mersenneforum.org - Algebraic factor issues base 24

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[quote=michaf;133684]I agree with you, Gary,

k*24^n-1

insert k=24*m^2 (k=base*perfect square)

24*m^2*24^n-1

insert n=2q-1 (n=uneven)

24*m^2*24^(2q-1)-1 = m^2*24^(2q)-1

which is exactly the same as we had with a perfect square.

Wonderful :)[/quote]

Very good. Now see if you can come up with the specific algebraic factors for k=6, which, I think, clearly has algebraic factors but doesn't fit the definitions that we've come up with so far.

Riesel base 24 is turning out to be a VERY strange base! :smile:

Gary

[QUOTE=gd_barnes;133729]Very good. Now see if you can come up with the specific algebraic factors for k=6, which, I think, clearly has algebraic factors but doesn't fit the definitions that we've come up with so far.

Riesel base 24 is turning out to be a VERY strange base! :smile:

Gary[/QUOTE]

For even n:

[TEX]6*24^n-1 = 2*3 * 8^n*3^n - 1 = 2*3 * 2^{3n} * 3^n - 1
= 2^{(3n+1)} * 3^{(n+1)} -1
[/TEX]

insert n=2q for even n:

[TEX]2^{(3*2q+1)} * 3^{(2q+1)} -1 = 2^{(6q+1)} * 3^{(2q+1)} -1[/TEX]

[TEX]2^{(6q+1)}[/TEX] is always even
[TEX]3^{(2q+1)}[/TEX] is always uneven
[TEX]2^{(6q+1)} * 3^{(2q+1)}[/TEX] is always uneven
[TEX]2^{(6q+1)} * 3^{(2q+1)} -1 [/TEX] is always even

But I can't get a grip on the uneven n's yet :)

[quote=michaf;133797]For even n:

[tex]6*24^n-1 = 2*3 * 8^n*3^n - 1 = 2*3 * 2^{3n} * 3^n - 1
= 2^{(3n+1)} * 3^{(n+1)} -1
[/tex]

insert n=2q for even n:

[tex]2^{(3*2q+1)} * 3^{(2q+1)} -1 = 2^{(6q+1)} * 3^{(2q+1)} -1[/tex]

[tex]2^{(6q+1)}[/tex] is always even
[tex]3^{(2q+1)}[/tex] is always uneven
[tex]2^{(6q+1)} * 3^{(2q+1)}[/tex] is always uneven
[tex]2^{(6q+1)} * 3^{(2q+1)} -1 [/tex] is always even

But I can't get a grip on the uneven n's yet :)[/quote]

I have more time tonight then I did the last couple of nights. I have to disagree here:

If 2^(6q+1) is always even and 3^(2q+1) is always uneven, then:

2^(6q+1)*3^(2q+1) should be always even (not uneven)

hence 2^(6q+1)*3^(2q+1) should always be uneven, i.e. odd

What we DO know about even n, even though it is always odd, is that it ALWAYS has a factor of 5!

So, like you said, it's the uneven (odd) n that we are concerned about. For me, when all else fails, then try reverse engineering, which is how I came up with quite a few of the algebraic factors that I did on the bases. Here goes:

6*24^1-1 = 11 * 13
6*24^3-1 = 287 * 289
6*24^5-1 = 6911 * 6913
6*24^7-1 = 165887 * 165889

Notice a pattern here? How about this:

(2^0*12^1-1)*(2^0*12^1+1)
(2^1*12^2-1)*(2^1*12^2+1)
(2^2*12^3-1)*(2^2*12^3+1)
(2^3*12^4-1)*(2^3*12^4+1)

-or- to generalize it:
{2^[(n-1)/2]*12^[(n+1)/2]-1}*{2^[(n-1)/2]*12^[(n+1)/2]+1}

-or- to make it easier, let n=2m+1 and you have:

[2^m*12^(m+1)-1]*[2^m*12^(m+1)+1]

Therefore k=6 can be eliminated as follows:

For even n: factor of 5

For odd n, let n=2m+1; factors to:
[2^m*12^(m+1)-1]*[2^m*12^(m+1)+1]

This makes me wonder if there are other k-values with this type of unusual algebraic factors. If you have any other k-values that ONLY have numeric factors of 2 and 3 in them, you might check Alperton's site [URL="http://www.alpertron.com.ar/ECM.HTM"]here[/URL] and look for patterns in the factors for each n-value.

k=6 will be removed from the web pages and the count remaining dropped by one.

:smile: Gary

I have renamed this thread to be "algebraic factor issues base 24". It includes all posts in this thread originally titled "sr2sieve / sierpinksi base 24" by Micha about Sierp base 24 issues and all posts in the primes thread related to Riesel base 24 issues that I moved here.

Gary

[QUOTE=gd_barnes;133812]
With THREE different sets of algebraic factors, this base has been some kind of challenge! :smile:
Gary[/QUOTE]

In short: WOW :smile:

In removing all the k's with algebraic factors, I noticed they were also very low weight, about 200-300 terms per k (as compared to about 2000-4000 for others

Well Micha, you're not going to believe this, but I've generalized the algebraic factors even further and was able to eliminate 11 more k-values in addition to what I've already listed!! :smile:

I erred when I jumped the gun by saying that k=24*m^2 gives k's with algebraic factors. That is true but I didn't take it far enough. Here are the conditions that not only eliminate all 13 of those k-values but they also eliminate the 2 k-values of 6 and 486 as stated in the last post plus 11 more. It technically eliminates a total of 26 k-values so my original k=24*m^2 condition was only half of the 'puzzle'. Here are the generalized conditions and algebraic factors:

All k=6*m^2 where k==(6 mod 10) contain factors as follows and can be eliminated from your search:

even n; factor of 5

odd n; let k=6*m^2 and let n=2*q-1; factors to:

[m*3^q*2^(3q-1) - 1] *
[m*3^q*2^(3q-1) + 1]

It seems so easy in retrospect but I got started on the wrong path to begin with and made it hard.

So now, in addition to the 13 previous k-values that I listed for k=24*m^2, these conditions also eliminate the following k-values:

6, 486, 726, 2166, 5046, 5766, 9126, 10086, 14406, 15606, 20886, 22326, and 28566.

This now drops us down to 142 k-values remaining at n=25K and also simplifies things quite a bit. Now, there are technically only 2 separate conditions that have algebraic factors. The web pages will be updated shortly.

Now you know why so many were so low weight. They actually had ZERO weight! :smile:

I've now closely inspected many of the remaining 142 k-values. There should be no more with algebraic factors.

OK, I'm done with this now! :smile:

Gary

Gee :smile:

That's a load of elimination in a short time now!
I'll be sieving the rest to an optimum level for n=50k, and then continue prp-ing.