20130926, 19:51  #353 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
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.

20131010, 00:41  #354  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×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:
Last fiddled with by science_man_88 on 20131010 at 00:55 

20131025, 20:11  #355 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
should work for any sequence of exponents (2^n*p+(2^n1)) 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 for steps. using pari:
Code:
? 2*(xy)^21 %9 = 2*x^2  4*y*x + (2*y^2  1) Code:
? %(2*x^2+4*x+1) %10 = (4*y  4)*x + (2*y^2  2) Last fiddled with by science_man_88 on 20131025 at 20:42 
20131025, 23:13  #356 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Code:
(2*m^2  2)*z^2 + (4*m*d  4)*z + (2*d^2  2) 
20131027, 14:25  #357 
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Jul 2009
Dumbassville
2^{6}·131 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 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.

20131115, 19:23  #358 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·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] Last fiddled with by science_man_88 on 20131115 at 19:24 
20131116, 00:06  #359 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
127*(y*x+z)+x+9
127*(y*x+z)+x+120 is what I get for conversion from mod to equations. 
20140325, 21:54  #360 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} 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));a1 } %1 = (x,p)>a=2;for(y=2,#binary(p),if(binary(p)[y]==1,a=(2*a^2)%x,a=(a^2)%x));a1 ? MMTF(47,23) %2 = 0 ? MMLL(x,p) = { s=4;for(y=1,p2,s=(s^22)%x);s } %3 = (x,p)>s=4;for(y=1,p2,s=(s^22)%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 Last fiddled with by science_man_88 on 20140325 at 22:01 
20140326, 15:19  #361  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20140409, 22:55  #362 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·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^p1 that could be prime through use of the binomial theorem ? because 4^(q1)==(2^2)^(p1) = (2^(2*p2)) 1 mod q and 14^(q1) = 1 mod q that is (4+10)^(p1) which can be expanded under the binomial theorem, or is this just over complicating things ?
Last fiddled with by science_man_88 on 20140409 at 22:57 
20140410, 00:04  #363 
If I May
"Chris Halsall"
Sep 2002
Barbados
2^{2}×23×103 Posts 

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