20110701, 01:28  #1 
"William"
May 2003
New Haven
2^{3}×5×59 Posts 
Other Bases?
Is there any plan for GPU trial factoring for a^n1 when a is not 2?
Oddperfect has a few long standing uses for this (a=41 a=127 a=157 and others) http://home.earthlink.net/~oddperfec...TrialFactoring William 
20110701, 02:04  #2 
Dec 2010
Monticello
5·359 Posts 
wb:
Since you are the supplier of our OBD page, the truthful answer is no, there is no such plan that anyone has discussed....but it sounds to me like there should be one. The large amount of GPU firepower is only going to take GIMPS proper, or even OBD, so far. I see a few bumps in the road, none of them unsurmountable: 1) I need to finish what I am doing in terms of automating responses to the Primenet server. 2) The sieving part that finds candidate factors needs to move from the CPU to the GPU. 3) Whoever is programming it needs to understand the algorithm, and how it differs from the current ones that decide on whether 2^n1 is divided by a potential factor. They then need to program it. 4) i see a need for a P1 program on GPUs. At least one of the potential people to carry out step 3 (myself) is interested in doing this. mfaktc, on the outer levels,is not a hard or large program. At the risk of distracting me, why don't you explain to me (and quite possibly Oliver) what it is we are attempting to TF, which candidates we can eliminate by theorem, and how to decide whether a candidate is a factor. If I read you correctly, the second line says we are trying to factor 41^(2 * 128159)1, and the third line says we are trying to factor 41^(2^2 * 128159)1 => 41^(4*128159)1, correct? 
20110704, 15:36  #3  
"William"
May 2003
New Haven
2^{3}·5·59 Posts 
Quote:
Like in the base 2, all factors will be k*exponent+1. When the exponent is odd, k must be even. Last fiddled with by wblipp on 20110704 at 15:38 

20110704, 18:15  #4 
Dec 2010
Monticello
1795_{10} Posts 
So, how do we efficiently determine if k*(4*128159)+1 divides 41^(4*128159)1?

20110704, 21:11  #5 
Jun 2011
Henlopen Acres, Delaware
7×19 Posts 

20110704, 21:46  #6  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
http://www.mersenne.org/various/math.php and it works for base 2 but i realized over night how to expand it to base 4 or use it to test 2^(2p+1)1 in less multiplications squaring and mods but higher numbers faster could we not change this to base 41 ? because it should work no ? so if 1 multiply by 41 on top of squaring. I can check it out for you if you want. 

20110704, 21:48  #7  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20110704, 22:25  #8  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
the general form for what that TF does: Quote:
Code:
Remove Optional Square top bit mul by b mod x     1*1 = 1 1 0111 1*b= b 2 2*2 = 4 0 111 no 4 4*4 = 16 1 11 16*b = 32 32 32*32 = 1024 1 1 1024*b = 2048 27 27*27 = 729 1 729*b = 1458 1 

20110705, 12:16  #9  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20110705, 17:28  #10 
Dec 2010
Monticello
5×359 Posts 
That's a weak statement. The general form, of which I am sure the odd perfect folks are aware, is as follows:
for any integer a >1, and any integer p>=1, (a+1)^p  1 is congruent to 0 mod a. To prove it, use the binomial expansion of (a+1)^p In our case, a=40, so 5 and 2^3 divide 41^p  1 I assume they want to find some nontrivial factors, too! 
20110705, 17:34  #11 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
none of you probably care. But ...
Code:
Remove Optional Square top bit mul by b=4 mod x =47     1*1 = 1 1 0111 1*4= 4 4 4*4 = 16 0 111 no 16 16*16 = 256 1 11 256*4 =1024 37 37*37 = 1369 1 1369*4 = 5476 24 24*24 = 576 1 576*4 = 2304 1 changing the last one up by a multiple of 2 brings it back to base 2: Code:
Remove Optional Square top bit mul by b=4 mod x =47     1*1 = 1 1 0111 1*4= 4 4 4*4 = 16 0 111 no 16 16*16 = 256 1 11 256*4 =1024 37 37*37 = 1369 1 1369*4 = 5476 24 24*24 = 576 1 576*8 = 4608 2 okay I know I'd be better formatting it first. 
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