[QUOTE=Dougal;196742]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]
IF (k % 5 == 1) THEN GOTO next_k
IF (k % 7 == 1) THEN GOTO next_k
[/code]

what do i need to change in this script?[/QUOTE]

Hi Dougal,

you should delete these two lines. These were there because the previous base. I've attached the same script with full notes, adapted for your base.

Good luck! Willem.

First thing, we factor [B]base-1[/B]; for 129, 128=2^7. But note that 2's are already built into the script. So we remove (or comment out) those lines.

E.g. for R161, 160=2^5*5, again 2's are taken care of, so the line
[FONT=Courier New]IF (k % [B]5[/B] == 1) THEN GOTO next_k[/FONT] is the line to use.
For R88, 87=3*29, then we modify the lines into
[FONT=Courier New]IF (k % [B]3[/B] == 1) THEN GOTO next_k [/FONT]
[FONT=Courier New]IF (k % [B]29[/B] == 1) THEN GOTO next_k[/FONT]
I.e. for every prime factor p of b-1, we use the line
[FONT=Courier New]IF (k % [B]p[/B] == 1) THEN GOTO next_k[/FONT]

For Sierp script, replace -1 things into +1 and the skip conditions should be of form
[FONT=Courier New]IF (k % [B]p[/B] == (p-1)) THEN GOTO next_k[/FONT]
[FONT=Courier New][/FONT]
Did I get this right?

[QUOTE=Batalov;196810]First thing, we factor [B]base-1[/B]; for 129, 128=2^7. But note that 2's are already built into the script. So we remove (or comment out) those lines.

E.g. for R161, 160=2^5*5, again 2's are taken care of, so the line
[FONT=Courier New]IF (k % [B]5[/B] == 1) THEN GOTO next_k[/FONT] is the line to use.
For R88, 87=3*29, then we modify the lines into
[FONT=Courier New]IF (k % [B]3[/B] == 1) THEN GOTO next_k [/FONT]
[FONT=Courier New]IF (k % [B]29[/B] == 1) THEN GOTO next_k[/FONT]
I.e. for every prime factor p of b-1, we use the line
[FONT=Courier New]IF (k % [B]p[/B] == 1) THEN GOTO next_k[/FONT]

For Sierp script, replace -1 things into +1 and the skip conditions should be of form
[FONT=Courier New]IF (k % [B]p[/B] == (p-1)) THEN GOTO next_k[/FONT]
[FONT=Courier New][/FONT]
Did I get this right?[/QUOTE]

For the Riesel part: absolutely correct.
For the Sierpinsky: don't ask me, I don't know first thing about those!

Willem.
PS ok, ok, I'll adapt my script to include Sierpinsky if it so easy.

Well guys the system about trivial factors is the same, no matter if we are talking Riesel or Sierpinski numbers. All primefactors of b-1 (both on the Riesel and Sierpinski bases), tells us which numbers of k has trivial factors. Example given:

Riesel base 128, has b-1=127 (127 is a prime, so no further factors), which means that for all k mod 127 = 1 (128, 255, 383 etc.) there is trivial factors for all n's.

Sierpinski base 128, has b-1=127 (127 is a prime, so no further factors), which means taht for all k mod 127 = 126 (126, 253, 381 etc.) there is trivial factors for all n's.

A final note, on the Riesel side, for some reason that I'm not aware of, k=1 is always excluded from any testings. However this is not the case on the Sierpinski side.

Hope this helped :smile:

Regards

KEP

ok,so i started riesel base 129,i found primes for all k up to the conjectured lowest riesel k (14) apart from k=4.all were k eliminated long before n=1000,but k=4 still stands at n=20000,is there a reason for this,or have i just not found a prime for it yet?

[QUOTE=Dougal;197003]ok,so i started riesel base 129,i found primes for all k up to the conjectured lowest riesel k (14) apart from k=4.all were k eliminated long before n=1000,but k=4 still stands at n=20000,is there a reason for this,or have i just not found a prime for it yet?[/QUOTE]

The reason is that you just havent found a prime for it yet. It will most likely not yield a prime for maybe quite some time, since it is a very low weight k, with <1 % of the initial candidates at sieve depth p=0 remaining at p=1G. So with paitence and a gigantic sieving, maybe of a higher n-value of 10 or 100M, you will maybe eventually find a prime for this base. But remember many bases that has 1 k remaining at n=25K also has 1 k remaining at n=100K or higher :smile: Chances is of course always that you missed the prime, however if your computer can pass a Prime95 torture test, this is most likely not the case :smile:

KEP

thanks for clearing that up.

[QUOTE=Dougal;197003]ok,so i started riesel base 129,i found primes for all k up to the conjectured lowest riesel k (14) apart from k=4.all were k eliminated long before n=1000,but k=4 still stands at n=20000,is there a reason for this,or have i just not found a prime for it yet?[/QUOTE]

You spotted it well, there is a reason for this. The tip off is that k = 4 and 4 is a square.
All entries with an odd exponent have a factor 5.
For all entries with an even exponent n = 2m the following holds:
4*129^2m-1 = 2^2*129^2m-1 = (2*129^m)^2-1 = (2*129^m+1)(2*129^m-1).
So there is always a factor for k = 4.

Willem.

so this base is proven,or not?

[QUOTE=Dougal;197013]so this base is proven,or not?[/QUOTE]

Indeed it is, congratulations.

Willem.

[quote=Dougal;197013]so this base is proven,or not?[/quote]

Hi Dougal,

I know it sounds a little confusing but by (bad) luck, you happened to encounter a base that has a k that contains partial algebraic factors that make a full covering set to eliminate the k.

I know that probably sounds a little confusing but it means that the k is proven composite for all n-values and so is not considered.

If you want to get into the math behind how and why this happens, see [url]http://www.mersenneforum.org/showthread.php?t=11143[/url].


Gary