[quote=henryzz;153011]the script that was attached in post 3 is the one i use
it is easier to understand IMO[/quote]
Thanks. I'd missed that one, I'll take a look. :smile:

[quote=Flatlander;152990]Okay, obviously more complicated than I thought!
I had noticed the trivial factors mentioned on your site. I had assumed that if I test every k the trivial ks would be illiminated by WinPFGW (with -f) because a factor would be found quickly. Then after testing to n=1000, say, any remaining k would be 'genuine'. If the conjectured k was just a few hundred then very little time would be wasted.
Am I misunderstanding something here?
[B][edit: yes I think I am!][/B]
[B][edit2: Yes, I know I am! We're looking for primes not factors!][/B]

Probably for the best! I'm still to grasp this 'covering set' stuff.

I hadn't noticed so I've learned [I]something [/I]today. :smile:

Maybe the best way would be if you choose a base for me and tell me exactly what to do. I know you've just come back from a business trip and have lots to do, so I'll wait patiently. :smile:

Thanks
Chris[/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 over-and-over for all n-values 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 odd-k 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

[quote=Flatlander;152990]Maybe the best way would be if you choose a base for me and tell me exactly what to do. I know you've just come back from a business trip and have lots to do, so I'll wait patiently. :smile:

Thanks
Chris[/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

Thanks Gary. :smile: I'll check them out.
Primes for n=1 count, yes?

[quote=Flatlander;153220]Thanks Gary. :smile: I'll check them out.
Primes for n=1 count, yes?[/quote]

Yes; n=1 counts. n=0 does not.

[quote=gd_barnes;153141]...
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[/quote]

For R base 154 I have just k=9 and k=144 left at n=10k.

[QUOTE=Flatlander;153423]For R base 154 I have just k=9 and k=144 left at n=10k.[/QUOTE]

That means that you have proven the conjecture for base 154: Both 9 and 144 are squares, so for even n = 2m: 9*154^2m-1 = (3*154^m + 1)(3*154^m - 1).
And for the odd n there is always factor 5.

Cheers, Willem.

[quote=Siemelink;153484]That means that you have proven the conjecture for base 154: Both 9 and 144 are squares, so for even n = 2m: 9*154^2m-1 = (3*154^m + 1)(3*154^m - 1).
And for the odd n there is always factor 5.

Cheers, Willem.[/quote]

Okay, thanks. :smile:
I'll start Riesel base 143.

[quote=Flatlander;153423]For R base 154 I have just k=9 and k=144 left at n=10k.[/quote]

[quote=Siemelink;153484]That means that you have proven the conjecture for base 154: Both 9 and 144 are squares, so for even n = 2m: 9*154^2m-1 = (3*154^m + 1)(3*154^m - 1).
And for the odd n there is always factor 5.

Cheers, Willem.[/quote]

[quote=Flatlander;153523]Okay, thanks. :smile:
I'll start Riesel base 143.[/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

[quote=gd_barnes;153525]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[/quote]
Yes, the only ks remaining.

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^n-1

IF (ISPRIME) THEN GOTO Prime_found
IF (n < max_n) THEN GOTO next_n

SETS tmpstr,%d*%d^n-1;k;base;
WRITE k_file,tmpstr
GOTO next_k

LABEL Prime_found
SETS tmpstr,%d*%d^%d-1;k;base;n;
WRITE p_file,tmpstr
GOTO next_k

END
[/code]

what do i need to change in this script?