[quote=MrOzzy;144710]I was playing around a little bit with everything and I got this when I ran srsieve using the output from pgfw: "removed candidate sequence 30*117^n-1 from the sieve"
.. but it doesn't say why ...[/quote]


k=30 has a trivial factor of 29 for all n-values for Riesel base 117. Hence any sieving quickly eliminates all possible candidates.

For this base, all k==(1 mod 2), which has a trivial factor of 2, and k==(1 mod 29), which has a trivial factor of 29, would not be considered in the conjecture.

Whenever srsieve eliminates a candidate sequence, there is always some trivial set of factors or factor that make all n-values composite. We don't consider the k's where only one factor make all n-values composite. But if it is more than one, the smallest k-value would be the conjectured k-value until proven.


Gary

[quote=CedricVonck;144717]Hi! can anyone please explain what covering does???

I plugin following numbers:

[code]
104
2
-1
500000
3 5 13
Checking k*2^n-1 sequence for exponent=104, bound for primes in the covering set
=500000, bound for k is 3
Examining primes in the covering set: 3,5,17,8191,2731,53,157,1613
And their orders: 2,4,8,13,26,52,52,52[/code]

Base = 104
2 = exponent
-1 = Riesel
500000 = k ??
??

Thank you![/quote]


I believe you are asking what the covering.exe program does. Correct?

If so, it attempts to determine the lowest k-value that makes all n-values composite, which is what we call the 'conjectured k-value'. k-values with a single trivial factor for all n-values are excluded.

I believe you have mixed up the entries for the program and so it was spitting out info. for base 2.

The first entry is the 'exponent', which represents the 'period' in which a covering set of factors could repeat. I would generally suggest using 24, 72, or 144 for that. There's no way to know specifically but any relatively low value < 200 with a lot of factors of 2's and 3's is a good choice.

The second entry is the base (b) in the form k*b^n-1 or k*b^n+1. You chose 2 in your example although it appears you intended base 104.

You have the 3rd entry correct as representing Riesel (-1) or Sierp (+1).

The 4th entry is the highest factor you wish the program to consider. 10e3 or 25e3 is a good starting point. The higher it is, the longer the program runs but if you make it too low, it might not find the lowest conjectured k-value that has a covering set of factors.

The 5th entry is the highest k-value you wish the program to consider as the conjectured k. For a base this high, 1e6 should be sufficient but if it doesn't find anything, try 10e6. You chose '3 5 13' and the program picked it up as '3' hence it did not find a k-value with a full covering set. Riesel base 2 should be k=509203, i.e. the former Riesel Sieve project.

BTW, for Riesel base 104, the conjecture would be k=4 with a covering set of {3 5}, which is easily proven because:
k=1 has a trivial factor of 103 for all n
k=2 has a prime at n=68
k=3 has a prime at n=1

See if you can make the program spit out k=4 as the conjecture for Riesel base 104 and k=509203 as the conjecture for Riesel base 2 by messing around with some of the above parameters. If you get that far, it'll be straightforward for most bases after that.


Gary

Gary,

Thank you very much!
I will try as soon as possible.

Regards
Cedric

[CODE]
72
104
-1
1000000 (10e6)
3 5 13
Checking k*104^n-1 sequence for exponent=72, bound for primes in the covering se
t=1000000, bound for k is 3
Examining primes in the covering set: 3,5,7,67,163,29,373,3571,17,1657,4153,19,1297,2791,18397,37,291853,73
And their orders: 2,2,2,3,3,4,4,6,8,8,8,9,9,9,9,18,36,72[/CODE]

Sorry for the stupid questions, but the "covering set" is this the list of number that I should test or definetly not???

Again sorry ... I am trying and will to understand it.
Regards
C.

[quote=CedricVonck;144877][code]
72
104
-1
1000000 (10e6)
3 5 13
Checking k*104^n-1 sequence for exponent=72, bound for primes in the covering se
t=1000000, bound for k is 3
Examining primes in the covering set: 3,5,7,67,163,29,373,3571,17,1657,4153,19,1297,2791,18397,37,291853,73
And their orders: 2,2,2,3,3,4,4,6,8,8,8,9,9,9,9,18,36,72[/code]

Sorry for the stupid questions, but the "covering set" is this the list of number that I should test or definetly not???

Again sorry ... I am trying and will to understand it.
Regards
C.[/quote]

Your input is not correct. Here is a good example (using the winxp dos console):
[code]
c:\>covering.exe
144 [enter]
121 [enter]
1 [enter]
100000 [enter]
10000000 [enter]

Checking k*121^n+1 sequence for exponent=144, bound for primes in the covering set=100000, bound for k is 10000000
Examining primes in the covering set: 61,7,19,37,7321,13,1117,17,10657,20113,51329,97,241,1777,73,40177,577,3457
And their orders: 2,3,3,3,4,6,6,8,12,12,16,24,24,24,36,36,48,144
**************** Solution found ****************
1300068
**************** Solution found ****************
720586
**************** Solution found ****************
312640
**************** Solution found ****************
4192
**************** Solution found ****************
360
[/code]

- 144 is the "exponent". It is used to calculate the different periods used for the primes considered. You get the best results by picking a number with a lot of small devisors. 144 is a very good one for this. In short, just pick 144 and you'll find a good solution in 99.9% of the cases.
- 121 is the base (b) in k*n^b+/-1
- 1 is the choice between sierpinski or riesel. It can be 1 or -1.
- 100000 is the upper boundary for the primes considered for the covering set. Only primes smaller as this number are considered for the covering set. Incresing this number slows down the covering.exe program significantly.
- 10000000 is the upper boundary of the k considered. If you can't find a solution this is the number you should increse (unless you are looking for a covering set for a k smaller as a current known one).

A covering set is a set of primes which makes all solutions for k*n^b+/-1 composite for a specific k (so no prime can be found).

One more question from me: How do you calculate the number of digits for a nubmer in the form k*n^b+/-1?

digits = integer(log(k)+log(b)*n+1)
for k*base^n+/-1

Now that I have the conjectures linked from here:
[URL]http://mersenneforum.org/showpost.php?p=152196&postcount=176[/URL]
is it just a matter of changing the script that Kep sent me and running WinPFGW?
[code]SCRIPT // autobodgified by scriptify.pl
: pre-declare scriptify's globals
DIMS PCtmpString
OPENFILEAPP noprimesfile,HiCands.rb3
OPENFILEAPP primefile,LoPRPs.rb3
DIM mink,600000002
DIM maxk,610000000
DIM minn,1
DIM maxn,1000
DIM bignum
DIM k
DIM n
SET k,mink
SET n,minn
LABEL label
IF !(k>maxk) THEN GOTO PCnotif_a
GOTO end
LABEL PCnotif_a
SET bignum,k*3^n-1
PRP bignum
IF !(ISPRIME) THEN GOTO PCnotif_b
: synthesise fprintf
SETS PCtmpString,%d*3^%d-1;k;n
WRITE primefile,PCtmpString
SET k,k+(2)
PRINT k
SET n,1
GOTO label
GOTO PCendelse_c
LABEL PCnotif_b
IF !(n<maxn) THEN GOTO PCnotif_d
SET n,n+(1)
GOTO label
GOTO PCendelse_e
LABEL PCnotif_d
: synthesise fprintf
SETS PCtmpString,%d*3^n-1;k
WRITE noprimesfile,PCtmpString
SET n,1
SET k,k+(2)
GOTO label
LABEL PCendelse_e

LABEL PCendelse_c

LABEL end
END[/code]Those GOTOs hurt my brain. :huh: But I can se the variables/constants I would need to change, and possible also the k increment.
Do I assume that if there is no reference to work on a conjecture in this forum (or Gary's stats) that that conjecture is available? (As advised, I would chose a low conjecture.)
If I am correct do you have any recommendations for a nice juicy Reisel that I could start that might be proved in days/weeks?

[quote=Flatlander;152972]Now that I have the conjectures linked from here:
[URL]http://mersenneforum.org/showpost.php?p=152196&postcount=176[/URL]
is it just a matter of changing the script that Kep sent me and running WinPFGW?
[code]SCRIPT // autobodgified by scriptify.pl
: pre-declare scriptify's globals
DIMS PCtmpString
OPENFILEAPP noprimesfile,HiCands.rb3
OPENFILEAPP primefile,LoPRPs.rb3
DIM mink,600000002
DIM maxk,610000000
DIM minn,1
DIM maxn,1000
DIM bignum
DIM k
DIM n
SET k,mink
SET n,minn
LABEL label
IF !(k>maxk) THEN GOTO PCnotif_a
GOTO end
LABEL PCnotif_a
SET bignum,k*3^n-1
PRP bignum
IF !(ISPRIME) THEN GOTO PCnotif_b
: synthesise fprintf
SETS PCtmpString,%d*3^%d-1;k;n
WRITE primefile,PCtmpString
SET k,k+(2)
PRINT k
SET n,1
GOTO label
GOTO PCendelse_c
LABEL PCnotif_b
IF !(n<maxn) THEN GOTO PCnotif_d
SET n,n+(1)
GOTO label
GOTO PCendelse_e
LABEL PCnotif_d
: synthesise fprintf
SETS PCtmpString,%d*3^n-1;k
WRITE noprimesfile,PCtmpString
SET n,1
SET k,k+(2)
GOTO label
LABEL PCendelse_e

LABEL PCendelse_c

LABEL end
END[/code]Those GOTOs hurt my brain. :huh: But I can se the variables/constants I would need to change, and possible also the k increment.
Do I assume that if there is no reference to work on a conjecture in this forum (or Gary's stats) that that conjecture is available? (As advised, I would chose a low conjecture.)
If I am correct do you have any recommendations for a nice juicy Reisel that I could start that might be proved in days/weeks?[/quote]


I'm not familiar with the language used here but it appears that it would only work for base 3. Do you know how to figure k's with trivial factors? As a random example, on Riesel base 67, k's where k==(1 mod 2), k==(1 mod 3), and k==(1 mod 11) would not be considered with factors of 2, 3, and 11 respectively. The above script only eliminates k==(1 mod 2), i.e. odd n, which are the only k's with a trivial factor for base 3.

BTW, in case you haven't noticed, to come up with the k's that have trivial factors, you subtract 1 from the base and take the prime factors of that. It then becomes all k's that are [1 mod (factor)] for all of the prime factors. For example on Riesel base 67:

67-1=66

prime factors of 66 = 2*3*11

Therefore Riesel base 67 has the following k's with trivial factors:
k==(1 mod 2)
k==(1 mod 3)
k==(1 mod 11)

For Sierp base 67, it's the same only the mod is slightly different, i.e.
k==(1 mod 2)
k==(2 mod 3)
k==(10 mod 11)


If you can get the k's with trivial factors down pat and change the above program to account for them, you'll be headed in the right direction.

I won't talk about k's with algebraic factors that make a full covering set right now. :smile: Base 3 doesn't have any that I'm aware of so it was not an issue in this program. Also on bases that are perfect squares, you need to refer to the square root of the base for work that would have effectively already been done. Riesel base 25 is an excellent example of how tricky it can be to avoid double-work on bases that are perfect square. Effectively a ton of work on it had already been done by the base 5 project.

If you still want to tackle a new base after reading this, I can likely give you a couple of good bases to go after that shouldn't take too long to prove. If they aren't perfect squares nor do they have k's with algebraic factors that make a full covering set, it's not bad.


Gary

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!]
[edit2: Yes, I know I am! We're looking for primes not factors!]
[/B]
[quote]I won't talk about k's with algebraic factors that make a full covering set right now.[/quote]Probably for the best! I'm still to grasp this 'covering set' stuff.

[quote]BTW, in case you haven't noticed, to come up with the k's that have trivial factors, you subtract 1 from the base and take the prime factors of that...[/quote]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

the script that was attached in post 3 is the one i use
it is easier to understand IMO