20140903, 02:51  #386  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
2) if you only consider this page it's a modular residue mod the mersenne number of a sequence and can't be said to be used like Si is on the wikipedia page about it. 3) as Si is very roughly 2^(2i) we can say i<(p/2) but that's all I know, as to you saying it's grade three math I didn't learn algebra until grade nine. Last fiddled with by science_man_88 on 20140903 at 02:54 

20140903, 03:45  #387  
Jun 2003
5×29×37 Posts 
Quote:
Code:
4 16 64 256 1024 4096 16384 65536 262144 1048576 

20140903, 11:58  #388 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
sorry I realize now it's roughly 2^(2^i)
Last fiddled with by science_man_88 on 20140903 at 11:59 
20140903, 12:11  #389 
Jun 2003
5×29×37 Posts 

20140903, 12:20  #390 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Last fiddled with by science_man_88 on 20140903 at 12:21 
20141016, 23:18  #391 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
if only it was simpler
I'm guessing if an easy way to use one result of the LL test to figure another it would already be in use, but is there a way to figure out using the multinomial theorem the multinomial connecting two values in the LL sequence separated by a certain difference if so y>a*x+b where y is in the LL sequence and x is the previous mersenne you're jumping from I'm thinking about the result formed when the multinomial is divided by the form z=wp+n; 2^p1=x; 2^z1 = 2^n*((x+1)^w)1 ( yes I eliminated part of a term). I'm guessing it's too complicated or some other flaw has come up ?

20150330, 17:44  #392  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
working backwards from Sieve_of_Atkin wikipedia on mersenne number factors
Quote:


20150517, 23:36  #393  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
moved on again this time to the sieve of sundaram
I started off with the sieve of sundaram using the form of value that created a composite and said well if a mersenne is composite the next one has such and such a form and tried to equate form there I was hoping this would lead me to a property of some sort to look for the exponent or a form to solve into simpler terms that I could evaluate easy for a prime exponent. and yes I went through multiple steps but only show one and mix * and not leaving a space again.
Quote:


20150519, 00:05  #394  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
I got to:
Quote:
Quote:
Last fiddled with by science_man_88 on 20150519 at 00:21 

20150602, 13:48  #395 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
I know I've talked about the original nonreduced LL tests with x^22 and the difference a bit to people but I can't remember if I've talked about how the reduced form using 2*x^21 matches up to within a sum that I believe is mostly powers of two must equal to 1 mod 2^p1 , p obviously being prime ( by convention only) for 2^p1 to be prime. I guess I would have to figure out the formula for the sum to be of any use, anyone else interested ? edit: I realize now it's 2 mod 2^p1 the reason this came to mind is outside the 1 this resembles the trail factoring example on the GIMPS math page. edit: 2 I keep thinking I would have to tack a 1 on the begging but I don't in the case of the one starting with 2 because it's already like it's done the first 1 in the binary.
Last fiddled with by science_man_88 on 20150602 at 14:03 
20150602, 17:26  #396 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
I just thought I'd back this up with code ( prettied up after pasting):
Code:
Mcode(p,{flag=1},{modulus=[p<<1+1,1<<p1][flag]},{zerocode=x^2},{onecode=[zerocode<<1,zerocode<<11][flag]},{bin=[p,1<<(p2)1][flag]},{result=[1,0][flag]},{x=[1,2][flag]})={ a=binary(bin); for(y=1,#a, if(a[y], x=eval(onecode)%modulus, x=eval(zerocode)%modulus ) ); if(x==result, if(flag==1, print("M"p" has a factor "modulus), if(flag==2, print("M"p" is prime") ) ) ) } Code:
addhelp(Mcode," if flag=1 proceed with checking if modulus is a divisor of Mp, if flag=2 proceeed with LL test of Mp") 
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