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2008-01-14, 21:17   #9
Jean Penné

May 2004
FRANCE

56010 Posts

Quote:
 Originally Posted by gd_barnes 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
In the definitions of these four conjectures, the k multipliers must be odd!
For example : 46902*2^n+1 is the same as 23451*2^(n+1)=+1 and if n is odd, n+1 is even, so, you are testing an even exponents candidate!
So, the 8 even k's remaining are relevant to the even n conjecture, and not to the odd n one...

Also, I tested k = 46623 up to n = 79553 and found a prime, so me are almost matching now... I am terminating to gather my results, and will send them to this thread as soon as possible.
Regards,
Jean