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Old 2014-09-03, 02:51   #386
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Quote:
Originally Posted by Batalov View Post
Homework for you:
1. Re-read http://mersenne.org/various/math.php#lucas-lehmer
2. Pop-quiz: write the definition of Sn in Lucas-Lehmer iteration
3. For bonus points: determine the first n at which lift(Sn)^2-2 > 2^p-1 and modulo will apply.
1) done
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.

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Old 2014-09-03, 03:45   #387
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Quote:
Originally Posted by science_man_88 View Post
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.
This is how 2^(2i) grows:
Code:
4
16
64
256
1024
4096
16384
65536
262144
1048576
Does that look even remotely like LL sequence?
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Old 2014-09-03, 11:58   #388
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Quote:
Originally Posted by axn View Post
This is how 2^(2i) grows:
Code:
4
16
64
256
1024
4096
16384
65536
262144
1048576
Does that look even remotely like LL sequence?
sorry I realize now it's roughly 2^(2^i)

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Old 2014-09-03, 12:11   #389
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Quote:
Originally Posted by science_man_88 View Post
sorry I realize now it's roughly 2^(2^i)
Right. So then, the answer to #3 is?
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Old 2014-09-03, 12:20   #390
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Quote:
Originally Posted by axn View Post
Right. So then, the answer to #3 is?
i= roughly log2(p)

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Old 2014-10-16, 23:18   #391
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Default 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^p-1=x; 2^z-1 = 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 ?
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Old 2015-03-30, 17:44   #392
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Default working backwards from Sieve_of_Atkin wikipedia on mersenne number factors

Quote:
4x^2+y^2-1 = 2kp
y= odd
4x^2+(2a+1)^2-1 = 2kp
4x^2+4a^2+4a = 2kp
2x^2+2a^2+2a = kp; k is even
x^2+a^2+a = bp
if( b=odd and x=odd , a = either,if(b==odd and x=even, a= impossible))
if(b= even and x=odd,a=impossible, if(b==even and x=even, a= either))

aka:

if (k mod 4=2 and x mod 4 = 1 or 3 , y mod 4 = 1 or 3)
if(k mod 4 =0 and x mod 4 = 2 or 0, y mod 4 = 1 or 3)


3x^2+y^2-1 = 2kp
1) x=even -> y=odd
2)x=odd -> y=even

1) 3x^2+(2a+1)^2-1= 2kp
3x^2+4a^2+4a = 2kp
6c^2+2a^2+2a = kp; k is even
3c^2+a^2+a = bp
if( b=odd and c=odd , a = either,if(b==odd and c=even, a= impossible));
( aka (2=x=k)mod 4; y^2=1 mod 4)
if(b= even and c=odd,a=impossible, if(b==even and c=even, a= either))
( aka (0=x=k)mod 4; y^2=1 mod 4;)

2) 3(2a+1)^2+4b^2-1 = 2kp
6a^2+6a+2b^2+1 =kp ; k=odd
6a^2+6a+2b^2+1=(2c+1)(2d+1)
6a^2+6a+2b^2+1 = 4cd+2c+2d+1
3a^2+3a+b^2 = 2cd+c+d
3a^2+3a+b^2 = (2d+1)c+d
if((b=c) mod 2, d=even;a=either)
if((b=d) mod 2, c=even;a=either)
if(a= (c or d) mod 2, (d or c) = b mod 2)

aka:

if((y=0 mod 4 && (k or p) mod 4 = 1) || ( y=2 mod 4 && (k or p)=3 mod 4) , (p or k)=1 mod 4 && x=1 or 3 mod 4)

etc.



3x^2-(y^2+1) = 2kp
1) x=even -> y=odd
2)x=odd -> y=even
1) 3x^2-((2a+1)^2+1)= 2kp
3x^2-4a^2-4a-2 = 2kp
6b^2-2a^2-2a-1 = kp; k is odd
6b^2-2a^2-2a-2+1 = (2c+1)(2d+1)
6b^2-2a^2-2a-2 = 4cd+2c+2d
3b^2-a^2-a-1 = 2cd+c+d
3b^2-a^2-a-1= (2d+1)c+d
if((b=c) mod 2, d=odd; a= either)
if((b=d) mod 2, c=odd; a= either)
if((a=(c or d)= even,(d or c)!=b mod 2)

2) 3(2a+1)^2-(4b^2+1) = 2kp
6a^2+6a-2b^2+1 =kp ; k is odd
3a^2+3a-b^2 = 2cd+c+d
(3a+3)a-b^2 = (2d+1)c+d
if( a=(c or d)=odd, ((d or c)!=b mod 2))
if(a=(c or d)=even, ((d or c)=b mod2))
I haven't completed working around on them but I'd like to see what people think and see if this can be turned into something fast to eliminate k.
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Old 2015-05-17, 23:36   #393
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Default 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:
2(4ij+2j+2i+1)+1
4ij+2j+2i+1 = 2kl+k+l
1 = (2l+1)*k+l-2*(2ij+j+i)
1=(3*k+1)*l+(2*k*l+k)*k-((4k-4)*i*j+(2k-2)*j+(2k-2)*i) /// assume 2k-2 is a variable m
1=(m+k+3)*l +(2*k^2*l+k^2)-(2*m*i*j+m*j+m*i)
1=(k+3)*l+(2*k^2*l+k^2) - m*(2*i*j+j+i -l)
m*(2*i*j+j+i -l)+1=(k+3)*l+(2*k^2*l+k^2)
2*m*i*j+m*j+m*i-m*l+1= k*l+3*l+2*k^2*l+k^2
-k^2 = k*l+3*l+2*k^2*l-2*m*i*j-m*j-m*i+m*l-1
-(2*l+1)*k^2= -(4k-4)*i*j-(2k-2)*j-(2k-2)*i+(3k+1)*l-1
(2l+1)*k^2 = 4*k*i*j-4*i*j +2*k*j-2*j +2*k*i-2*i-3kl-l+1
(2*k*l+k-4*i*j-2*j-2*i+3*l)*k+l = -4*i*j-2*j-2*i+1 /// assume 4*i*j+2*j+2*i-1 is a variable n
(2*k*l+k-4*i*j-2*j-2*i+3*l)*k+l = -n+2
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Old 2015-05-19, 00:05   #394
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I got to:

Quote:
k^2-7 +3*k = r*l /// r*l must be odd regardless of what parity k is and therefore so is l;
before noting something in A002808

Quote:
An integer is composite if and only if it is the sum of strictly positive integers in arithmetic progression with common difference 2: 4 = 1 + 3, 6 = 2 + 4, 8 = 3 + 5, 9 = 1 + 3 + 5, etc. - Jean-Christophe Hervé, Oct 02 2014
which when you show that 2*k*p+1 is just a subset of 2n+1 you can eliminate all k that are certain polite numbers( those that are a sum of (2*j*p+1) numbers) + j in theory because those k happen when you sum up 2*j*p+1 consecutive 2*k*p+1 values. I'm always so slow to catch on to simple things.

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Old 2015-06-02, 13:48   #395
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I know I've talked about the original non-reduced LL tests with x^2-2 and the difference a bit to people but I can't remember if I've talked about how the reduced form using 2*x^2-1 matches up to within a sum that I believe is mostly powers of two must equal to 1 mod 2^p-1 , p obviously being prime ( by convention only) for 2^p-1 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^p-1 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.

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Old 2015-06-02, 17:26   #396
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I just thought I'd back this up with code ( prettied up after pasting):

Code:
Mcode(p,{flag=1},{modulus=[p<<1+1,1<<p-1][flag]},{zerocode=x^2},{onecode=[zerocode<<1,zerocode<<1-1][flag]},{bin=[p,1<<(p-2)-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")
        )
     )
  )
}
here's an addhelp:

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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