113, 04:21  #5 
Jun 2003
11000011110_{2} Posts 
Jean,
I looked at the parity thing you mentioned. It sounds interesting, perhaps we can go to higher bases once we are done with 256. I also looked at all k's 132, that are supper fast for all possible parities up to 2^100. There were some really low weight sequences and some high weight sequences. Do you think we might have similar luck like RPS or 15K with some of these high weight or low weight sequences and possibly find a 10M prime? What do you think? If you or any one else is interested please let me know. Thanks Last fiddled with by Citrix on 20080113 at 04:22 
20080113, 05:08  #6 
Jun 2003
2·3^{3}·29 Posts 
Low weight Reisel
K=29 not included as it is already low weight. Code:
k*2^m*(2^i)^n1 // last column is the candidates left, approx estimate of weight 19 9 17 163 19 9 34 332 15 8 35 244 15 28 35 243 19 9 51 496 21 15 55 480 19 2 63 549 21 45 65 416 23 39 65 554 19 43 68 664 21 47 69 613 27 15 69 623 15 1 70 456 15 7 70 384 15 15 70 481 15 21 70 453 15 29 70 483 15 35 70 483 15 43 70 482 15 49 70 483 15 57 70 388 15 63 70 485 15 31 74 457 
20080113, 07:26  #7  
May 2004
FRANCE
560_{10} Posts 
Quote:
Presently, I am more interested in trying to prove one or more of these four mathematical conjectures, than to find large primes (although it may be a subproduct). Moreover, I hope we will not need to find a 10M prime before proving at least one of them! So, perhaps it would be better to restrict this project to base 4 for now, and not to dissipate our efforts too much... But, indeed there is place for other similar projects! Regards, Jean 

20080114, 19:19  #8  
May 2007
Kansas; USA
2^{6}·3·53 Posts 
Quote:
Jean, Is there any reason that we are not testing even k's (i.e. multiples of the base) with these conjectures? If a k is even but is not divisible by 4, it yields a different set of factors and prime than any other odd k. I am testing the Sierp oddn conjecture of k=95283. Can you tell me how you arrived at 21 kvalues remaining at n=32K? I have now tested up to n=56K. I just now finished sieving up to n=200K and am starting LLRing now. At n=56K, I show 22 odd k's and 8 even k's remaining that are not redundant with other k's remaining; for a total of 30 k's. At n=32K, I showed 26 odd k's and 9 even k's remaining; for a total of 35 k's. I checked the top5000 site for previous smaller primes and there were none for these k's so I wonder why you have less k's remaining than me. Here are the k's that I show remaining at n=32K, both odd and even, and primes that I found for n=32K56K for the Sierp oddn conjecture: Code:
k comments/prime 2943 9267 17937 prime n=53927 24693 26613 29322 even 32247 35787 prime n=36639 37953 38463 39297 43398 even 46623 46902 even 47598 even 50433 53133 60357 60963 61137 61158 even; prime n=48593 62307 prime n=44559 67542 even 67758 even 70467 75183 prime n=35481 78753 80463 83418 even 84363 85287 85434 even 91437 93477 93663 Thanks, Gary Last fiddled with by gd_barnes on 20080114 at 19:25 

20080114, 21:17  #9  
May 2004
FRANCE
2^{4}×5×7 Posts 
Quote:
For example : 46902*2^n+1 is the same as 23451*2^(n+1)=+1 and if n is odd, n+1 is even, so, you are testing an even exponents candidate! So, the 8 even k's remaining are relevant to the even n conjecture, and not to the odd n one... Also, I tested k = 46623 up to n = 79553 and found a prime, so me are almost matching now... I am terminating to gather my results, and will send them to this thread as soon as possible. Regards, Jean 

20080114, 22:42  #10  
May 2007
Kansas; USA
2^{6}×3×53 Posts 
Quote:
Gary 

20080115, 18:37  #11 
May 2007
Kansas; USA
2^{6}×3×53 Posts 
I have now run the Sierp oddn conjectures up to n=115K and will be continuing on to n=200K sometime next week after completing some sieving for conjectures team drive #1 and a couple of other things. Below are the k's left at n=32K with primes found for n=32K115K.
I decided to leave the even k's in because in effect it is testing the even conjecture for all k < 95282/2=47641 and I had already sieved them. That should save a lot of effort on that side. Code:
k comments/prime 2943 prime n=108041 9267 17937 prime n=53927 24693 26613 prime n=89749 29322 even; prime n=91367 32247 35787 prime n=36639 37953 38463 prime n=58753 39297 43398 even; prime n=72873 46623 prime n=79553 46902 even 47598 even; prime n=105899 50433 53133 60357 60963 prime n=73409 61137 61158 even; prime n=48593 62307 prime n=44559 67542 even 67758 even 70467 75183 prime n=35481 78753 prime n=63761 80463 83418 even; prime n=80593 84363 85287 85434 even 91437 93477 prime n=63251 93663 prime n=82317 Total of 14 odd k's and 4 even k's remaining. So...based on this effort by itself, here are the statuses of the base 2 Sierp oddn and evenn cojectures: Oddn: 14 k's remaining at n=115K from odd k's above. k's remaining: 9267 24693 32247 37953 39297 50433 53133 60357 61137 70467 80463 84363 85287 91437 Evenn: 47641<k<66741: still needs to be tested. k<=47641: 4 k's remaining at n=115K from even k's above. k's remaining converted to oddk: 23451 33771 33879 42717 Edit: I just now realized that it was already stated that only k=23451 and 60849 are remaining on the evenn side as a result of the Sierp base 4 project. OK, NEXT time I'll remove the even k's from my testing. Ergh! Gary Last fiddled with by gd_barnes on 20080115 at 18:47 
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