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 2016-11-28, 17:10 #12 sweety439   "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 1110011000102 Posts https://www.utm.edu/staff/caldwell/preprints/2to100.pdf There is another researcher for the Sierpinski numbers base b (for 2<=b<=100), but he or she thinks the k's with full or partial algebraic factors (e.g. 2500*16^n+1) as Sierpinski number, but excluding the GFNs in the research. (The list in the final page has many errors, e.g. k=4 is remaining in the Sierpinski base 53 problem, but the list only lists {1816, 1838, 1862, 1892})
2016-11-29, 07:39   #13
gd_barnes

"Gary"
May 2007
Overland Park, KS

2×19×317 Posts

Quote:
 Originally Posted by sweety439 https://www.utm.edu/staff/caldwell/preprints/2to100.pdf There is another researcher for the Sierpinski numbers base b (for 2<=b<=100), but he or she thinks the k's with full or partial algebraic factors (e.g. 2500*16^n+1) as Sierpinski number, but excluding the GFNs in the research. (The list in the final page has many errors, e.g. k=4 is remaining in the Sierpinski base 53 problem, but the list only lists {1816, 1838, 1862, 1892})
I am familiar with Prof. Caldwell's work and I printed off a copy of his paper from long ago shortly after this project started. In case you were not aware, his name is Chris Caldwell and he is the administrator of the prime database or better known as the top-5000 primes site. I read much of the paper. He and I briefly corresponded about the inclusions and exclusions of the conjectures. He only looked at the Sierp side. CRUS chose to exclude k's with all or partial algebraic factors from becoming the conjectures of various bases. This only had minimal impact on the Sierp side but it had a huge impact on the Riesel side. We felt that the two sides should be consistent in their approach but the Riesel bases are the one main reason that we excluded algebraic's like we did. If you were to allow k's with partial or full algebraic factors to become a Riesel conjecture many of such conjectures would be k=4 or k=9. More specifically please take a look at this page here:
http://www.noprimeleftbehind.net/cru...es-powers2.htm

If we were to allow partial or algebraic factors to become the conjecture of each base on this page, bases 2 even-n, 4, 16, 64, 256, and 1024 would have a conjecture of k=9 and all would be quickly proven. The power-of-2 bases are some of our most interesting and to have nearly half of them quickly proven would be mathematically uninteresting. Also if you look at the main Riesel conjectures page you will see a relatively high percentage of bases that have algebraic exclusions. Many of these conjectures, especially among the lowest bases, would become uninteresting. A second reason that we chose to do this is that we had an individual early in our project who created a program called covering.exe. The program was very effective and efficient at coming up with the lowest k for each base that contained a full "numeric" covering set.

To answer a question from another post that you had: The base 2 even-n and odd-n searches are technically a completely different effort and are referred to as the Liskovets-Gallot conjectures. See here:
http://www.primepuzzles.net/problems/prob_036.htm

CRUS chose to include them in our project not long after our project started for three reasons: (1) They were interesting and appeared provable in our lifetimes. [1 of the 4 bases was proven just in the last year with some huge primes and the other three have <= 4 k's remaining.] (2) They were relatively similar to our project and involve our lowest base 2, which is the easiest for us to search. (3) We could combine the sieving effort with base 4 and other powers-of-2 bases. The conjectures are not the same as base 4. They only include k where k==3 mod 6 and the trivial k's are different. There are many overlapping primes and k's remaining but if you look closely at our pages you will see how different that they are.

Last fiddled with by gd_barnes on 2016-11-29 at 08:01

2016-11-29, 07:53   #14
gd_barnes

"Gary"
May 2007
Overland Park, KS

2·19·317 Posts

Quote:
 Originally Posted by sweety439 72590 > 269^2. Thus, n^(2^23)+1 is composite for all 2<=n<=269. 269 > 16^2. Thus, n^(2^24)+1 is composite for all 2<=n<=16. Therefore, the test limit for the GFNs are: b=2: 2^32 (https://web.archive.org/web/20151125...et/fermat.html) b=4: 2^31 (the same as b=2) b=6: 2^27 (https://web.archive.org/web/20151122...net/GFN06.html) b=8: algebra factorization b=10: 2^23 (https://web.archive.org/web/20151122...net/GFN10.html), but now 2^24 b=12: 2^23 (https://web.archive.org/web/20151122...net/GFN12.html), but now 2^24 b=14: 2^24 b=16: 2^30 (the same as b=2) 18<=b<=268: 2^23 270<=b<=72588: 2^22
Thank you for the detail. I will update my GFN page to show all CRUS bases as having been searched to n=2^22.

In doing this research I think you can see by how high the lowest base with a GFN prime is as a reason why GFNs must be excluded from our searches. Although it cannot be proven, there are unlikely to be any more GFN primes for b<=1030. Even if one did somehow pop up sometime in the distant future for one of our over 500 bases, for each specific base the chances are extremely small and mathematicians believe that the number of GFN primes for each base is finite.

Last fiddled with by gd_barnes on 2016-11-29 at 08:03

 2016-11-29, 13:56 #15 sweety439   "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2×7×263 Posts However, there may be infinitely many primes of the form 1*2^n+1, just as that there may be infinitely many primes of the form 3*2^n+1. I think the number of both type of primes are infinite. For example, 2^(2^34)+1 may be prime. Last fiddled with by sweety439 on 2016-11-29 at 14:02
2016-11-29, 18:46   #16
gd_barnes

"Gary"
May 2007
Overland Park, KS

2×19×317 Posts

Quote:
 Originally Posted by sweety439 However, there may be infinitely many primes of the form 1*2^n+1, just as that there may be infinitely many primes of the form 3*2^n+1. I think the number of both type of primes are infinite. For example, 2^(2^34)+1 may be prime.
Consider doing more research before making such statements.

 2016-12-06, 20:52 #17 henryzz Just call me Henry     "David" Sep 2007 Liverpool (GMT/BST) 10111101010012 Posts Should CRUS consider the situation where there is a full covering set for a k even if odd or even ns are also eliminated with a partial algebraic factorization? As long as there is a full covering set surely the partial algebraic factorization shouldn't matter when selecting a Reisel/Sierpinski k.
2016-12-07, 06:47   #18
gd_barnes

"Gary"
May 2007
Overland Park, KS

2×19×317 Posts

Quote:
 Originally Posted by henryzz Should CRUS consider the situation where there is a full covering set for a k even if odd or even ns are also eliminated with a partial algebraic factorization? As long as there is a full covering set surely the partial algebraic factorization shouldn't matter when selecting a Reisel/Sierpinski k.
The short answer to your question is no. Here is the best way that I can put it: The conjectured k's must be full fixed numeric covering sets. No GFN's or algebraic factors can be considered when determining the conjectured k. We have many k's on many bases with even n having algebraic factors and odd n having a single trivial factor or a covering set of factors meaning that they are proven as composite. (Some k's have even n with a trivial factor and odd n with algebraic factors but they are relatively rare.) Such k's are shown on the pages as not being considered in testing but they do not constitute the conjecture.

Last fiddled with by gd_barnes on 2016-12-07 at 06:48

 2016-12-07, 08:15 #19 henryzz Just call me Henry     "David" Sep 2007 Liverpool (GMT/BST) 32×673 Posts It is possible to have a full fixed numeric covering set for a k as well as having having algebraic factors. I suppose the algebraic factors mean that it is more likely that there is a covering set due to two composites having potential factors.
 2016-12-08, 03:24 #20 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23·439 Posts The covering factors of the algebraic composites will "leak through" into the "normal" covering set. This happens in the CRUS set frequently. Take the meta-series R14, R44, R74, R104, etc -- R(30q+14) -- they all have CK=4. So, you could say that for each of them, even n's are eliminated algebraically, or you can say that they are eliminated by factor 3; and that odd n's are eliminated by factor 5. Potae-toe, po-tah-toh.
2016-12-10, 13:01   #21
henryzz
Just call me Henry

"David"
Sep 2007
Liverpool (GMT/BST)

32·673 Posts

Quote:
 Originally Posted by Batalov The covering factors of the algebraic composites will "leak through" into the "normal" covering set. This happens in the CRUS set frequently. Take the meta-series R14, R44, R74, R104, etc -- R(30q+14) -- they all have CK=4. So, you could say that for each of them, even n's are eliminated algebraically, or you can say that they are eliminated by factor 3; and that odd n's are eliminated by factor 5. Potae-toe, po-tah-toh.
I had begun thinking similarly. We are effectively finding a covering set that covers three sequences, 4*14^(2n-1)-1, 2*14^n-1, 2*14^n+1. In this sort of form, 3 will always divide either even or odd ns in 2*14^n-1 and the opposite in 2*14^n+1 since 3 doesn't divide 2*14^n. This produces a covering set for the algebraic factors.
ns in 4*14^(2n-1)-1 are divisible by 5.
It feels a little artificial to me for the covering set for the even ns be based upon the covering sets for the two algebraic factors overlapping like that. It would be much nicer if there was a full covering set for all the three forms above.
How many bases would have their CK extended if that logic was used?

 2016-12-15, 19:45 #22 sweety439   "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2·7·263 Posts Another question: Why you does not consider 8 and 32 as Sierpinski numbers base 128? All numbers of the form 8*128^n+1 and 32*128^n+1 are composite, and they have no algebra factors. Last fiddled with by sweety439 on 2016-12-15 at 19:46

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