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Old 2008-01-14, 21:17   #9
Jean Penné
 
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May 2004
FRANCE

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Quote:
Originally Posted by gd_barnes View Post
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
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