20140415, 05:16  #375 
Apr 2014
5×17 Posts 
While we're at it, I'd like to propose a nice little conjecture, just based on some patterns I noticed of known contiguous Mersenne primes when I was trying to find limitations on the exponent (excluding p = 2):
exponents = 1 mod 3 = 22/41 = 53.7% exponents = 2 mod 3 = 29/41 = 46.3% exponents = 1 mod 4 = 24/42 = 57.1% exponents = 3 mod 4 = 18/42 = 42.9% exponents = 1 mod 5 = 10/41 = 24.4% exponents = 2 mod 5 = 12/41 = 29.3% exponents = 3 mod 5 = 11/41 = 26.8% exponents = 4 mod 5 = 8/41 = 23.4% See the pattern? For any class modulo p that isn't ruled out by primality, the distribution across classes is pretty close to equal proportions. This behavior comes from the fact that there are no real group restrictions enforced by the form of 2^p1 on p when you look at the order calculations. This conjecture would be false if there existed a composite number with a prime order of 2... I want to say it's impossible for such a number to exist... I think I always took this for granted to be true. At the very least it should be really rare. If the order p is prime for composite n, then n divides 2^p1... But that would require order of all prime factors of n to be equal to p, and I believe multiplicative order of a composite number is strictly greater than atleast the individual multiplicative orders which compose it, if nothing else enforced by the totient calculation for composite n. 
20140415, 11:38  #376  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
since I know is greater than 1 I can subtract 2 from this new exponent value and turn most powers of four in the sum into which we know is 1 mod q then we just have to deal with the binomial coefficients times 4^{(nk)*log_4(10)(nk)} for each value k. 

20140604, 00:20  #377 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
I looked into this:
Code:
s=4;q=10;for(x=1,10,s=s^22;q=q^22;print(qs)) 
20140604, 02:35  #378 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Actually I realized since both 4 and 10 divide by 2 we can use the form 2*x^21 for both. I realized this by finding someone's video's partial error on Youtube and continuing. The difference without this alteration is always the difference between two squares, the difference between these in 2*x^21 form is twice the difference between two squares. just pondering things right now I know the difference between two squares is the sum of the difference between two pairs of triangular number to which mersennes can relate. anyways time to be off for the night.

20140616, 12:26  #379 
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 
I've been looking into prime(n) and the totient function:
a mersenne number with y factors has: totatives assume 2kp+1 is prime since prime(n)= we can say you can break this down to a sum involving a mersenne number, but without knowing the properties needed of 2^n1 I can't say that it's prime etc. if I could this is one way to prove the infinitude of the mersenne primes since there are infinitely many prime 2kp+1 so if 2^n1 had to be prime they would also be infinite , am I just looking at this wrong ? I just wish I could find something that was useful for once. Last fiddled with by science_man_88 on 20140616 at 12:28 
20140903, 00:02  #380 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
I just found something new that intrigued me:
for n>=2; and then we have the already show math of Double Mersenne Numbers ? is it so hard to connect when these two make the math work for LL with X or Y changing respectively? Last fiddled with by science_man_88 on 20140903 at 00:06 
20140903, 00:11  #381 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10001110100101_{2} Posts 
[Shatnercommamode]
What, are, you, talking, about? [/Shatnercommamode] 
20140903, 01:02  #382 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
I basically mean each Sn n>=2 is of form 16x^2+16x+2 with x values working in a specific pattern. the Double mersenne numbers can be connected with 2y^2+4y+1 with y taking on the previous value in the double mersenne numbers. so I was asking since each of these sequences can be linked to a polynomial with a certain pattern for there variable in each is it harder/easier to connect the two with these polynomials or the modular connection used in LL.
Last fiddled with by science_man_88 on 20140903 at 01:03 
20140903, 02:01  #383  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
5^{3}·73 Posts 
Quote:
For 127, S_{n} take these values: S_{0}=4, S_{1}=14, S_{2}=67, S_{3}=42, S_{4}=111, S_{5}=0. Explain how you claim that "each Sn n>=2 is of form 16x^2+16x+2" ? As soon as (nonmodulo value) S_{n} > 2^p1 (which is at n ~ log_{2}p), you will take a modulo, or else you are talking about third grade arithmetic (p=3, 5, 7...). Even for the largest testable candidates, the range where your "each Sn n>=2 is of form 16x^2+16x+2" holds is less than 2<=n<=30. For the rest of the millions of iterations, you've got nothing. For doubleMersennes, explain how are you going to do the iterations, whether with modulo >= 2^2^{61}1 or ...maybe without? Elaborate. 

20140903, 02:25  #384  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:
16*3^2 +16*3+1 = 194; 16*48^2+16*48+1 = 37634; etc. as to the modulo I was thinking of having something provable about the relation ship of x and y, and possibly a formula outside of 4*x^2+4*x for the new values of x, which are 3,48,9408,..... ( which I swear I've seen before but can't remember where) edit:I realized that x follows edit2: "one less than "the square of the terms of A002812 which is the reduced LL sequence. Last fiddled with by science_man_88 on 20140903 at 02:36 

20140903, 02:38  #385 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
5^{3}×73 Posts 
Homework for you:
1. Reread http://mersenne.org/various/math.php#lucaslehmer 2. Popquiz: write the definition of Sn in LucasLehmer iteration 3. For bonus points: determine the first n at which lift(Sn)^22 > 2^p1 and modulo will apply. 
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