Bases 251-500 reservations/statuses/primes

[rogue]

Originally Posted by Mini-Geek

From looking at the lists in the FactorDB, I'm inclined to think the odd n's are eliminated due to a factor of 5. Since I fail to see a mathematical reason for all this, I'll just leave it at that.

But that isn't an algebraic factorization.

[axn]

Originally Posted by rogue

But that isn't an algebraic factorization.

That's why it is a "partial" algebraic factorization, I guess.

[Mini-Geek]

Originally Posted by axn

That's why it is a "partial" algebraic factorization, I guess.

Yeah, I think the "partial" algebraic factorization covers the even n's, and the factors of 5 (not algebraic factorizations) cover the odd n's.

[rogue]

Originally Posted by Mini-Geek

Yeah, I think the "partial" algebraic factorization covers the even n's, and the factors of 5 (not algebraic factorizations) cover the odd n's.

Here is the case. If b%10==4 and n%2==1, then b^n%10==4. Note that when n is odd, that b^n ends with a 4. When n is even b^n ends with a 6. 4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1024, etc. If k%10==4 or k%10==9, then k*b^n%10==6, thus k*b^n%5=1, thus (k*b^n-1)%5=0.

Obviously if k is a square and n is even, then k*b^n-1 has the algebraic factorization of sqrt(k)*b^(n/2)-1 * sqrt(k)*b^(n/2)+1.

Gary already shows these on the Riesel Conjectures page.

Would it be helpful for people to list the algrebraic/partial algebraic factorizations when they submit their results?

[gd_barnes]

Originally Posted by rogue

Here is the case. If b%10==4 and n%2==1, then b^n%10==4. Note that when n is odd, that b^n ends with a 4. When n is even b^n ends with a 6. 4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1024, etc. If k%10==4 or k%10==9, then k*b^n%10==6, thus k*b^n%5=1, thus (k*b^n-1)%5=0.

Obviously if k is a square and n is even, then k*b^n-1 has the algebraic factorization of sqrt(k)*b^(n/2)-1 * sqrt(k)*b^(n/2)+1.

Gary already shows these on the Riesel Conjectures page.

Would it be helpful for people to list the algrebraic/partial algebraic factorizations when they submit their results?

To clarify for the layman: You only gave an example above for b==(4 mod 10). Similar factorizations, only in reverse, can happen where b==(9 mod 10). So technically such factorizations can occur for all b==(4 mod 5).

I feel that this statement:

"Obviously if k is a square and n is even, then k*b^n-1 has the algebraic factorization of sqrt(k)*b^(n/2)-1 * sqrt(k)*b^(n/2)+1."

needs a little clarification to avoid confusion. Yes, all even n are eliminated by algebraic factors when k is squared on the Riesel side. But most of the time, odd n will not have a trivial factor and so the k must still be searched. I wasn't clear if that para. was made in conjunction with the first para.

As an example, the above is why Ian still had to search k=64, 81, 100, etc. on his recently posted base. For bases where b==(4 mod 5), partial algebraic factorizations to make a full covering set can only occur where:

k=m^2 and m==(2 or 3 mod 5)

To be specific, only k=2^2, 3^2, 7^2, 8^2, 12^2, 13^2, etc, or k=4, 9, 49, 64, 144, 169, etc. are partially covered by algebraic factorization to make a full covering set on Ian's base.

Reference wouldn't it be helpful for people to list such factorizations with their results: I think Ian is already showing k's partially covered by algebraic factors to make a full covering set when he submits his primes and k's remaining. I think that works well. Also, whenever someone reserves a base, when putting it on the page, I "usually" state all k's with partial or full algebraic factorizations. Although if it is a large conjecture or a complex situation, I may not "see" them until someone lists their k's remaining at a certain limit.

These get even more involved for bases where b==(12 mod 13), (16 mod 17), (28 mod 29), etc. For more info. see the "generallizing algebraic factors for Riesel bases" thread.

Does anyone have better wording than "covered by partial algebraic factors"? Since we're referring to the entire set of n's on each k here when making that statement, that's what I came up with. Obviously for the even n's, those are covered fully by algebraic factors. But in the universe of n's for those k's, only part of them are covered by algebraic factors.

Perhaps "partially covered by algebraic factors" would be better wording when referring to all n's for each of the covered k's. It's not easy to put it in words clearly.

I figured at some point, the higher math types would come in and pick apart the pages. I'm surprised it's taken this long. I welcome better wording and better ways of showing these "tricky to state" situations where odd n's have a trivial "numeric" factor (factor of 5 in this case) and even n's have "full" algebraic factors.

Axn or Mark, any thoughts about how to more clearly state this mathwise?


Gary

[rogue]

Originally Posted by gd_barnes

I figured at some point, the higher math types would come in and pick apart the pages. I'm surprised it's taken this long. I welcome better wording and better ways of showing these "tricky to state" situations where odd n's have a trivial "numeric" factor (factor of 5 in this case) and even n's have "full" algebraic factors.

Axn or Mark, any thoughts about how to more clearly state this mathwise?

Where Bob Silverman when you need him?

I agree with everything you have written and no, I can't quickly think of a better way to state it.

[Mini-Geek]

Originally Posted by gd_barnes

Perhaps "partially covered by algebraic factors" would be better wording when referring to all n's for each of the covered k's. It's not easy to put it in words clearly.

I figured at some point, the higher math types would come in and pick apart the pages. I'm surprised it's taken this long. I welcome better wording and better ways of showing these "tricky to state" situations where odd n's have a trivial "numeric" factor (factor of 5 in this case) and even n's have "full" algebraic factors.

Axn or Mark, any thoughts about how to more clearly state this mathwise?

You may not want me to reply, but I will anyway.
I'd suggest:
something along the lines of "proven composite by a combination of algebraic and trivial factors" and/or
label it (the thing already listed on every such base explaining this, "All k where k = m^2 ...") Condition 1 (even though it's the only condition for most/all applicable bases) and say "proven composite by condition 1", and/or
instead of listing the Condition on every such base and saying "by condition 1", just link the "proven composite by ..." (or similar) text to somewhere, like an anchor to elsewhere on the page, that has a base-generic form of that statement listed, i.e. one that includes the b==4 mod 5 condition and says b throughout instead of a specific number.

[gd_barnes]

Originally Posted by MyDogBuster

Riesel Base 424
Conjectured k = 69
Covering Set = 5, 17
Trivial Factors k == 1 mod 3(3) and k == 1 mod 47(47)

Found Primes: 40k's File attached

Remaining k's: Tested to n=25K
9*424^n-1 <------ Proven composite by partial algebraic factors
18*424^n-1
21*424^n-1
44*424^n-1 <------ Proven composite by partial algebraic factors

Trivial Factor Eliminations: 23k's

Base Released

k=44 is not eliminated by partial algebraic factors so I'll show it as remaining. Just to confirm: You did test it to n=25K; correct?

[gd_barnes]

Originally Posted by rogue

I have a question that I hope someone could answer regarding algebraic factorizations. I understand where you can find one (easily) if n is even, but what is the algebraic factorization if n is odd or when n is prime?

I don't think this ever clearly got answered if someone new just happened to see it. The web pages specify this, although their clarity is probably not the best it could be for many people. This may be beating a dead horse for many of you but I think it needs one final clarification:

Even n's have full algebraic factorization.

Odd n's have a factor of 5. (no algebraic factorization)

This "combines" to what I call "proven composite by partial algebraic factors" to make a full covering set for the entire universe of n-values.

For anyone who hasn't followed the context of the subsequent discussion, this only refers to bases (b) where b==(4 mod 5) and where k=m^2 and m==(2 or 3 mod 5).

To drive the point home just a little more: See Ian's recent work on Riesel base 444. There, k=36 remains whereas k=4, 9, 49, 64, 144, and 169 are eliminated by the above condition. For k=36, although even n have the above algebraic factorization, odd n do not have a trivial factor of 5 so a prime must be found. It just so happens that primes n<25K have already been found for other squares such as k=16, 25, 81, 100, and 121.

FYI, I am mulling over some of the subsequent discussion about rewording the "proven composite by partial algebraic factors". I agree that it should be stated differently but will likely get to it after updating the PFGW new bases script for the latest version of PFGW -and- splitting this thread up into multiple threads like I promised.


Gary

[gd_barnes]

Originally Posted by MyDogBuster

If no one objects, I'll be attacking the following base from n=100K to n=200K.

Sierp 252

I now have this officially reserved for you.

[MyDogBuster]

k=44 is not eliminated by partial algebraic factors so I'll show it as remaining. Just to confirm: You did test it to n=25K; correct?

Yes, I checked the output, it's tested.