20071219, 13:31  #1 
May 2004
FRANCE
2^{2}×5×29 Posts 
Bases 2 & 4 reservations/statuses/primes
Hello Gary,
Congrats for organizing this search! I am working alone on the Riesel base 4 candidates for a while, so I wish to reserve all the divisible by 3 k's candidates, up to n = 131072 base 4 for now... The remaining k's I wish to reserve are : 9519, 13854, 14361, 16734, 19401 and 20049, which are marked as available on your reservation page. I already eliminated all the others... Regards, Jean 
20071219, 14:34  #2  
May 2007
Kansas; USA
2×5,261 Posts 
Quote:
Hi Jean, Thank you and welcome to the effort! It's nice to have you on board. The k's are available and I will show them reserved by you today. It looks like you're now the official base 4 person; both Riesel and Sierpinski! If you have the time, come vote in our poll in the lounge here. Gary 

20071220, 03:32  #3  
Sep 2004
UVic
46_{16} Posts 
Quote:
64494 is at 1770506 and still no prime 

20071220, 03:54  #4  
May 2007
Kansas; USA
2×5,261 Posts 
Quote:
Thanks for the update Tcadigan. I'll put you down for testing completion of n=885.2K base 4. I had previously noticed that you were searching one k for Sierp base 4 so sorry about the omission there. If you'd like to pick up a smaller search effort that also LLR's as fast as base 2, we've got 55 k's for Sierp base 16, all tested to just n=25K that are ready for testing. At some future time, we also want to start on Riesel base 16 from the beginning. Thanks, Gary 

20080108, 06:27  #5  
May 2007
Kansas; USA
24432_{8} Posts 
Riesel/Sierp base 2/4 reservations/statuses/primes
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  #6 
May 2004
FRANCE
244_{16} 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  #7  
May 2007
Kansas; USA
2×5,261 Posts 
Quote:
I responded in the "Conjecture 'R Us" thread to this. Gary 

20080110, 19:29  #8  
May 2004
FRANCE
1001000100_{2} Posts 
Quote:
k = 9519 is up to n = 427816, no prime. k = 14361 is up to n = 318510, no prime. k = 19401 is up to n = 262578, no prime. k = 20049 is up to n = 265144, no prime. I thought to test these k's up to 524288 base 2 if no prime found below... However, to day, only 9519 is running... If you have more available computing power, you may continue on the last three k's, but if so, let me know about your progress. Regards, Jean 

20080110, 21:42  #9  
May 2007
Kansas; USA
2·5,261 Posts 
Quote:
Here is what I show for the test limits for the k's that you're unreserving: k=13854 and 16734; I show no work so they are still at n=100K base 4. k=19464 per your status report of 1/7/08 is at n=137112 base 2. Is that correct? I'll coordinate getting the 3 unreserved k's tested. If they aren't reserved in the new few days, I'll reserve and test them. Since you did some testing on k=19464, do you have a sieve file for that one or for any of the k's? If so, I will post them on the website. Gary 

20080110, 21:54  #10 
Mar 2006
Germany
3^{2}×5^{2}×13 Posts 
got sievefiles from jean for 14361, 19401 and 20049 and sieving further!

20080110, 23:21  #11  
May 2007
Kansas; USA
2·5,261 Posts 
Quote:
You can ignore my last post. Edit: It's all correct now. You might check the Riesel base 4 and base 16 reservations to make sure I have it correct. Gary Last fiddled with by gd_barnes on 20080111 at 03:08 Reason: Figured it out; will straight out res. 

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