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Old 2013-09-26, 19:51   #353
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8{p^2}{k^2} + 16{p}{k} + 7 But now I found my mistake. It's not 2kp+1 form it's 2j(2lp+1)+1 form which is only of 2kp+1 form if j divides by p.
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Old 2013-10-10, 00:41   #354
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Code:
 
? a=[2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281,3217,4253,4423,
 9689,9941,11213,19937,21701,23209,44497,86243,110503,132049, 216091,756839,
 859433,1257787,1398269,2976221,3021377,6972593,13466917,20996011,24036583,
 25964951,30402457,32582657,37156667,42643801,43112609,57885161];
b=vector(#a,n,n);for(y=1,#a,print((sum(x=1,y,a[x]*b[x])%120)%2","y))
Quote:
0,1
0,2
1,3
1,4
0,5
0,6
1,7
1,8
0,9
0,10
1,11
1,12
0,13
0,14
1,15
1,16
0,17
0,18
1,19
1,20
0,21
0,22
1,23
1,24
0,25
0,26
1,27
1,28
0,29
0,30
1,31
1,32
0,33
0,34
1,35
1,36
0,37
0,38
1,39
1,40
0,41
0,42
1,43
1,44
0,45
0,46
1,47
1,48
I'm guessing this is just random but the dot product of these two vectors ( at least by the Wikipedia's example of dot product) appear to follow 0,0,1,1,... (x (mod 120) mod 2) what are the odds it continues like this ? EDIT: I realized that the pattern is only because odd * odd index = odd and that switches the value, then odd * even index = even keeping the previous value.

Last fiddled with by science_man_88 on 2013-10-10 at 00:55
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Old 2013-10-25, 20:11   #355
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2x^2+4x+1 should work for any sequence of exponents (2^n*p+(2^n-1)) for some p. I have a few examples outside of the Double Mersenne numbers (edit:though I'll admit I can't prove it with my mind right now). Most probably know of the LL version using 2x^2-1 for steps. using pari:

Code:
? 2*(x-y)^2-1
%9 = 2*x^2 - 4*y*x + (2*y^2 - 1)
lets say y is the number needed to get to the residue in that step the resulting value for the next residue can subtract 2x^2+4x+1 and we get:

Code:
? %-(2*x^2+4*x+1)
%10 = (-4*y - 4)*x + (2*y^2 - 2)
now I know this will be a negative value (edit: most of the time I think), but I'm hoping to be able to generalize to residue Mx+y or at least 2x+y ( edit:doh that was quick for a 2x+y solution) eventually then we can continue to find residues for new Mersennes possibly faster this way (2,5,11,23,47,... is an exponent sequence like what I described above, using this we can use the residue from one step on one of them to find residues for them all at other steps). What does everyone think ?

Last fiddled with by science_man_88 on 2013-10-25 at 20:42
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Old 2013-10-25, 23:13   #356
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Code:
 
(2*m^2 - 2)*z^2 + (4*m*d - 4)*z + (2*d^2 - 2)
seems to be the mz+d ( since I used x and y before they had values already) case.
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Old 2013-10-27, 14:25   #357
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one thing I've realized is this mz+d form shows for the next residue to be 0 (because z mod z=0 mod z) the only part that needs to be shown to be 0 is 2d^2-2 which is 0 any time d^2=1 mod the next one in the sequence ( and d has to be less than the current one in the list). I guess I should look more into the catalan sequence.
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Old 2013-11-15, 19:23   #358
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Code:
 
? factormod(2*x^2+4*x+1,127)
%4 =
[Mod(1, 127)*x + Mod(9, 127) 1]
[Mod(1, 127)*x + Mod(120, 127) 1]
If it's the same x (which a closer look using y as the variable says it is) I'm wondering what does this tell us about things ? All factors have to be 1 mod the exponent when the exponent is prime. What k's could this eliminate ?

Last fiddled with by science_man_88 on 2013-11-15 at 19:24
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Old 2013-11-16, 00:06   #359
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127*(y*x+z)+x+9
127*(y*x+z)+x+120

is what I get for conversion from mod to equations.
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Old 2014-03-25, 21:54   #360
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Code:
? MMTF(x,p) = {
a=2;for(y=2,#binary(p),if(binary(p)[y]==1,a=(2*a^2)%x,a=(a^2)%x));a-1
}
%1 = (x,p)->a=2;for(y=2,#binary(p),if(binary(p)[y]==1,a=(2*a^2)%x,a=(a^2)%x));a-1
? MMTF(47,23)
%2 = 0
? MMLL(x,p) = {
s=4;for(y=1,p-2,s=(s^2-2)%x);s
}
%3 = (x,p)->s=4;for(y=1,p-2,s=(s^2-2)%x);s
? MMLL(2047,11)
%4 = 1736
? MMTF(15,7)
%5 = 7
? MMTF(15,127)
%6 = 7
? MMTF(255,127)
%7 = 127
? MMLL(255,127)
%8 = 194
? MMLL(15,7)
%9 = 14
In reality, this is more a question. If I didn't subtract one in the MMTF script would the difference between the scripts for Catalan Mersennes stay at floor(2/3*10^ ((number of digits in p)-1))

Last fiddled with by science_man_88 on 2014-03-25 at 22:01
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Old 2014-03-26, 15:19   #361
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Quote:
Originally Posted by science_man_88 View Post
Code:
? MMTF(x,p) = {
a=2;for(y=2,#binary(p),if(binary(p)[y]==1,a=(2*a^2)%x,a=(a^2)%x));a-1
}
%1 = (x,p)->a=2;for(y=2,#binary(p),if(binary(p)[y]==1,a=(2*a^2)%x,a=(a^2)%x));a-1
? MMTF(47,23)
%2 = 0
? MMLL(x,p) = {
s=4;for(y=1,p-2,s=(s^2-2)%x);s
}
%3 = (x,p)->s=4;for(y=1,p-2,s=(s^2-2)%x);s
? MMLL(2047,11)
%4 = 1736
? MMTF(15,7)
%5 = 7
? MMTF(15,127)
%6 = 7
? MMTF(255,127)
%7 = 127
? MMLL(255,127)
%8 = 194
? MMLL(15,7)
%9 = 14
In reality, this is more a question. If I didn't subtract one in the MMTF script would the difference between the scripts for Catalan Mersennes stay at floor(2/3*10^ ((number of digits in p)-1))
Sorry wrong value, the point is that for x=15,p=7 I get a difference of 6 once I add the one back in. Doing the same for x=255,p=127 I get a difference between the scripts of 66 , what I want to know is is this just random or does it follow a pattern we can use to figure out the values for (2^127-1) etc.
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Old 2014-04-09, 22:55   #362
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Okay this time I'm just grasping at things, but could Fermat's little theorem be used to figure out values for q=2^p-1 that could be prime through use of the binomial theorem ? because 4^(q-1)==(2^2)^(p-1) = (2^(2*p-2)) 1 mod q and 14^(q-1) = 1 mod q that is (4+10)^(p-1) which can be expanded under the binomial theorem, or is this just over complicating things ?

Last fiddled with by science_man_88 on 2014-04-09 at 22:57
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Old 2014-04-10, 00:04   #363
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Quote:
Originally Posted by science_man_88 View Post
Okay this time I'm just grasping at things...
In all honesty, I can't tell if you are profoundly stupid, or profoundly brilliant.

I suspect the former.
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