20140203, 04:49  #1 
Sep 2002
Database er0rr
2^{2}×883 Posts 
Why NeRDs_360360?
Why did I choose 10^36036010^k1?
360360 = 2*2*2*3*3*5*7*11*13 For small primes p, 10^((p1)*a)==1 (mod p), and so 10^36036010^k1 is not divisible by p. Consequently, after sieving, there is about 15% of the range left and we expect to find about 3 primes in the provable range k=90090360360. 
20140203, 08:19  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{3}·7^{3} Posts 
It would have been better to chose a>1290000*log_{10}2 for 10^a10^k1. With a=360360, the found primes will be swept away in about a year by the TwinGenial deluge. a=17#, perhaps?

20140203, 08:33  #3 
Feb 2003
773_{16} Posts 

20140203, 17:08  #4 
Sep 2002
Database er0rr
2^{2}·883 Posts 
42320 candidates were left in the range 90000360360.
Chuck Lasher is crunching 3/19 of this. Thomas, you are crunching 1/19. I crunched some. The rest was put up, ready for others to crunch  1 or 2 weeks per file folks. 
20140203, 17:10  #5  
Sep 2002
Database er0rr
2^{2}·883 Posts 
Quote:
I have exponents 388080 and 471240 sieved. Last fiddled with by paulunderwood on 20140203 at 17:29 

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