Starting new bases - Page 4

Siemelink · 2009-11-23, 22:11

Quote:

Originally Posted by Dougal

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

what do i need to change in this script?

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.

Batalov · 2009-11-23, 22:35

First thing, we factor base-1; 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
IF (k % 5 == 1) THEN GOTO next_k is the line to use.
For R88, 87=3*29, then we modify the lines into
IF (k % 3 == 1) THEN GOTO next_k
IF (k % 29 == 1) THEN GOTO next_k
I.e. for every prime factor p of b-1, we use the line
IF (k % p == 1) THEN GOTO next_k

For Sierp script, replace -1 things into +1 and the skip conditions should be of form
IF (k % p == (p-1)) THEN GOTO next_k

Did I get this right?

Siemelink · 2009-11-23, 23:00

Quote:

Originally Posted by Batalov

First thing, we factor base-1; 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
IF (k % 5 == 1) THEN GOTO next_k is the line to use.
For R88, 87=3*29, then we modify the lines into
IF (k % 3 == 1) THEN GOTO next_k
IF (k % 29 == 1) THEN GOTO next_k
I.e. for every prime factor p of b-1, we use the line
IF (k % p == 1) THEN GOTO next_k

For Sierp script, replace -1 things into +1 and the skip conditions should be of form
IF (k % p == (p-1)) THEN GOTO next_k

Did I get this right?

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.

KEP · 2009-11-24, 15:53

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

Regards

KEP

Dougal · 2009-11-25, 18:19

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?

KEP · 2009-11-25, 19:03

Quote:

Originally Posted by Dougal

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?

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

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

KEP

Dougal · 2009-11-25, 19:07

thanks for clearing that up.

Siemelink · 2009-11-25, 19:33

Quote:

Originally Posted by Dougal

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?

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.

Dougal · 2009-11-25, 19:41

so this base is proven,or not?

Siemelink · 2009-11-25, 20:42

Quote:

Originally Posted by Dougal

so this base is proven,or not?

Indeed it is, congratulations.

Willem.

gd_barnes · 2009-11-27, 09:34

Quote:

Originally Posted by Dougal

so this base is proven,or not?

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 http://www.mersenneforum.org/showthread.php?t=11143.

Gary

2009-11-25, 19:41	#42
Dougal Jan 2009 Ireland 272₈ Posts	so this base is proven,or not? Last fiddled with by Dougal on 2009-11-25 at 19:59 Reason: grammar

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2009-11-23, 22:35	#35
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2⁴×593 Posts	First thing, we factor base-1; 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=2^55, again 2's are taken care of, so the line IF (k % 5 == 1) THEN GOTO next_k is the line to use. For R88, 87=329, then we modify the lines into IF (k % 3 == 1) THEN GOTO next_k IF (k % 29 == 1) THEN GOTO next_k I.e. for every prime factor p of b-1, we use the line IF (k % p == 1) THEN GOTO next_k For Sierp script, replace -1 things into +1 and the skip conditions should be of form IF (k % p == (p-1)) THEN GOTO next_k Did I get this right?

2009-11-24, 15:53	#37
KEP Quasi Admin Thing May 2005 1706₈ Posts	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 Regards KEP

2009-11-25, 18:19	#38
Dougal Jan 2009 Ireland 10111010₂ Posts	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?

2009-11-25, 19:07	#40
Dougal Jan 2009 Ireland 2·3·31 Posts	thanks for clearing that up.