Reserving the following bases (all new) to n=25K.

Reisel 729

Sierp 729

The above bases will all be complete in 4 days. I will report them 1 per day so Gary doesn't kill me.

Primes: (by b then k)
[code]2*509^1+1
2*524^1+1
3*524^2+1
2*539^7+1
2*554^1+1
3*554^1+1
2*569^29+1
2*584^111+1
3*584^1+1
2*599^13+1
2*614^1+1
3*614^18+1
2*629^1+1
2*644^1+1
3*644^1+1
2*659^1+1
2*674^5+1
3*674^3+1
2*689^3+1
2*704^1+1
3*704^1+1
2*719^1+1
2*734^3+1
3*734^1+1
2*749^1+1
2*764^1189+1
3*764^1+1
2*779^1+1
2*794^3+1
3*794^1+1
2*809^1+1
2*824^7+1
3*824^1+1
2*839^5+1
2*854^1+1
3*854^4+1
2*884^5+1
3*884^3+1
3*914^12+1
2*929^99+1
2*944^1+1
3*944^1+1
2*959^5+1
2*974^1+1
3*974^7+1
2*989^1+1
3*1004^19+1
2*1019^1+1
[/code]Sequences remaining:
[code]2*869^n+1
2*899^n+1
2*914^n+1
2*1004^n+1
[/code]GFN primes: (by b)
[code]554^4+1
584^2+1
614^256+1
644^2+1
674^2+1
704^2+1
764^2+1
824^1024+1
884^8+1
914^4+1
1004^2+1
[/code]GFNs with no primes to 32768 (2^15):
[code]524^n+1
734^n+1
794^n+1
854^n+1
944^n+1
974^n+1[/code]

Thanks Tim for the explanation on GFN's.

Are you testing the 4 k's left from your ck=4 post. If not, I have a PRPNET run going for n=2.5K-25K on a bunch of k's and I could add those to it.

They were:
2*869^n+1
2*899^n+1
2*914^n+1
2*1004^n+1

[quote=MyDogBuster;199130]Are you testing the 4 k's left from your ck=4 post. If not, I have a PRPNET run going for n=2.5K-25K on a bunch of k's and I could add those to it.

They were:
2*869^n+1
2*899^n+1
2*914^n+1
2*1004^n+1[/quote]
Nope I wasn't planning on working them, feel free to take those higher. Just be aware that taking such high bases to a large n value is rather slow. The tests grow large very quickly when you add 3 digits with each n! 1004^25K > 2^250K

Sierp
 

Proven with k=8:

primes:

2*1022^727+1
3*1022^1+1
4*1022^6+1
5*1022^5+1
6*1022^1+1
7*1022^36+1

Let me know if I'm missing anything.

reserving riesel base 589
it is currently at ~3k with 8 ks remaining
i will post details when i have taken it as far as i can

it seems riesel base 589 is a slightly interesting base
i have found interesting properties about two of the remaining ks

216*589^n-1
216 is 6^3
216*589^(3*n)-1=(6*589^n-1)*(36*589^(2*n)+6*589^n+1)
this should mean that i can remove all n that are 0 mod 3 for this k
does anyone have any hints on how to do this?

648*589^n-1
648 is 2^3*3^4
i dont think we can learn anything from this but someone may know better

edit:
i have already removed ks 144 and 324 because they are squares so even ns have algebraic factors and odd ns have a trivial factor of 5

[quote=henryzz;199317]216*589^n-1
216 is 6^3
216*589^(3*n)-1=(6*589^n-1)*(36*589^(2*n)+6*589^n+1)
this should mean that i can remove all n that are 0 mod 3 for this k[/quote]
Sounds right to me.
[quote=henryzz;199317]does anyone have any hints on how to do this?[/quote]
Here's what I'd probably do: load the sieved file into a spreadsheet program, add a column next to each one that is equal to n mod 3, sort by that column, delete all rows that are 0, and then resave it back with everything how it was. (OpenOffice.org Calc's Text to Columns and advanced settings for saving as a .csv would be very helpful here) Or you could make a Perl script to do it.

[quote=Mini-Geek;199324]Or you could make a Perl script to do it.[/quote]
Like this one: :smile:[code]#!/usr/bin/perl
#usage: file k a b
#e.g. 'mod 589 216 0 3' looks in 589.txt, removes values with k=216 and n=0 mod 3, and writes to 589-out.txt

open(IN, $ARGV[0].'.txt');
open(OUT, '>' .$ARGV[0].'-out.txt');
$k = $ARGV[1];
$a = $ARGV[2];
$b = $ARGV[3];
$removecount = 0;

while(<IN>)
{
$line = $_;
chomp($line);
@linearray = split(/ /,$line);
$thisk = @linearray[0];
#if this isn't the right k, then we don't want to remove it, so print it out
if ($thisk != $k) {
print OUT $line."\n";
next;
}
$thisn = @linearray[1];
#if there is nothing after a space on this line (usually the first line of a NewPGen-format file), then we don't want to remove it, so print it out
if ($thisn == '') {
print OUT $line."\n";
next;
}
$thisa = $thisn % $b;

#if the modular value isn't right, then we don't want to remove it, so print it out
if ($thisa != $a) {
print OUT $line."\n";
next;
}

#if we've got here, then everything matches, and we remove it by not doing anything, we also increment a counter so we can see how many were removed
$removecount++;
}
print "removed $removecount line(s)";

close(IN);
close(OUT);[/code]To use it, (say it's named mod.pl and the NewPGen-format file is 589.txt) run 'perl mod.pl 589 216 0 3', and it will write everything except k=216 values whose n=0 mod 3 to 589-out.txt. It's had only minimal testing, but I don't see any reason for it not to work as long as you use a NewPGen format file.

[quote=henryzz;199317]it seems riesel base 589 is a slightly interesting base
i have found interesting properties about two of the remaining ks

216*589^n-1
216 is 6^3
216*589^(3*n)-1=(6*589^n-1)*(36*589^(2*n)+6*589^n+1)
this should mean that i can remove all n that are 0 mod 3 for this k
does anyone have any hints on how to do this?

648*589^n-1
648 is 2^3*3^4
i dont think we can learn anything from this but someone may know better

edit:
i have already removed ks 144 and 324 because they are squares so even ns have algebraic factors and odd ns have a trivial factor of 5[/quote]


Ah, I see you happen to have chosen a tricky base. It's what keeps the conjectures interesting. The Sierp side is usually easier with few k's that have algebraic factors. Technically you don't need to worry about testing GFNs.

That is correct on removing n==(0 mod 3) for k=216.

To generalize your statements, for ANY k that is a perfect square on ANY base, you can remove n==(0 mod 2), that is even n's. For any k that is a perfect cube on any base, you can remove n==(0 mod 3). The same pattern applies to k's that are 5th powers, 7th powers, 11th powers, etc. for any prime power. Those are the k's that have algebraic factors to make a PARTIAL covering set for the k. If the other n's (i.e. odd n's on squared k's) can be eliminated by a "numerical" factor such as 3 or 5, that's when you have a k that has partial algebraic factors to make a FULL covering set, which is what you found with k=144 and 324.

As for your squared k's being removed, that is correct and see the algebraic factors thread. Any Riesel base that is b==(4 mod 5) will have SOME squared k's (not all) where even n's have algebraic factors and odd n's where there is a factor of 5. It's only the squared k's where k=m^2 and m==(2 or 3 mod 5). That is k=2^2, 3^2, 7^2, 8^2, 12^2, 13^2, 17^2, 18^2, etc. Your k=144 and 324 fit that criteria. You didn't encounter many of the others listed here because they were eliminated by a trivial factor.

As for k=648 remaining, you are correct. Nothing can be inferred from its factorization other than it is a very low weight k that should eventually have a prime. All odd n's have a factor of 59. The even n's have many small factors but not a covering set. The first n-value without a factor < 2000 is at n=22, which has a 21-digit smallest factor. So clearly no covering set there.

I also do exactly what Tim does on removing the even n's or the n's that are n==(0 mod 3). Personally, I don't worry about it for n<=10K. It's not worth my personal time. But above that, the CPU time savings is worth it.

(caps for emphasis)


Gary

[quote=Mini-Geek;199328]NewPGen format file.[/quote]
do u mean a .prp file as .npg files are only single k values?:smile:
might be worth editing that in in the script thread