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 2008-01-08, 20:33 #2 Jean Penné     May 2004 FRANCE 24·5·7 Posts Liskovets-Gallot 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^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 first 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 Last fiddled with by LaurV on 2019-06-14 at 16:51 Reason: fixed few typos