Riesel Base 898
 

The conjectured k is 30.

Primes found:

2*898^6-1
3*898^1-1
5*898^16-1
6*898^1-1
8*898^2-1
9*898^1-1
11*898^2-1
12*898^2-1
15*898^1-1
17*898^54-1
18*898^45-1
20*898^1-1
21*898^2-1
23*898^6-1
26*898^15-1
29*898^1-1

The other k have trivial factors. This conjecture is proven.

Riesel Base 531
 

2*531^1-1
4*531^5-1
8*531^5-1
10*531^1-1
12*531^2-1
14*531^1-1
18*531^6-1

k=6 and k=16 have trivial factors. With conjectured k=20, this conjecture is proven.

Riesel Base 821
 

This has a conjectured k=958. I will reserve it to n=25000.

S637 and S1011 are proven
 

S637 and S1011 are proven.

R1011 has 3 k remaining.

R931
 

R931 with conjectured k=3960 has a very smooth [I]b[/I]-1.
At n=2000, there are only 11 [I]k[/I]'s left. Reserving to 25K.
Initial files are attached.

[SIZE=1]Another one is R361. It is a square which allows for some trivial eliminations, plus some primes can be borrowed from R19. Will reserve in the other thread.[/SIZE]

Riesel Base 709
 

Riesel Base 709
Conjectured k = 924
Covering Set = 5, 71
Trivial Factors k == 1 mod 2(2) and k == 1 mod 3(3) and k == 1 mod 59(59)

Found Primes: 289k's File attached

Remaining k's: 13k's - Tested to n=25k
144*709^n-1 <<<< Proven composite by partial algebraic factors
170*709^n-1
174*709^n-1
218*709^n-1
234*709^n-1
324*709^n-1 <<<< Proven composite by partial algebraic factors
354*709^n-1
408*709^n-1
606*709^n-1
746*709^n-1
774*709^n-1
776*709^n-1
834*709^n-1

Trivial Factor Eliminations: 159k's

Base Released

[quote=MyDogBuster;208107]Riesel Base 999 k = 1776
k=1776 even n's trivial - odd n's difference of squares[/quote]

Hah! Saved ME some testing time. Thanks Ian. :smile:

I was able to generalize this across all of base 999. It would happen where:
k=111*m^2 and m==(1 or 4 mod 5)

k=1776=111*4^2 is the lowest one.
The next one would be k=111*6^2=3996. Since k=3996 is above the conjecture, k=1776 is the only k on this base with this condition.

I saw the pattern where even-n has a factor of 5 as consistent with bases 24 and 54. It is:
k=f*m^2 and k==(6 mod 10)

Where f is the product of all prime factors of the base that are not raised to an even power with powers removed. Hence for base 999, since b=999=3^3*37 then f=3*37=111. For base 24, b=24=2^3*3 and hence f=2*3=6. You can also see that for base 54, f would also be 6. Also, for base 294, b=2*3*7^2, hence you would eliminate the 7^2 and would end up with f=2*3=6.

Seeing this, I'm getting close to generalizing it across all bases where even-n gives a factor of 5 and odd-n is the algebraic product of squares. Since that is the most common of this type, I hope to eliminate a large percentage of the problem from many bases by Friday. I'm also aware there are some where even-n has a factor of 17 and 41 at this point bases on your guys analysis.


Gary

[quote=kar_bon;208116]what about this:

64*741^n-1 got a divisor of 11 when n=7,17,27,37,47,57,...

a sieve-file for 25000<n<100000 contains no n ending in 7! any hint why?[/quote]

A divisor of 11 always occurs every 5th or 10th n-value if it is going to occur at all. In this case, all n==(7 mod 10) are divisible by 11, hence the reason why no remaining n-values end in a 7.

Also, since k=64 is a perfect square and cube, you could remove all even n-values and n's divisible by 3.

But neither of those help a lot. Most n's==(1 or 5 mod 6) still remain. The fact that n=19 has an 18-digit smallest factor means there is virtually no chance for a full covering set.


Gary

[quote=MyDogBuster;208118]First of all, my covering set was wrong. s/b 7, 37 not 5, 37

Other than that, I don't have a clue. I am NOT a math person. I'm sure k=64 is probably algebraic, but don't ask me why.[/quote]

Your covering set is still wrong. lol

It should be 7, 53 -not- 7, 37 or 5, 37.

I corrected your original post where you reported the status.

Alas, the covering set of the conjectured k=160 makes no difference for k=64.

[quote=Batalov;208600]S637 and S1011 are proven.

R1011 has 3 k remaining.[/quote]

Serge,

I have to have test limits on any k's remaining. This is 4-5 bases now without them in different threads. For R1011, what are its remaining k's testing limits and are you keeping the base reserved?

In the future, if I don't have a test limit, I'm just going to ignore the work and it won't be reflected anywhere.


Gary

[quote=gd_barnes;208972]Serge,

I have to have test limits on any k's remaining. This is 4-5 bases now without them in different threads. For R1011, what are its remaining k's testing limits and are you keeping the base reserved?

In the future, if I don't have a test limit, I'm just going to ignore the work and it won't be reflected anywhere.


Gary[/quote]
[URL]http://www.mersenneforum.org/showthread.php?p=208969#post208941[/URL] :smile: