20120229, 02:18  #166 
Jan 2005
Sydney, Australia
5×67 Posts 
I'm having a problem with the PRPnetclient.exe
It started with an Intel E8400 Wolfdale CPU based system. Now it has occurred on an Intel Q8200 quadcore system also. Both are running Windows 7 Ultimate 64bit edition. Both computers were happily running NPLB prpnet on port 2000. Now they throw a message "prpclient.exe has stopped working" and below this heading it says "A problem caused the program to stop working correctly. Windows will close the program and notify you if a solution is available." It doesn't. This occurs when the start batch file is initiated. On the quad only 2 of the cores crash, so it leaves 2 instances (in separate subdirectories as per Lennart's batch file) running. I tried reinstalling the client using the install batch file right over the top of the existing files. Didn't help. I'm stuck. Any ideas? 
20120229, 02:27  #167  
"Lennart"
Jun 2007
2^{5}×5×7 Posts 
Quote:
Run install # core then update # core then start and it should work. This is only a fast way to get it running again. You don't have to save the resultfiles if you don't need them. Lennart 

20120229, 06:56  #168 
Jan 2005
Sydney, Australia
5·67 Posts 
Thanks Lennart, that worked for both machines.

20121119, 13:55  #169 
Jan 2005
Sydney, Australia
5·67 Posts 
Question: Do all Prime Numbers occur next to a number divisible by 6?
Eg 17 is next to 18. 
20121119, 13:59  #170 
Apr 2010
Over the rainbow
5·11·43 Posts 
yep, 1 mod 6 , 1 mod 6 or 5 mod 6.
meaning a prime will be of tthe form (x*6) plus or minus one 
20121120, 08:47  #171 
Feb 2012
Prague, Czech Republ
A2_{16} Posts 

20121120, 08:51  #172 
Romulan Interpreter
Jun 2011
Thailand
5×11×157 Posts 

20121120, 15:46  #173 
Jan 2005
Sydney, Australia
335_{10} Posts 

20121120, 23:14  #174 
Jan 2006
deep in a whileloop
2×7×47 Posts 
just for giggles I tested the first 100,000 primes and this is true for
3 < prime <= 239737 However there is no discernible pattern for whether the mod6 'seed' is 1 above or 1 below. I sent you the worksheet 
20121121, 00:21  #175 
May 2007
Kansas; USA
2·3^{2}·5·113 Posts 
There is no need to test. It can be easily proven that all primes > 3 must be either p==(1 mod 6) or p==(5 mod 6). Proof:
p==(0 mod 6) is always divisible by 2 and 3 p==(2 mod 6) is always divisible by 2 p==(3 mod 6) is always divisible by 3 p==(4 mod 6) is always divisible by 2 This leaves p==(1 mod 6) and p==(5 mod 6) as the only possible primes mod 6. 
20121121, 00:53  #176 
"Daniel Jackson"
May 2011
14285714285714285714
599 Posts 
If you convert the primes into base6, you'll notice that, except for 2 and 3, they all end in 1 or 5. That's easy proof.
Last fiddled with by Stargate38 on 20121121 at 00:53 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Some old stuff  Batalov  Miscellaneous Math  1  20170127 04:56 
Stuff for sale  fivemack  Lounge  12  20110612 11:28 
useful stuff  paulunderwood  Linux  3  20051205 22:18 
Free stuff...  Xyzzy  Software  6  20041006 13:35 
Extra Stuff...  Xyzzy  Lounge  11  20030915 23:22 