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Old 2008-01-14, 19:19   #8
gd_barnes's Avatar
May 2007
Kansas; USA

100111111010112 Posts

Originally Posted by Jean Penné View Post

On the 23 May 2006, Citrix warned us, in the Sierpinski base 4 thread, about this problem :

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...


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:

  k     comments/prime
17937   prime n=53927
29322   even
35787   prime n=36639
43398   even
46902   even
47598   even
61158   even; prime n=48593
62307   prime n=44559
67542   even
67758   even
75183   prime n=35481
83418   even
85434   even


Last fiddled with by gd_barnes on 2008-01-14 at 19:25
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