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Old 2010-06-08, 02:03   #551
gd_barnes
 
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Quote:
Originally Posted by rogue View Post
Reserving
Mark,

I just finished S878 to n=25K. See my recent post. If you've started on it, feel free to doublecheck it though.


Gary
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Old 2010-06-08, 11:04   #552
rogue
 
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Quote:
Originally Posted by gd_barnes View Post
Mark,

I just finished S878 to n=25K. See my recent post. If you've started on it, feel free to doublecheck it though.
I haven't started it yet, so I'll get rid of it.

In case people are curious, I am dumping all of these into a PRPNet server trying to keep two computers "fed" through the next week as I will be on vacation. Right now I have about six days of work queued up, but if k fall, then that six days gets reduced. I'm only reserving whenever I have the time to sieve. I've been taking conjectures on the Sierspinski side because it hasn't had as much "love" as the Riesel side. There will be (unfortunately) an number of new single k conjectures when I'm done.

And speaking of such, I'll reserve S702, conjectured k of 75.

Last fiddled with by rogue on 2010-06-08 at 11:07
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Old 2010-06-08, 13:05   #553
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Default Sierpinski results

Base 548

Code:
2*548^1+1
3*548^6+1
4*548^2+1
5*548^1+1
6*548^115+1
7*548^4+1
8*548^5311+1
9*548^1+1
10*548^12+1
11*548^1+1
12*548^1+1
13*548^22+1
14*548^1+1
15*548^1+1
Conjecture proven.

Base 679

Code:
6*679^4+1
10*679^1+1
12*679^10+1
k=4 remains at n=25000. Releasing.

Base 740

Code:
2*740^1+1
3*740^1+1
5*740^1+1
6*740^1+1
7*740^2+1
8*740^83+1
9*740^1+1
10*740^12+1
12*740^5+1
k=4, 11, and 13 remain at n=25000. Releasing.

Base 812

Code:
2*812^1003+1
3*812^1+1
4*812^26+1
5*812^5+1
6*812^19+1
7*812^2+1
8*812^3461+1
9*812^1+1
10*812^18+1
11*812^1+1
12*812^6+1
13*812^2+1
14*812^1+1
15*812^31+1
Conjecture proven.

Base 866

Code:
2*866^1+1
3*866^7+1
5*866^5+1
6*866^1+1
7*866^2+1
10*866^2+1
11*866^35+1
12*866^531+1
13*866^1492+1
15*866^8+1
k=8 remains at n=25000. Releasing.
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Old 2010-06-08, 13:08   #554
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Default Sierpinski results

Base 934

Code:
3*934^1+1
6*934^4+1
7*934^6+1
9*934^429+1
10*934^1+1
12*934^44+1
13*934^1+1
15*934^1+1
k=4 remains at n=25000. Releasing.

Base 935

Code:
2*935^1+1
4*935^2+1
6*935^8+1
8*935^1+1
12*935^3+1
k=10 remains at n=25000. Releasing.

Base 968

Code:
2*968^917+1
3*968^2+1
4*968^90+1
5*968^3+1
6*968^40+1
7*968^8+1
8*968^7+1
9*968^1+1
10*968^162+1
12*968^1+1
13*968^2+1
14*968^1+1
15*968^20+1
k=11 remains at n=25000. Releasing.

Base 1013

Code:
2*1013^1+1
4*1013^2+1
6*1013^1+1
12*1013^1+1
k=8 remains at n=25000. Releasing.

Note that any other k that appear to be "missing" have trivial factors.
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Old 2010-06-08, 14:49   #555
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Default Sierpinski Bases 542, 743, 747, and 893

Reserving
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Old 2010-06-08, 17:10   #556
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Default Sierpinski Bases 879, 924, 993, and 846

Reserving
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Old 2010-06-09, 02:46   #557
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R788 is proven
CK=14

Largest prime

7*788^1663-1

k=9 is removed by algebraic factors

attached are the results
Attached Files
File Type: zip R788.zip (24.2 KB, 61 views)
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Old 2010-06-09, 03:26   #558
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Quote:
k=9 is removed by algebraic factors
Factor 3 can never eliminate k's due to algebraic factors. Gary will have to explain the math. k=9 will have to be tested. Ian

Last fiddled with by MyDogBuster on 2010-06-09 at 03:26
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Old 2010-06-09, 04:22   #559
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Quote:
Originally Posted by Mathew Steine View Post
R788 is proven
CK=14

Largest prime

7*788^1663-1

k=9 is removed by algebraic factors

attached are the results
Algebraic factors only remove the even n. There is no common factor for the odd n. As Ian said, you'll need to test k=9. If there was a common factor for the odd n, then you could make the statement: k=9 is removed by partial algebraic factors. You could never make the statement that k=9 is removed by algebraic factors unless the base was also a perfect square. In that case, k's that are perfect squares are removed by (full) algebraic factors because the algebraic factors occur on all n; not just the even n.

To determine which common factor(s) for the odd n that there could be for a k that is a perfect square on a base, prime factor the base + 1 and only consider factors (f) that are f==(1 mod 4). Here:

789 = 3 * 263

Since 3 and 263 are both f==(3 mod 4), there can be no common factor for any odd n on squared k's for base 788. Therefore all squared k's must be tested unless they are eliminated by trivial factors, which the script would do automatically.

We'll wait to show this on the pages until you let us know that you've tested k=9.


Gary

Last fiddled with by gd_barnes on 2010-06-09 at 04:29
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Old 2010-06-09, 05:21   #560
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Default Round 2

R788 is proven

9*788^11325-1 is 3-PRP! (44.1679s+0.0027s)
Primality testing 9*788^11325-1 [N+1, Brillhart-Lehmer-Selfridge]
Running N+1 test using discriminant 7, base 1+sqrt(7)
Calling Brillhart-Lehmer-Selfridge with factored part 79.21%
9*788^11325-1 is prime! (458.7986s+0.0052s)

Edit: Thanks Mathew

Last fiddled with by MyDogBuster on 2010-06-09 at 05:54
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Old 2010-06-09, 13:51   #561
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Default Riesel 908

Riesel 908 the last k (8*908-n-1) tested n=25K-50K. Nothing found

Results attached - Base released

Last fiddled with by MyDogBuster on 2014-09-02 at 09:15
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