20081212, 11:34  #23 
I quite division it
"Chris"
Feb 2005
England
31×67 Posts 

20081212, 12:11  #24  
May 2007
Kansas; USA
27A2_{16} Posts 
Quote:
On your "edit2", it looks like you figured it out. If we leave k's with trivial factors in for PFGW to search, it will just keep finding the same factor for them overandover for all nvalues for those k's. It will never eliminate them. It takes little CPU time but makes for some huge results files and accomplishes nothing since the k's with trivial factors can be easily determined ahead of time. It would be like searching oddk for base 3. All would remain forever because all have a factor of 2. OK, I can come up with an easier one for you and simply explain which k's have trivial factors and hence should not be searched. Going to bed now so I'll round up something for you later today. Unfortunately I've already done most of the very easy Riesel bases < 125. But there should be plenty for Riesel bases 125 thru 200 that will be doable for you. We could do a Sierp base but Prof. Caldwell's group did a lot of work for bases <= 100 on it already. Of course we could do bases > 100 on the Sierp side if you'd like. Many bases with large conjectures are not any more difficult to start, they just take forever to prove. lol Gary Last fiddled with by gd_barnes on 20081212 at 12:17 

20081213, 07:16  #25  
May 2007
Kansas; USA
23642_{8} Posts 
Quote:
OK, here is what I have for you from easiest to hardest: Easy to get your feet wet: Riesel base 140 with a conjecture of k=46. The only trivial factors here are k==(1 mod 139), so effectively only k=1 is eliminated. So you can test k=2 thru 45. I'd suggest a PFGW script going up to n=3000. I suspect it will be proven by that point but it may not. Medium: Riesel base 154 with a conjecture of k=216. This has trivial factors where k==(1 mod 3) and (1 mod 17). I suggest 2 PFGW scripts: (1) One where k==(0 mod 3). (2) One where k==(2 mod 3). You can do this by a statement in the script: (1) b: from 0 thru 215 step 3 (2) b: from 2 thru 215 step 3 Instructions: 1. Run the above scripts up to n=3000 using the f100 and l parameters. 2. Look at the end of the results files, i.e. pfgw.out, for the k's that are remaining. You will see that all k==(1 mod 17) will remain. That's because they all have a trivial factor of 17. Those should be removed. What is left over are the TRUE k's that are remaining. 3. Use the true k's remaining from #2 to run PFGW up to n=5000 or 10000. Optionally you can sieve the k's at this point and start using LLR or Phrot. (That's it.) Hard: Riesel base 143 with a conjecture of k=1226. This has trivial factors where k==(1 mod 2) and (1 mod 71). I would suggest only one script that eliminates all of the odd k's for this one. After running it, then look in the results file and eliminate the k's==(1 mod 71) before continuing on. If you have any questions about this one, first refer to the process for base 154. If you're still unsure, get back with me. I cannot guarantee that there are not algebraic factors for some k's on these bases. If you have some stubborn k's that are perfect squares, then that is a good possibility, although still not a probability. Let me know you if you have any k's that are perfect squares remaining. I can step you through the process to check for them. Gary Last fiddled with by gd_barnes on 20081213 at 07:19 

20081213, 21:46  #26 
I quite division it
"Chris"
Feb 2005
England
4035_{8} Posts 
Thanks Gary. I'll check them out.
Primes for n=1 count, yes? 
20081214, 08:45  #27 
May 2007
Kansas; USA
2×3×19×89 Posts 

20081215, 13:53  #28  
I quite division it
"Chris"
Feb 2005
England
31·67 Posts 
Quote:
Last fiddled with by Flatlander on 20081215 at 13:54 

20081215, 19:26  #29 
Jan 2006
Hungary
2^{2}×67 Posts 
That means that you have proven the conjecture for base 154: Both 9 and 144 are squares, so for even n = 2m: 9*154^2m1 = (3*154^m + 1)(3*154^m  1).
And for the odd n there is always factor 5. Cheers, Willem. 
20081216, 00:32  #30 
I quite division it
"Chris"
Feb 2005
England
31×67 Posts 

20081216, 00:56  #31  
May 2007
Kansas; USA
2×3×19×89 Posts 
Quote:
Very good. Thanks for helping with the algebraic factors that make a full covering set Willem. Chris, just post all the primes here and I'll update the web pages to show the base proven. Good job. Edit: I just looked a little closer at your post. Were k=9 and 144 the ONLY k's remaining at n=10K? If so, then it is proven. Thanks, Gary Last fiddled with by gd_barnes on 20081216 at 00:57 

20081216, 01:11  #32  
I quite division it
"Chris"
Feb 2005
England
31·67 Posts 
Quote:


20091123, 12:47  #33 
Jan 2009
Ireland
2·3·31 Posts 
i was thinking about starting a new base,and i found that riesel base 129 has a conjectured smallest riesel k of 14.i though it would be very easy to prove and it would give me an idea of what im doing.so i start with this
Code:
SCRIPT DIM base, 129 DIM min_k, 1 DIM max_k, 13 DIM max_n, 1000 OPENFILEAPP k_file,pl_remain.txt OPENFILEAPP p_file,pl_prime.txt DIMS tmpstr DIM n DIM k_step, 1 DIM k IF (base % 2 == 1) THEN SET k_step, 2 IF (base % 2 == 1) && (min_k % 2 == 1) THEN SET min_k, min_k + 1 SET k,min_k  k_step LABEL next_k SET k, k + k_step IF (k > max_k) THEN GOTO END IF (k % 5 == 1) THEN GOTO next_k IF (k % 7 == 1) THEN GOTO next_k SET n, 0 LABEL next_n SET n, n + 1 PRP k*base^n1 IF (ISPRIME) THEN GOTO Prime_found IF (n < max_n) THEN GOTO next_n SETS tmpstr,%d*%d^n1;k;base; WRITE k_file,tmpstr GOTO next_k LABEL Prime_found SETS tmpstr,%d*%d^%d1;k;base;n; WRITE p_file,tmpstr GOTO next_k END 
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