science_man_88

[CRGreathouse]

Originally Posted by science_man_88

like you did for k if we can relate s p and q then we replace s and k with p and q equalities of them then simply to a relation of p and q ? sound possible ?

I don't know what s is, so I can't relate it nontrivially to p and q. But even if I did, why would this speed up L-L?

[science_man_88]

e.g if I find a relation between s and k maybe I can come up with something like
p*(k/s) = q if k and s can be written as p and q rewrite and simplify as close to a relation between p and q.

[CRGreathouse]

Originally Posted by science_man_88

e.g if I find a relation between s and k maybe I can come up with something like
p*(k/s) = q if k and s can be written as p and q rewrite and simplify as close to a relation between p and q.

You're welcome to waste time on this. I won't.

[flouran]

Originally Posted by science_man_88

judge and jury I understand he thinks everything I write is pure BS

He's not the only one who things at least most of what you have written is BS.

Originally Posted by science_man_88

yet he has presented no evidence to support such a claim hence his failure to present the evidence clearly undermines his whole entire claim.

It has been presented to you clearly in past threads (cf. Misc. Math), and a noted forum-goer posted above that he was wasting his time with these "ideas" of yours.

Of course, you may continue to post your formulas and methods here (since I have no control over that), just don't be taken aback if someone thinks your ideas are nonsensical.

[science_man_88]

Of course, you may continue to post your formulas and methods here (since I have no control over that)

good thing ,

just don't be taken aback if someone thinks your ideas are nonsensical.

I'm not I expect on here as:

1)attempts by non-elites to see if things are found before or asking of a proof etc. always seem to be signs of a "crank"

2) anyone non-elite is reportedly full of "BS".

[science_man_88]

Originally Posted by science_man_88

I've been taught a bit about these a little could these fit

(( primes = -1 mod 8 ) + 1)/8. <---A005122
or
(( Primes = -1 mod 4 ) + 1)/4. <---A005099


these work for 3 and 2 for n I think checked a range of k under 100 the general formula looks like

((Primes=-1 mod 2^n)+1)/2^n does this work for the knowns if so could we use this on the higher exponents to check for k = 4^a which would show the possible mersenne exponents ? I'm almost looking like this at http://www.mersenneforum.org/showthr...725#post212725 I think what is your oppinion ? that's basically what I think I know I had a pari code to check low k and n but I bet it's slow and not worth posting ( I don't even know if what i have posted is worth it)

lets see if I can go through this step by step for you:

1)

(13:37) gp > for(n=2,3,for(k=1,100,if(isprime(k*2^n-1),print(k"*"2"^"n"-1"))))
1*2^2-1
2*2^2-1
3*2^2-1
5*2^2-1
6*2^2-1
8*2^2-1
11*2^2-1
12*2^2-1
15*2^2-1
17*2^2-1
18*2^2-1
20*2^2-1
21*2^2-1
26*2^2-1
27*2^2-1
32*2^2-1
33*2^2-1
35*2^2-1
38*2^2-1
41*2^2-1
42*2^2-1
45*2^2-1
48*2^2-1
50*2^2-1
53*2^2-1
56*2^2-1
57*2^2-1
60*2^2-1
63*2^2-1
66*2^2-1
68*2^2-1
71*2^2-1
77*2^2-1
78*2^2-1
83*2^2-1
87*2^2-1
90*2^2-1
92*2^2-1
95*2^2-1
96*2^2-1
1*2^3-1
3*2^3-1
4*2^3-1
6*2^3-1
9*2^3-1
10*2^3-1
13*2^3-1
16*2^3-1
19*2^3-1
21*2^3-1
24*2^3-1
25*2^3-1
28*2^3-1
30*2^3-1
33*2^3-1
34*2^3-1
39*2^3-1
45*2^3-1
46*2^3-1
48*2^3-1
54*2^3-1
55*2^3-1
58*2^3-1
60*2^3-1
61*2^3-1
63*2^3-1
75*2^3-1
76*2^3-1
79*2^3-1
81*2^3-1
90*2^3-1
91*2^3-1
93*2^3-1
94*2^3-1

for n = 2 look up 17,18,20,21,26,27,32,33,35,38,41,42,45,48,50,53,56,57,60,63,66,68,71,77,78,83,87,90,92,95,96 in the OEIS
A005099 formula given (( Primes = -1 mod 4 ) + 1)/4.

n = 3 look up
1,3,4,6,9,10,13,16
in the OEIS
A005122 formula given (( Primes = -1 mod 8 ) + 1)/8.

general formula of these looks like it might fit ((Primes=-1 mod 2^n)+1)/2^n

if this works out to be true (plug in n = 4 and n = 5 to check) we could use it to predict all k then.

2) take those k values and check known mersenne exponents for ? k=4^a ? plug these in you should get something like:

4*2^3-1 = 1*2^5-1 since the new k value is 1 (and n is prime) we can then plug it in for further mersenne prime testing.

though in the case of n = 2 k = 2 works because 4^a is used for gaps of 2 (most of the primes would fit that)

NOTE: I have NJA sloane's email if he needs to be contacted to make sense (not that he'd listen to me so I should give it to a mod to talk to him).

[science_man_88]

((Primes=-1 mod 2^n)+1)/2^n

I too don't understand the = in the parentheses (of course someone going to say that's a given)

[Beyond]

And this helps the Riesel Prime Search project, how?

[science_man_88]

k for n =2 fits the ((Primes=-1 mod 2^n)+1)/mod2^n simplify and solve all k for 2

plug in n=3 solve to try and get all k for n = 3

plug in 4 ? solve for all k for n=4 ?

etc. up to n = infinity, if this works out


I've checked until k=100 for 2 and 3

[em99010pepe]

E a merda continua....

[Beyond]

Tell you what, when you get up to n => 1e6 or better yet 2e6 and you have positive results, call us. Then we might(?) be interested.