[QUOTE=MyDogBuster;209158]
k=4, 9, 49, 64, 144, 169, & 289 proven composite by partial algebraic factors
[/QUOTE]

Am I missing something here? Based upon the hiddenpowers.pl perl script, I see

4*74^n-1 n=0 mod 2 factors due to 2^2
9*74^n-1 n=0 mod 2 factors due to 3^2
49*74^n-1 n=0 mod 2 factors due to 7^2
64*74^n-1 n=0 mod 2 factors due to 8^2
64*74^n-1 n=0 mod 3 factors due to 4^3
144*74^n-1 n=0 mod 2 factors due to 12^2
169*74^n-1 n=0 mod 2 factors due to 13^2
289*74^n-1 n=0 mod 2 factors due to 17^2

What about n=1 mod 2? There is no albegraic factorization for it. These k are not always composite.

For example, I see these with Riesel base 928:

1521*928^n-1 n=0 mod 2 factors due to 39^2
1728*928^n-1 n=0 mod 3 factors due to 12^3

1521*928^11273-1 and 1728*928^12796-1 are prime. These cases are no different than yours, so I don't see how you can eliminate all of those k.

[quote]
These k are not always composite.
[/quote]

They are if n==(1 mod 2) always has a factor of 5, which they do for R744. :-)

[quote]
These cases are no different than yours,
[/quote]

They are different. See the "generalizing algebriac factors for Riesel bases" thread. n==(1 mod 2) always has a factor of 5 if the following 2 conditions are BOTH met:

1. The base is b==(4 mod 5).
2. k=m^2 and m==(2 or 3 mod 5), i.e. k=2^2, 3^2, 7^2, 8^2, etc.

R928 is b==(3 mod 5) so does not have such factorization.

There are other conditions where n==(1 mod 2) has a factor of 13, 41, 53, etc. that allow k's to be eliminated but a factor of 5 is by far the most common.

Sorry, R928 is just plain a tough base. There are no k's that I'm personally aware of that can be eliminated due to partial algebraic factorization unless something new comes out that we haven't observed yet.

The script written by Serge (or Tim; I'm not sure), while helpful and useful, can be misleading. It will tell you the partial (or full) algebraic factorization of a k-value. It will not necessarily tell you whether the k can be eliminated or not. To be eliminated, the "other side", i.e. odd n in this case, has to always have a covering set or single factor; most of the time the latter.

The main value of the script is to allow you to manually remove n-values from a sieve; not to tell you whether a k can be eliminated completely from testing. Sometimes it will have you remove all remaining n-values from the sieve (as would happen in the above situation for base 744 and would allow you to remove the k from testing) but much more frequently, it will not (as would be the case for base 928).

BTW, one last thing: Although logically it makes no difference, you used base 74 but it was Ian's base 744 testing that you were referring to. That is why I refer to base 744 here. It makes no logical difference because if base 74 had a higher conjecture than k=4, the same situation would apply since it is also b==(4 mod 5).


Gary

[QUOTE]4*74^n-1 n=0 mod 2 factors due to 2^2
9*74^n-1 n=0 mod 2 factors due to 3^2
49*74^n-1 n=0 mod 2 factors due to 7^2
64*74^n-1 n=0 mod 2 factors due to 8^2
64*74^n-1 n=0 mod 3 factors due to 4^3
144*74^n-1 n=0 mod 2 factors due to 12^2
169*74^n-1 n=0 mod 2 factors due to 13^2
289*74^n-1 n=0 mod 2 factors due to 17^2[/QUOTE]

I ran base 744 not 74

[quote=rogue;209162]The conjectured k is 958.

106 k have trivial factors.
351 k have primes with 140*821^24442-1 as the largest found (so far).
21 k have no primes.

The hiddenpowers script gave this message:

144*821^n=0 mod 2 factors due to 12^2

Clearly that removes k=144 when n is even, but uncertain about when n is odd.

This base is searched to n=25000 and is released.[/quote]

k=144 still remains. There is no single factor or covering set for odd n.

[QUOTE=MyDogBuster;209221]I ran base 744 not 74[/QUOTE]

Oops. A typo on my part. When I have a chance I'll look again at the correct base. :redface:

And yes, I continue to test n for k=144 for R928, although I was able to use the script to identify values with algebraic factorizations and then removed them from my local PRPNet server.

R999 is complete to n=25K; 13 primes found for n=10K-25K; 73 k's remaining; base released

Sierpinski Base 713
 

Primes found:

2*713^1+1
4*713^26+1
6*713^9+1

With a conjectured k of 8, this one is proven.

I'll take 2*1004^n+1 to 100K.

KEP is releasing bases R900 and S955.

Sierpinski Bases 965 and 923
 

Primes found:

2*965^1+1
4*965^62+1
6*965^1+1

2*923^1+1
4*923^10+1
6*923^41+1

Both have a conjectured k of 8, these conjectures are proven.

Riesel base 548
 

Riesel base 548 has one k remaining at n = 25,000. I won't pursue this.
k n
2 4
3 14
4 45
5 8
6 2
7 k > 25000
8 2
9 1
10 1
11 2
12 14
13 Conjecture

Willem.