Bases 501-1030 reservations/statuses/primes

[rogue]

Primes found:

2*581^1+1
6*581^2+1
8*581^1+1
10*581^2+1
12*581^2+1
16*581^24+1
18*581^1+1
20*581^1+1
22*581^54+1
26*581^1+1
30*581^1+1
32*581^1+1
36*581^8+1
38*581^1+1
40*581^4+1
42*581^2+1
46*581^120+1
48*581^37+1
50*581^533+1
52*581^4+1
56*581^1+1
58*581^8+1
60*581^2+1
62*581^5+1
66*581^12+1
68*581^1+1
70*581^6+1
72*581^2+1
76*581^48+1
78*581^1+1
80*581^3+1
82*581^1494+1
88*581^30+1
90*581^1+1
92*581^1+1
96*581^3+1

The other k have trivial factors. With a conjectured k of 98, this conjecture is proven.

[Siemelink]

These are the primes I found for Riesel base 812:
2 10
3 3
4 k > 25000
5 50
6 1
7 1
8 8
9 1
10 1575
11 2
12 1
13 Conjecture.

as you can see there is one k remaining with n > 25,000. I won't take this further.
Willem.

[MyDogBuster]

Reserving R319 & R504 as new to n=25K

[KEP]

Reserving following 30 Sierpinski bases to n=100K (as new):

272, 278, 293, 335, 356, 398, 437, 440, 473, 482, 503, 545, 566, 587, 608, 632, 650, 668, 671, 692, 722, 755, 776, 797, 818, 827, 860, 863, 881, 902

+ Sierpinski base (as old)

230 to n=100K

Hopes this evens out the balance between untested Riesel and Sierpinski conjectures

Many of them is already started and proven on my Dual Core, so I think that it will be a great contribution to complete the remaining untested k=8 and the previously started k=8 conjectures to n=100K.

KEP

Ps. Plans to hand over each conjecture on e-mail as they completes completes to n=100K

[gd_barnes]

Originally Posted by KEP

Reserving following 30 Sierpinski bases to n=100K (as new):

272, 278, 293, 335, 356, 398, 437, 440, 473, 482, 503, 545, 566, 587, 608, 632, 650, 668, 671, 692, 722, 755, 776, 797, 818, 827, 860, 863, 881, 902

+ Sierpinski base (as old)

230 to n=100K

Hopes this evens out the balance between untested Riesel and Sierpinski conjectures

Many of them is already started and proven on my Dual Core, so I think that it will be a great contribution to complete the remaining untested k=8 and the previously started k=8 conjectures to n=100K.

KEP

Ps. Plans to hand over each conjecture on e-mail as they completes completes to n=100K

2 bases at a time please KEP. I've kindly been asking that of everyone that so that others have an opportunity at new bases and so that I'm not innundated with these things.

I'll reserve the 2 lowest bases for you for now. Please stick with testing only those first. Then migrate on to the next 2. Don't worry, there will still be plenty available when you're done with the first 2. Testing 2 bases to n=100K will take quite a bit of time if there are any k's remaining at n=25K.

Thank you,
Gary

[MyDogBuster]

Riesel Base 504
Conjectured k = 201
Covering Set = 5, 101
Trivial Factors k == 1 mod 503(503)

Found Primes: 188k's - File attached

Remaining k's: 3k's - Tested to n=25K
94*504^n-1
100*504^n-1
116*504^n-1

k=4, 9, 49, 64, 144, 169 proven composite by partial algebraic factors
k=56 and 126 proven composite by a difference of squares

Base Released

[MyDogBuster]

Riesel Base 319
Conjectured k = 1526
Covering Set = 5, 17, 41
Trivial Factors k == 1 mod 2(2) and k = 1 mod 3(3) and k == 1 mod 53(53)

Found Primes: 488k's - File attached

Remaining: 8k's - Tested to n=25K
276*319^n-1
614*319^n-1
626*319^n-1
1244*319^n-1
1266*319^n-1
1356*319^n-1
1496*319^n-1
1506*319^n-1

k=144 & 324 proven composite by partial algebraic factors

Trivial Factor Eliminations: 263 k's

MOB Eliminations:
638

Base Released

[gd_barnes]

Originally Posted by MyDogBuster

Riesel Base 504
Conjectured k = 201
Covering Set = 5, 101
Trivial Factors k == 1 mod 503(503)

Found Primes: 188k's - File attached

Remaining k's: 3k's - Tested to n=25K
94*504^n-1
100*504^n-1
116*504^n-1

k=4, 9, 49, 64, 144, 169 proven composite by partial algebraic factors
k=56 and 126 proven composite by a difference of squares

Base Released

Well...

Wouldn't you know it. Right when you think you have it all figured out, something new comes along. We have our first factor of 101 that combines with partial algebraic factors to make a full covering set for k=100.

Conditions:
b==(100 mod 101)
all k = m^2
m==(10 or 91 mod 101)

for even n, let k=m^2 and n=2q
factors to:
(m*504^q-1)*(m*504^q+1)
for odd n:
factor of 101


This is one of the rare bases that we've found that have 3 different "kinds" of algebraic factors and I missed the final one when showing them on the pages after the reservation. We have the "old" standby for a factor of 5 on odd n and the "new" kind with a factor of 5 on even n. I showed those. But we now have the "old" kind but with a brand new factor of 101 on odd n. I missed that one, which knocks out k=100 in this case.

This is pretty amazing. There are now only 2 k's remaining after having a total of 9 k's knocked out by the 3 different kinds of algebraic factors.


Gary

[Batalov]

R637 is proven, conj. k=144 (largest prime 32*637^18096-1)

[unconnected]

Riesel base 911, k=20
Primes:
2*911^14-1
4*911^1-1
10*911^1-1
12*911^2-1
18*911^2-1

Trivially factors: 6,8,14,16
Base proven.

[MyDogBuster]

Reserving Sierp 829 and 851 as new to n=25K