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 2013-09-26, 19:51 #353 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26×131 Posts $8{p^2}{k^2} + 16{p}{k} + 7$ 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.
2013-10-10, 00:41   #354
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

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

 2013-10-25, 20:11 #355 science_man_88     "Forget I exist" Jul 2009 Dumbassville 203008 Posts $2x^2+4x+1$ 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
 2013-10-25, 23:13 #356 science_man_88     "Forget I exist" Jul 2009 Dumbassville 203008 Posts 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.
 2013-10-27, 14:25 #357 science_man_88     "Forget I exist" Jul 2009 Dumbassville 203008 Posts 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.
 2013-11-15, 19:23 #358 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts 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
 2013-11-16, 00:06 #359 science_man_88     "Forget I exist" Jul 2009 Dumbassville 838410 Posts 127*(y*x+z)+x+9 127*(y*x+z)+x+120 is what I get for conversion from mod to equations.
 2014-03-25, 21:54 #360 science_man_88     "Forget I exist" Jul 2009 Dumbassville 203008 Posts 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
2014-03-26, 15:19   #361
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by science_man_88 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.

 2014-04-09, 22:55 #362 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts 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
2014-04-10, 00:04   #363
chalsall
If I May

"Chris Halsall"
Sep 2002

3×3,181 Posts

Quote:
 Originally Posted by science_man_88 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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