Algebraic factor issues base 24

[gd_barnes]

Originally Posted by michaf

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 :)

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!


Gary

[michaf]

Originally Posted by gd_barnes

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!


Gary

For even n:



insert n=2q for even n:



is always even
is always uneven
is always uneven
is always even

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

[gd_barnes]

Originally Posted by michaf

For even n:



insert n=2q for even n:



is always even
is always uneven
is always uneven
is always even

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

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 here 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.


Gary

[gd_barnes]

Originally Posted by michaf

For even n:



insert n=2q for even n:



is always even
is always uneven
is always uneven
is always even

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

We need to generalize this even further because it appears that k=486, whose k-value only has prime factors of 2 and 3, also has algebraic factors. Analysis:

For all k=6*3^(4q) where q>=0 (hence inclusive of k=6 and k=486 and no others < the Riesel #):

For even n, always a factor of 5.

For odd n, let k=3^(4q) and let n=2m+1 and you have:

[3^(2q)*2^m*12^(m+1)-1]*
[3^(2q)*2^m*12^(m+1)+1]


I have now analyzed all of the rest of the remaining k-values for Riesel base 24. None only have prime factors of 2 and 3 so HOPEFULLY, this is all of the remaining k's with algebraic factors.

With the elimination of k=6 and k=486, the pages will now be reduced by 2 k's to 153 k's remaining at n=25K.

With THREE different sets of algebraic factors, this base has been some kind of challenge!


Gary

[gd_barnes]

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

[michaf]

Originally Posted by gd_barnes

With THREE different sets of algebraic factors, this base has been some kind of challenge!
Gary

In short: WOW

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

[gd_barnes]

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!!

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!

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!


Gary

[michaf]

Gee

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.