Bases 501-1030 reservations/statuses/primes - Page 17

MyDogBuster · 2010-02-13, 23:59

Reserving Riesel 639, 709 and 744 as new to n=25K

rogue · 2010-02-14, 04:19

Primes found:

Code:

2*803^48-1
4*803^89-1
6*803^1-1
8*803^4-1
10*803^3-1
12*803^14-1
16*803^31-1
18*803^2-1
20*803^4-1
24*803^4-1
26*803^10-1
28*803^1-1
30*803^48-1
32*803^56-1
34*803^119-1
36*803^7-1
38*803^328-1
40*803^1-1
42*803^3-1
44*803^12372-1
46*803^21-1
48*803^1-1
50*803^8-1
54*803^3-1
56*803^2-1
58*803^1-1
60*803^1-1
62*803^14-1
66*803^2-1

Remaining k at n=25000:

Code:

14*803^n-1
22*803^n-1
52*803^n-1
64*803^n-1

Since 64 = 8^2, I know that there are algebraic factorizations, but I don't know if they cover all n or if they only cover even n. Can someone answer that question?

Base released.

rogue · 2010-02-14, 04:20

Primes found

Code:

2*803^1+1
6*803^9+1
8*803^1243+1
10*803^6+1
12*803^13+1
14*803^1+1

k = 4 remains at n=25000. The base is released. As far as I understand it, k = 4 does not have algebraic factorizations for all n.

Mini-Geek · 2010-02-14, 12:09

Quote:

Originally Posted by rogue

Since 64 = 8^2, I know that there are algebraic factorizations, but I don't know if they cover all n or if they only cover even n. Can someone answer that question?

Even.

Where n being replaced by 2m shows that n is even...
$64*803^n-1=8^2*803^{2m}-1=(8*803^m)^2-1=(8*803^m+1)*(8*803^m-1)$
(simple manipulations followed by the difference of squares rule)

64*803^n-1 with even n also trivially has a factor of 3. (visible at http://factordb.com/search.php?query=64*803^%282*n%29-1)

I don't see any reason that the odd n's can be eliminated. They are extremely low-weight, though, so the lack of a prime by 25K is not surprising. (look at them here)

I don't think Sierp 803 k=4 has any algebraic factorizations, because b^n+1 with n=2 doesn't have an algebraic factorization. See http://www.mersenneforum.org/showpos...&postcount=814. In short, n must be at least 3 and not a power of 2. (3, 5, 6, 7, 9, etc.)
k=8...hm...8*803^n+1...has algebraic factors where n is a multiple of 3.
$8*803^n+1=2^3*803^{3m}+1=(2*803^m)^3+1=(2*803^m+1)*(4*803^{2m}-2*803^m+1)$
(simple manipulations followed by the addition of cubes rule)
I doubt if it has a full covering set. Looking at its list of factorizations in the DB, it seems unlikely.
The algebraic factorization isn't really necessary though, as a few small factors cover all n's divisible by 3.
http://factordb.com/search.php?query=8*803^(3*n)%2B1

rogue · 2010-02-14, 23:10

I have attached the list of primes.

1 is a GFN, which has not been tested.

k=122 is the only k remaining and has been tested up to n=25000.

The other k have trivial factors.

I am releasing this base.

gd_barnes · 2010-02-15, 11:58

Quote:

Originally Posted by Mini-Geek

Even.

Where n being replaced by 2m shows that n is even...
$64*803^n-1=8^2*803^{2m}-1=(8*803^m)^2-1=(8*803^m+1)*(8*803^m-1)$
(simple manipulations followed by the difference of squares rule)

64*803^n-1 with even n also trivially has a factor of 3. (visible at http://factordb.com/search.php?query=64*803^%282*n%29-1)

I don't see any reason that the odd n's can be eliminated. They are extremely low-weight, though, so the lack of a prime by 25K is not surprising. (look at them here)http://factordb.com/search.php?query=8*803^(3*n)%2B1

BTW, I factored 64*803^n-1 a little more on the DB factors page...always entertains me for a few mins. :-)

The reason why k=64 is so low weight on many different Riesel bases (if it's not eliminated fully by trivial or algebraic factors) is that it is both a perfect square and perfect cube. Both n==(0 mod 2) and n==(0 mod 3) are eliminated by algebraic factors.

Here:
Factor of 3 is n==(0 mod 2) leaves n==(1 mod 2)
Cubed algebraic is n==(0 mod 3) leaves n==(1, 5 mod 6)
Factor of 17 is n==(1 mod 4) leaves n==(7, 11 mod 12)

So...same conclusion...

No clear set of factors take out n==(7 or 11 mod 12) so the search must go on.

Interestingly...8*803^n+1 also has all n==(7 or 11 mod 12) without a clear set of factors.

mdettweiler · 2010-02-16, 06:28

Polished off a few of the untested k=6 Sierp. conjectures:

S559: conjectured k 6, proven, primes:
4*559^1+1

S594: conjectured k 6, proven, primes:
2*594^4+1
3*594^1+1
4*594^1+1
5*594^1+1
k=1 is a GFN.

S664: conjectured k 6, proven, primes:
3*664^1+1
4*664^1+1
k=1 is a GFN; k=2 and k=5 eliminated by trivial factors.

S699: conjectured k 6, proven, primes:
2*699^1+1
4*699^1+1

S769: conjectured k 6, proven, primes:
4*769^3+1
k=2 eliminated by trivial factors.

S804: conjectured k 6, proven, primes:
2*804^1+1
3*804^4+1
4*804^1+1
5*804^1+1
k=1 is a GFN.

S874: conjectured k 6, proven, primes:
3*874^2+1
4*874^77+1
k=1 is a GFN, k=2 and k=5 eliminated by trivial factors.

S909: conjectured k 6, proven, primes:
2*909^10+1
4*909^1+1

S979: conjectured k 6, proven, primes:
4*979^1+1
k=2 eliminated by trivial factors.

S1014: conjectured k 6, proven, primes:
2*1014^1+1
3*1014^3+1
4*1014^1+1
5*1014^3+1
k=1 is a GFN.

That should be the last of the untested Sierpinski k=6 conjectures.

rogue · 2010-02-16, 16:28

The primes are attached.

k=78 and k=87 do not have primes for n < 25000.

The other k have trivial factors.

I am releasing this base.

rogue · 2010-02-16, 20:22

I discovered that I made some errors in the bases that I had reported in the past week.

I went back through my logs and found these primes:

8*803^1243+1

95*516^726+1
120*516^647+1
121*516^531+1

I double-checked my work for Sierpinski and Riesel bases 322, 328, 516, and 803. These are the only discrepancies. I re-ran the script then looked at the primes I submitted verses the contents of pl_remain.txt.

I will rerun the remaining k for those conjectures up to 25000 to ensure that I didn't make any further mistakes. I'm not expecting anything to show up, but one never knows...

This is what I get for breaking up the work across multiple computers.

gd_barnes · 2010-02-16, 20:35

Quote:

Originally Posted by rogue

I went back through my logs and found this prime:

8*803^1243+1

OK...added to the 1k remaining list.

rogue · 2010-02-16, 23:23

Quote:

Originally Posted by rogue

I discovered that I made some errors in the bases that I had reported in the past week.

I went back through my logs and found these primes:

8*803^1243+1

95*516^726+1
120*516^647+1
121*516^531+1

I double-checked my work for Sierpinski and Riesel bases 322, 328, 516, and 803. These are the only discrepancies. I re-ran the script then looked at the primes I submitted verses the contents of pl_remain.txt.

I will rerun the remaining k for those conjectures up to 25000 to ensure that I didn't make any further mistakes. I'm not expecting anything to show up, but one never knows...

This is what I get for breaking up the work across multiple computers.

I just realized that I only need to retest for n < 10000 because I see the other remaining k in my sieve file for 10000 < n. That will save me a few days.

2010-02-14, 04:19	#178
rogue "Mark" Apr 2003 Between here and the 2⁴×397 Posts	Riesel base 803 Primes found: Code: 2803^48-1 4803^89-1 6803^1-1 8803^4-1 10803^3-1 12803^14-1 16803^31-1 18803^2-1 20803^4-1 24803^4-1 26803^10-1 28803^1-1 30803^48-1 32803^56-1 34803^119-1 36803^7-1 38803^328-1 40803^1-1 42803^3-1 44803^12372-1 46803^21-1 48803^1-1 50803^8-1 54803^3-1 56803^2-1 58803^1-1 60803^1-1 62803^14-1 66803^2-1 Remaining k at n=25000: Code: 14803^n-1 22803^n-1 52803^n-1 64803^n-1 Since 64 = 8^2, I know that there are algebraic factorizations, but I don't know if they cover all n or if they only cover even n. Can someone answer that question? Base released. Last fiddled with by rogue on 2010-02-14 at 04:22*

2010-02-14, 04:20	#179
rogue "Mark" Apr 2003 Between here and the 2⁴×397 Posts	Sierpinski base 803 Primes found Code: 2803^1+1 6803^9+1 8803^1243+1 10803^6+1 12803^13+1 14803^1+1 k = 4 remains at n=25000. The base is released. As far as I understand it, k = 4 does not have algebraic factorizations for all n. Last fiddled with by gd_barnes on 2010-02-17 at 00:14 Reason: add k=8 prime and remove from remaining

2010-02-16, 20:22	#185
rogue "Mark" Apr 2003 Between here and the 2⁴·397 Posts	I discovered that I made some errors in the bases that I had reported in the past week. I went back through my logs and found these primes: 8803^1243+1 95516^726+1 120516^647+1 121516^531+1 I double-checked my work for Sierpinski and Riesel bases 322, 328, 516, and 803. These are the only discrepancies. I re-ran the script then looked at the primes I submitted verses the contents of pl_remain.txt. I will rerun the remaining k for those conjectures up to 25000 to ensure that I didn't make any further mistakes. I'm not expecting anything to show up, but one never knows... This is what I get for breaking up the work across multiple computers. Last fiddled with by rogue on 2010-02-16 at 20:42

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2010-02-13, 23:59	#177
MyDogBuster May 2008 Wilmington, DE 2²·23·31 Posts	Reserving Riesel 639, 709 and 744 as new to n=25K

2010-02-16, 06:28	#183
mdettweiler A Sunny Moo Aug 2007 USA (GMT-5) 3·2,083 Posts	Polished off a few of the untested k=6 Sierp. conjectures: S559: conjectured k 6, proven, primes: 4559^1+1 S594: conjectured k 6, proven, primes: 2594^4+1 3594^1+1 4594^1+1 5594^1+1 k=1 is a GFN. S664: conjectured k 6, proven, primes: 3664^1+1 4664^1+1 k=1 is a GFN; k=2 and k=5 eliminated by trivial factors. S699: conjectured k 6, proven, primes: 2699^1+1 4699^1+1 S769: conjectured k 6, proven, primes: 4769^3+1 k=2 eliminated by trivial factors. S804: conjectured k 6, proven, primes: 2804^1+1 3804^4+1 4804^1+1 5804^1+1 k=1 is a GFN. S874: conjectured k 6, proven, primes: 3874^2+1 4874^77+1 k=1 is a GFN, k=2 and k=5 eliminated by trivial factors. S909: conjectured k 6, proven, primes: 2909^10+1 4909^1+1 S979: conjectured k 6, proven, primes: 4979^1+1 k=2 eliminated by trivial factors. S1014: conjectured k 6, proven, primes: 21014^1+1 31014^3+1 41014^1+1 5*1014^3+1 k=1 is a GFN. That should be the last of the untested Sierpinski k=6 conjectures.