View Single Post
2008-01-14, 19:19   #8
gd_barnes

May 2007
Kansas; USA

11×929 Posts

Quote:
 Originally Posted by Jean PennĂ© 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^n-1 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^n-1 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^n-1 if n is odd. Almost immediately after, Yves Gallot discovered the firt four Liskovets-Gallot 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^n-1=composite for all n=even: k=39939 k*2^n-1=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^n-1 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^n-1) 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

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 odd-n conjecture of k=95283. Can you tell me how you arrived at 21 k-values 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 top-5000 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=32K-56K for the Sierp odd-n 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 2008-01-14 at 19:25