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 [url]http://www.mersenneforum.org/showthread.php?p=212725#post212725[/url] 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)

You have an equals sign inside parentheses. I have no idea what you are trying to say, and no concept of how to fix your parens so that you have an equation that makes sense.

How does this relate to low-weight k's? Or even Riesel numbers?
-Curtis

[quote=VBCurtis;212851]You have an equals sign inside parentheses. I have no idea what you are trying to say, and no concept of how to fix your parens so that you have an equation that makes sense.[/quote]
I think that when he says:[quote=science_man_88;212847](( primes = -1 mod 8 ) + 1)/8.[/quote]
He means:[quote]For a prime p where [tex]p \equiv -1 \text{ (mod 8)}[/tex],
(p + 1)/8[/quote]Or, in other words, n in p+1=8*n (where p is a prime).
[quote=VBCurtis;212851]How does this relate to low-weight k's? Or even Riesel numbers?[/quote]
Not even gonna guess. :smile:

vb and mini I copied that equation straight from the OEIS so it's not me who messed it up.

the way i related it is that I made a pari code to find some k<100 when n = 2 or 3 I looked them up in oeis and found the 2 sequences above. I'll post a pm I sent someone if you want what I think.

I see some that k and f do slightly different like:

1) f can count down from a maximum k seems to count up from a minimum.

but they both count groups of 2 if I understand accurately

we might be able to use this fact to solve both.

e.g

(14:00) gp >
forstep(n=5,6,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(k","n))))
1,5
4,5
256,5
4096,5
16384,5

first k>0 we get that fits 4^a = 1 a = 0 0 set of 2 up from 5 is 5 ?

then it's 4^1 = 4 a = 1 1 set of 2 up from 5 = 7 next exponent

then it's 4^4 = 256 a = 4 4 set of 2 up from 5 = 13 next exponent

then it's 4^6 = 4096 a = 6 6 set of 2 up from 5 = 17 next exponent

then it's 4^7 = 16384 a = 7 7 set of 2 up from 5 = 19 next exponent

see what I mean ? f is doing this decreasing haven't found a minimum for f.
Quote:
Originally Posted by mdettweiler
Quote:
Originally Posted by science_man_88
sorry wrong thing lol
First of all, I think you may have gotten the wrong guy--I haven't actively particated in the Riesel Prime Search project for quite a while. However, that aside, I'm afraid I can't make heads or tails of the math you're trying to present in your post--not that it's necessarily incorrect, just that there's probably others on this forum who can apply it much better than I could. My knowledge of such things is not that great.

Since you made your post only about 45 minutes ago, I'd suggest to give it a day or two--somebody else with more knowledge than I will probably spot it and give it the mathematical analysis you're looking for.

Max

[quote=science_man_88;212857]I see some that k and f do slightly different like:

1) f can count down from a maximum k seems to count up from a minimum.

but they both count groups of 2 if I understand accurately

we might be able to use this fact to solve both.

e.g

(14:00) gp >
forstep(n=5,6,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(k","n))))
1,5
4,5
256,5
4096,5
16384,5

first k>0 we get that fits 4^a = 1 a = 0 0 set of 2 up from 5 is 5 ?

then it's 4^1 = 4 a = 1 1 set of 2 up from 5 = 7 next exponent

then it's 4^4 = 256 a = 4 4 set of 2 up from 5 = 13 next exponent

then it's 4^6 = 4096 a = 6 6 set of 2 up from 5 = 17 next exponent

then it's 4^7 = 16384 a = 7 7 set of 2 up from 5 = 19 next exponent

see what I mean ? f is doing this decreasing haven't found a minimum for f.
Quote:
Originally Posted by mdettweiler
Quote:
Originally Posted by science_man_88
sorry wrong thing lol
First of all, I think you may have gotten the wrong guy--I haven't actively particated in the Riesel Prime Search project for quite a while. However, that aside, I'm afraid I can't make heads or tails of the math you're trying to present in your post--not that it's necessarily incorrect, just that there's probably others on this forum who can apply it much better than I could. My knowledge of such things is not that great.

Since you made your post only about 45 minutes ago, I'd suggest to give it a day or two--somebody else with more knowledge than I will probably spot it and give it the mathematical analysis you're looking for.

Max[/quote]
As the author of the PM you quoted, I think I've got a vague idea of what you're trying to do here right now (or at least how to apply it using PARI/GP). As you mentioned before it's rather slow going like this since PARI has to primality test each of the numbers itself, but I suppose if you modified the command like this:

[I]forstep(n=5,6,[2],for(k=1,10000000,if((log(k)/log(4))%1==0,print(k","n))))[/I]

it should simply print out the list of k/n pairs matching the formula--then those can be run through a much faster primality testing application, like LLR or PFGW, to see whether they were accurately predirected.

Please pardon me if I'm misunderstanding what you're trying to express here--I'm sort of taking a stab in the dark here. :smile:

neither n or k will be prime but log(k)/log(4) shows k = 4^a just like in my equation in my newest idea of an idiot post a*2 + n if prime in this case should be a mersenne prime exponent therefore it's the a that matters but that only gets found knowing it from finding a k=4^a.


by the way as 2* a implies a is counting the number of times 2 inbetween two prime n values that should both be mersenne if log(k)/log(4) is a whole number and works as a solution to k*2^n-1 of course.

the reason for posting the OEIS stuff is because all my found k values for n =2 or n=3 fit the respectively sequences i gave they have a general formula of
(( Primes = -1 mod 2^n ) + 1)/2^n.

as posted above

if this holds plug in a n value and get a list of all k values ( somehow I think this might not hold).

okay not all are prime:


[CODE](17:25) gp > forstep(n=5,6,[2],for(k=1,10000000,if((log(k)/log(4))%1==0,print(k","n))))
1,5
4,5
16,5
64,5
256,5
1024,5
4096,5
16384,5
65536,5
262144,5
1048576,5
4194304,5
(17:31) gp > forstep(n=5,3,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(k","n))))
(17:49) gp > forstep(n=5,6,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(k","n))))
1,5
4,5
256,5
4096,5
16384,5
(17:53) gp >[/CODE]

if only we could find a faster way to find the variable a.

[CODE](18:20) gp > forstep(n=5,6,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(log(k)/log(4)))))
0.E-28
1.000000000000000000000000000
4.000000000000000000000000000
6.000000000000000000000000000
7.000000000000000000000000000
(18:25) gp > forstep(n=3,4,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(log(k)/log(4)))))
0.E-28
1.000000000000000000000000000
2.000000000000000000000000000
5.000000000000000000000000000
7.000000000000000000000000000
8.000000000000000000000000000
(18:30) gp > forstep(n=7,8,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(log(k)/log(4)))))
0.E-28
3.000000000000000000000000000
5.000000000000000000000000000
6.000000000000000000000000000
(18:33) gp > forstep(n=13,14,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(log(k)/log(4)))
0.E-28
2.000000000000000000000000000
3.000000000000000000000000000
9.000000000000000000000000000
(18:37) gp > forstep(n=17,18,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(log(k)/log(4)))
0.E-28
1.000000000000000000000000000
7.000000000000000000000000000
(18:41) gp > forstep(n=5,6,[2],for(k=1,10000000,if((log(k)/log(4))%1==0,print(log(k)/log(4)))))
0.E-28
1.000000000000000000000000000
2.000000000000000000000000000
3.000000000000000000000000000
4.000000000000000000000000000
5.000000000000000000000000000
6.000000000000000000000000000
7.000000000000000000000000000
8.000000000000000000000000000
9.000000000000000000000000000
10.00000000000000000000000000
11.00000000000000000000000000
(18:53) gp > forstep(n=3,4,[2],for(k=1,10000000,if((log(k)/log(4))%1==0,print(log(k)/log(4)))))
0.E-28
1.000000000000000000000000000
2.000000000000000000000000000
3.000000000000000000000000000
4.000000000000000000000000000
5.000000000000000000000000000
6.000000000000000000000000000
7.000000000000000000000000000
8.000000000000000000000000000
9.000000000000000000000000000
10.00000000000000000000000000
11.00000000000000000000000000
(18:57) gp > forstep(n=2,3,[2],for(k=1,10000000,if((log(k)/log(4))%1==0,print(log(k)/log(4)))))
0.E-28
1.000000000000000000000000000
2.000000000000000000000000000
3.000000000000000000000000000
4.000000000000000000000000000
5.000000000000000000000000000
6.000000000000000000000000000
7.000000000000000000000000000
8.000000000000000000000000000
9.000000000000000000000000000
10.00000000000000000000000000
11.00000000000000000000000000
(19:01) gp > forstep(n=2,3,[2],for(k=1,10000000,if((log(k)/log(4))%1==0 && isprime(k*2^n-1),print(log(k)/log(4)))))
0.E-28
(19:05) gp >[/CODE]

wonder if I can find out a way to solve for a with this

the problem is 2*a + n isn't going to cut it as I'm almost positive that all primes can fit this.

[QUOTE=science_man_88;212861]neither n or k will be prime but log(k)/log(4) shows k = 4^a just like in my equation in my newest idea of an idiot post a*2 + n if prime in this case should be a mersenne prime exponent therefore it's the a that matters but that only gets found knowing it from finding a k=4^a.


by the way as 2* a implies a is counting the number of times 2 inbetween two prime n values that should both be mersenne if log(k)/log(4) is a whole number and works as a solution to k*2^n-1 of course.[/QUOTE]

I now retract this partially as it could just be interpreted as all prime exponents would give a mersenne prime

I always thought the 8 and 4 represented 2^n I then worked out what mini proposed to n*2^n-1=p if p must be prime I did the code


(19:48) gp > for(n=1,100,if(isprime(n*2^n-1),print(n)))
2
3
6
30
75
81


so i think if what mini said is accurate these are values of n for which we can calculate k values by putting them in the formula from the sequences

I think this all belongs to my idea of and idiot type post though k=4^f=4^a if i realized right if so can we rework the f=p-1-(q-1)/2 ? if k=4^f


k=4^(p-1-(q-1)/2)

if 2^q-1 = k*2^p-1

2^q-1 = 4^(p-1-(q-1)/2)*2^p-1
2^q-1 = 2^2*(p-1-(q-1)/2)*2^p-1

now if only i could get the q to the other side and simplify I might get a direct relation between p and q I've had dreams of this day I think I think it came out to something like q^5=p^6 in my dream.

2^q-1 = 2^(2*(p-1-(q-1)/2)+p)-2^2*(p-1-(q-1)/2) I think

[url]http://oeis.org/classic/A002234[/url]

2^q-1 = 2^((2p-2-(2q-2)/4)+p)-2^(2p-2-(2q-2)/4)


someone else finish it

or correct it if it's been done wrong ( probably more likely)

[QUOTE=CRGreathouse;212876][url]http://oeis.org/classic/A002234[/url][/QUOTE]

thanks you set me straight

[QUOTE=science_man_88;212877]2^q-1 = 2^((2p-2-(2q-2)/4)+p)-2^(2p-2-(2q-2)/4)


someone else finish it

or correct it if it's been done wrong ( probably more likely)[/QUOTE]

I believe q = ((2*p-2-(2*q-2)/4)+p)-(2*p-2-(2*q-2)/4)

I messed it up because I get q = p

1/2^(q-1) = 4^f/2^(2p-2) if k =4^f

k=1 q-1=2p-2;q=2p-1 like with 7 and 13
k=2 q-1=4p-4;q=4p-3
k=3 q-1 =6p-6;q=6p-5
k=4 q-1 =8p-8;q=8p-7

etc.

are these accurate ?

also I suck as this proves nothing beyond common knowledge right ?

though p and q both have to be prime and k must be a power of 4 if I deduced correctly.

k=1 q=2p-1
k=4 q=8p-7
k=16 q=32p-31
k=64 q=128p-127

is what I get but it must be wrong as it was in the newest idea of an idiot thread.

that is unless someone read my reduction wrong

too bad you can't merge threads between forums but leave them in their own forums because this relates to riesel to an extent but also to my newest idea of an idiot post in misc. math the only area I thought I could post in ( in which case good enforcing to the people who basically told me I'm a crank and to stay in misc. math).

oh and if my q=_p-_ are accurate the hard part is testing all k values(or f values) possible. I know the way to set the limit on both k and f but the hard part is still going to be finding the minimum ( k could be 1/4 if f = -1 ( found a few negative f values already))

science_man_88
 

Please state clearly what's your objective?

And kindly stop replying to your own posts. If nobody else replies it means people are not interested.

Thank you.

since it was only 1 hour between all that's not nobody is interested especially since i saw 3 members on the whole forum at one point during those. 1) if you didn't want people posting why make a forum. 2) if you don't like "cranks" put a anti crank test in the registration this would be able to solve it.

Post your BS elsewhere...

point to the bs then maybe i'll take your advice.

em99010pepe vs. science_man_88 the rest of the forum presiding

judge and jury I understand he thinks everything I write is pure BS yet he has presented no evidence to support such a claim hence his failure to present the evidence clearly undermines his whole entire claim.

Kosmaj my objective is to find k such that we could apply them to mersenne exponents and test for primality of the resulting exponents if we can narrow the search with q=2p-1 etc. where p and q are prime we could decrease # of checks needed to prove up to a new mersenne possibly increasing the speed of the search.

a or f max = p-1-(p-1)/2 I think
no wait it was f = (2*(p-1)-(p-1))/2

so k has a max of 4^f max or 4^a max

if we find a f minimum we can search k such that log(k)/log(4) fits a range of fmin-fmax

not to mention only ones to create a prime q and if we know a formula for min k maybe we can apply it to f and find f min to get it started.

so basically I'm trying to decrease the needed checks to find a new mersenne so far it's not looking good.

The point people are trying to make is that you have found *nothing* yet. Think on your own, not in a forum that has zero content dedicated to theory. When you find an actual result that *does* something, then post it. Posting 50 items in 24 hrs discourages anyone who might be slightly tempted to suggest help from wading through the BS to try to find anything useful.

I, for one, was interested for about 5 posts. Then I realized you had no idea what you were doing, but pretended you understood it all (e.g. "I quoted that equation, so it's not my error" rather than any explanation of what it meant. You didn't know, so you quoted and hid behind the quote.)

Please go away until you have a FINAL conclusion that actually does cut the number of tests- that is, a PROOF that certain forms can never be prime.

When you do find such a thing, post it in theory or misc math where people with high-level knowledge can check it. This forum is used to coordinate searches, and contains no theory.

-Curtis

I'll admit I messed up one thing. I'll also admit you didn't thoroughly look at the first post as I clearly stated before you claimed [QUOTE][B]You[/B] have an equals sign inside parentheses[/QUOTE] when in fact I told you outright by posting the sequence number in the OEIS they aren't mine.Therefore you were sinking from the start. you also said [QUOTE]I have no idea what you are trying to say, and [B]no concept of how to fix your parens[/B] so that you have an equation that makes sense.[/QUOTE] which shows that even you don't know what I'm saying hence how do you figure I will as I'm told I basically have the lowest iq of anyone on here or to ever have lived. I'll admit I talk a lot of crap some days but I've proven so do you by complaining I don't know anything but admitting in your very first response that you didn't / don't either since you acted like a king in the last post you made here I'm guessing you should feel dethroned(if not check for super glue ( aka friends) on your chair they just might of saved your behind from dethroning.

[QUOTE] or to ever have lived.[/QUOTE] wtf I didn't write that who the ______ is messing with my posts if you're willing to do that why not "organize my ideas" for me as you complain like that as well

science_man, thanks for posting your objective. I have some questions, if I may:

[QUOTE=science_man_88;212938]Kosmaj my objective is to find k such that we could apply them to mersenne exponents and test for primality of the resulting exponents if we can narrow the search with q=2p-1 etc. where p and q are prime we could decrease # of checks needed to prove up to a new mersenne possibly increasing the speed of the search.[/QUOTE]

Your objective is:
To find k such that[list][*] You can apply them to Mersenne exponents (what does this mean?)[*] You can test the primality of Mersenne exponents (is this what you mean, or do you mean to test the primality of the Mersenne numbers themselves?)[/list]to the end that you can[list][*] Narrow the search (for what?)[*] Decrease the number of checks needed to prove a Mersenne (do you mean decreasing the number of steps in the Lucas-Lehmer test, or do you mean a new algorithm for proving primality of Mersenne numbers, or a new algorithm to prove compositeness of Mersenne numbers?)[/list]
Can you clarify these points? Also, if possible, would you define the variables k, p, and q in your post? You do give an equation relating p and q, but it seems more relevant to Sophie Germain primes than Mersennes.

q is the next mersenne exponent if possible p is the one you know now for mersenne exponent k is the notation I was given in k*2^n-1 ( n=p)

basically it uses 2^q-1 = k*2^p-1

the search is for mersenne primes/ there exponents if possible

possibly we could incorporate this into lucas lehmer not sure (if I knew a relation for s p k q etc. [SIZE="7"]maybe[/SIZE] I'd be able to shorten a check(and that's a big maybe))

[QUOTE=science_man_88;213074]q is the next mersenne exponent if possible p is the one you know now for mersenne exponent k is the notation I was given in k*2^n-1 ( n=p)

basically it uses 2^q-1 = k*2^p-1[/QUOTE]

[TEX]2^q-1=k\dot2^p-1[/TEX]
[TEX]2^q=k\dot2^p[/TEX]
[TEX]k=2^{q-p}[/TEX]
[TEX]\lg k=q-p[/TEX]

So you're really just looking for the difference between successive Mersenne exponents. For what it's worth, that's Sloane's [url=http://oeis.org/classic/A134458]A134458[/url].

without a formula incorporating what I know I can't try to extend it though.

can we simplify the lucas-lehmer test this same way?

[QUOTE=science_man_88;213119]can we simplify the lucas-lehmer test this same way?[/QUOTE]

In the same way as what?

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 ?

[QUOTE=science_man_88;213150]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 ?[/QUOTE]

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?

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.

[QUOTE=science_man_88;213162]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.[/QUOTE]

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

[QUOTE=science_man_88;212936]judge and jury I understand he thinks everything I write is pure BS
[/QUOTE]
He's not the only one who things at least most of what you have written is BS.
[QUOTE=science_man_88;212936]
yet he has presented no evidence to support such a claim hence his failure to present the evidence clearly undermines his whole entire claim.[/QUOTE]
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.

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

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".

[QUOTE=science_man_88;212847]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 [url]http://www.mersenneforum.org/showthread.php?p=212725#post212725[/url] 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)[/QUOTE]


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

1)[CODE](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[/CODE]


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)

[B][U]NOTE[/U][/B]: 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).

((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)

And this helps the Riesel Prime Search project, how?

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

E a merda continua....

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.

give me a simplified version of the equation and I will try to go as high as possible

for 2e6 it should fit ((Primes = -1 mod 2^2e6)+1)/2^2e6 simplify and solve (or check)

[QUOTE=science_man_88;213231]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)[/QUOTE]

These give sequences. But what do we do with these sequences?

[QUOTE=science_man_88;213231][B][U]NOTE[/U][/B]: 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).[/QUOTE]

He's very busy, so I hope that won't be needed. I should be able to answer questions about the OEIS. [SIZE="1"](I am one of the associate editors, after all...)[/SIZE]

If you're wondering about A005122: I agree that the definition is badly worded. (Yes, it even happens to Neil some times...) The intent is this: For each prime p with [TEX]p\equiv-1\pmod8[/TEX], the associated sequence element is (p+1)/8. So the primes of that form are 7,23,31,47,71,79,103, ... = [url=http://oeis.org/classic/A007522]A007522[/url] and they become (7+1)/8 = 1, (23+1)/8 = 3, etc.

could you try the equivalents for n=4 plugged into 2^n ? does this come to to the k values known or do I need to rethink it a bit.

now that you've put it in that form even i can understand it. ((primes=-1 mod 8)+1)/8 so (A007522(n) + 1)/8 if I read it all correct. now I understand enough to start checking more.

got it working for n = 4 just as I said ((Primes=-1 mod (2^n))+1)/2^n worked until 100 should I now try 2e6 ? or still start low. I think my sister will be annoyed if my computer makes a sound as when it works it makes noise.

okay didn't work for 5 wonder if i can change it slightly to make it work.

[QUOTE=science_man_88;213269]got it working for n = 4 just as I said ((Primes=-1 mod (2^n))+1)/2^n worked until 100[/QUOTE]

Please tell us what you mean by this. The sequence you refer to is, I imagine, primes equivalent to -1 mod 16, plus one, divided by 16 -- that is,
2,3,5,8,12,14,15,17,23,24,27,29,30,38,45,47,54,57,62,...

But what do you mean when you say it "worked"? (Edit: missed your last post. What do you mean it worked up to 100 other than 5?)

basically I'm checking if ((primes=-1 mod 2^n)+1)/2^n is equal to k<100 for n= some value.

so far up to 100 it's worked for 2,3,4 but not 5 and not 6 as far as i can tell.

and if the numbers are in a sequence in the OEIS I can't tell which one as 2,3,4 gives over 9100 sequences.

[QUOTE=science_man_88;213272]basically I'm checking if ((primes=-1 mod 2^n)+1)/2^n is equal to k<100 for n= some value.[/QUOTE]

What does this mean? Please give numerical examples. The sequence members are
2,3,5,8,12,14,15,17,23,24,27,29,30,38,45,47,54,57,62,...
so give an example for 2, an example for 3, and example for 5, and example for 8, and an example for 62 (say).

[CODE](20:28) gp > forprime(x=1,1700,if(x%16==15,print((x+1)/16)))
2
3
5
8
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62
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104[/CODE]

compare k with those values

[CODE](20:17) gp > for(n=4,4,for(k=1,100,if(isprime(k*2^n-1),print(k"*"2"^"n"-1"))))
2*2^4-1
3*2^4-1
5*2^4-1
8*2^4-1
12*2^4-1
14*2^4-1
15*2^4-1
17*2^4-1
23*2^4-1
24*2^4-1
27*2^4-1
29*2^4-1
30*2^4-1
38*2^4-1
45*2^4-1
47*2^4-1
54*2^4-1
57*2^4-1
62*2^4-1
65*2^4-1
68*2^4-1
69*2^4-1
72*2^4-1
77*2^4-1
80*2^4-1
83*2^4-1
89*2^4-1
90*2^4-1
92*2^4-1
93*2^4-1
98*2^4-1
99*2^4-1[/CODE]

So you're taking primes p that are 15 mod 16 and applying the following:
[TEX]\frac{p+1}{16}\cdot2^4-1[/TEX]
and noticing that the results are prime?

if i read it right yes.

same with:

[TEX]\frac{p+1}{8}\cdot2^3-1[/TEX]

and

[TEX]\frac{p+1}{4}\cdot2^2-1[/TEX]

Precisely. Would you simplify those? For example, 2^3 = 8, and you can then bring the 8 into the fraction.

I realize they give p but if n is known we get k = p-1/2^n possible n's so far haven't really been checked but the ones I know work with this are 2,3,4 which like I said occur in over 9100 sequences in the OEIS.

[QUOTE=science_man_88;213282]I realize they give p but if n is known we get k = p-1/2^n possible n's so far haven't really been checked but the ones I know work with this are 2,3,4 which like I said occur in over 9100 sequences in the OEIS.[/QUOTE]

They all simplify in like fashion (for all n).* In all cases, you're just wrapping a prime, unwrapping it again, and saying, "Look, it's a prime!".

So take A000040 and multiply its members by 3. Now if you just divide by 3, you'll see the result is prime!


* You can use other sequences beside. If you take the primes that are -3 mod 2^n, add 3, divide by 2^n, multiply by 2^n, and subtract 3 you'll get a prime.

I get that but that's the form k took in these circumstances. worked for 2,3,4 that I can see might try more today.

I realize I made a mistake with 5 all n should work. get a hunting lol.

so with all my idiotic post I've got:

for(n=5,5,for(k=1,100,if(isprime(k*2^n-1),print(k"*"2"^"n"-1"))))

into

forprime(x=1,10000,if(x%32==31,print((x+1)/32)))

mind you forprime may need to be changed to look for higher primes

or turned into a prime finder like I tried to get help with on bytes.com forum just change that into a way to search a certain way for a given n and you've got what you need to put it to your own test.

oh and I think if I knew how to get it there completely i might be able to apply it to mersenne primes (a subset)


4^f*2^n-1 = 2^q-1
k*2^n-1 = 2^q-1
p+1/2^n = 2^q-1/2^n-1


CRG care to simplify it more ?

[QUOTE=science_man_88;213312]so with all my idiotic post I've got:

for(n=5,5,for(k=1,100,if(isprime(k*2^n-1),print(k"*"2"^"n"-1"))))

into

forprime(x=1,10000,if(x%32==31,print((x+1)/32)))

mind you forprime may need to be changed to look for higher primes[/QUOTE]


[CODE](12:28) gp > gettime();forstep(x=2^7-1,2^13,[2^7],if(isprime(x) && x%2^7==(2^7-1),print((x+1)/2^7)));gettime()
1
3
9
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31
40
43
48
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64
%3 = 0[/CODE]

n=7 k = 64 64*2^7-1 = 1*2^13-1 13 is prime next mersenne prime found get my drift ?

[CODE](12:29) gp > gettime();forstep(x=2^13-1,2^17,[2^13],if(isprime(x) && x%2^13==(2^13-1),print((x+1)/2^13)));gettime()
1
9
10
16
%4 = 0[/CODE]

n=13 k = 16 16*2^13-1 = 1*2^17-1 17 is prime next mersenne prime found.


[CODE](12:32) gp > gettime();forstep(x=2^17-1,2^19,[2^17],if(isprime(x) && x%2^17==(2^17-1),print((x+1)/2^17)));gettime()
1
4
%5 = 0[/CODE]

n=17 k = 4 4*2^17-1=1*2^19-1 19 is prime next mersenne prime found.

[CODE](12:37) gp > gettime();forstep(x=2^19-1,2^31,[2^19],if(isprime(x) && x%2^19==(2^19-1),print((x+1)/2^19)));gettime()
1
28
34
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51
61
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91
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114
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391
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579
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1021
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1123
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1126
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2115
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2143
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2173
2184
2200
2205
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2245
2250
2260
2266
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2284
2311
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2409
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2800
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2896
2901
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2980
3003
3004
3010
3039
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3054
3078
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3085
3108
3114
3120
3138
3150
3166
3174
3184
3186
3208
3223
3229
3234
3235
3244
3261
3265
3268
3271
3285
3303
3304
3310
3358
3361
3376
3384
3391
3396
3403
3409
3424
3444
3454
3463
3480
3493
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3510
3523
3526
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3583
3600
3606
3631
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3655
3661
3669
3705
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3778
3793
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3816
3825
3858
3864
3879
3901
3904
3913
3916
3936
3946
3988
3999
4005
4009
4018
4026
4035
4066
4074
4084
4086
4093
4096
%6 = 31[/CODE]

n=19 k = 4096 4096*2^19-1 = 1*2^31-1 31 is prime next mersenne prime found.

gave up on the 31 to 61 test took a hour or 2 to get to 40,000,000 + and i know it would have to check until 1,000,000,000 +

[QUOTE=science_man_88;213318]4^f*2^n-1 = 2^q-1
k*2^n-1 = 2^q-1[/QUOTE]

Add 1, take the base-2 log. But what are these supposed to do?

[QUOTE=science_man_88;213332]n=7 k = 64 64*2^7-1 = 1*2^13-1 13 is prime next mersenne prime found get my drift ?[/QUOTE]

No, not in the least way. Please say what you mean rather than hinting at it.

[QUOTE=CRGreathouse;213370]Add 1, take the base-2 log. But what are these supposed to do?[/QUOTE]

try for some link between k n and q very closely.

[QUOTE=CRGreathouse;213370]
No, not in the least way. Please say what you mean rather than hinting at it.[/QUOTE]


in the example n = 7 the first 4^f I see is 4^3 or 64 4^3 = 2^6 *2^7
and you get 2^13 then minus 1 2^13-1 the next mersenne after 2^7-1

[QUOTE=science_man_88;213371]in the example n = 7 the first 4^f I see is 4^3 or 64 4^3 = 2^6 *2^7
and you get 2^13 then minus 1 2^13-1 the next mersenne after 2^7-1[/QUOTE]

So your strategy for finding the next Mersenne prime after M is to test
4^1*(M+1) - 1, 4^2*(M+1) - 1, 4^3*(M+1) - 1, ...?

if it's a k generated and for odd primes this will work try 3:


[CODE](19:30) gp > gettime();forstep(x=2^3-1,2^13,[2^3],if(isprime(x) && x%2^3==(2^3-1),print((x+1)/2^3)));gettime()
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735
738
741
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781
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864
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978
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1005
1011
1014
1021
1024[/CODE]


first 4^f that shows up is 4=4^1 =2^2 2^2*2^3-1 = 2^5-1
second one 16 2^4*2^3-1=2^7-1
third one 1024 2^10*2^3-1 = 2^13-1


hard part is limiting k so we only find the ones we need to be prime.

I still don't know what you're doing.

look at the list 4 16 and 1024 appear these * 2^3-1 give a mersenne

like I said the hard part is narrowing ranges of k

[QUOTE=science_man_88;213380]look at the list 4 16 and 1024 appear these * 2^3-1 give a mersenne[/QUOTE]

You're still being evasive. What is your general procedure? Let's say you were looking for the Mersenne prime following 2^127 - 1. What would be the first, second, etc. number would check for primality?

1)find k=4^f using my codes
2)simplify k*2^127-1 if the resulting exponent is prime (it should be)then 2^new n-1 should be prime.

So you're saying to check 4^1*2^127 - 1, 4^2*2^127 - 1, 4^3*2^127 - 1, etc.?

well I was hoping to find suitable f to plug into k=4^f but pretty much I'm an idiot I know.

OK, so then that's the process I'm asking about.

see testing each f won't help as that's like checking 2^129-1,2^131-1, etc. every gap of 2. so it's most important to find suitable f values to test.

[QUOTE=science_man_88;213387]see testing each f won't help as that's like checking 2^129-1,2^131-1, etc. every gap of 2. so it's most important to find suitable f values to test.[/QUOTE]

Ah. So your idea is to find an idea.

I know k = p+1/2^n but without already knowing p that's pointless if I find good f values that I limit to a point we would only have to check those f that could work which would limit k and more easily choose which k to select so basically now i need a pattern to f maybe i can do what i did for k and find them in the OEIS lol

[QUOTE=science_man_88;213389]if I find good f values that I limit to a point we would only have to check those f that could work which would limit k and more easily choose which k to select[/QUOTE]

So tell us when you find these f.

A139481 seems to work for n=3 (no surprise it's related to p of mersennes A000043) the first difference of this starting at a(3) works for n=7 so if this is calculated in a way other than through A000043 this may be able to help.

[QUOTE=science_man_88;213393]so if this is calculated in a way other than through A000043 this may be able to help.[/QUOTE]

Once again, your idea is to find an idea.

[QUOTE=CRGreathouse;213395]Once again, your idea is to find an idea.[/QUOTE]

At least I'm willing to.

figures adding 1 to every term gives me the one OEIS sequence I've been able to add something to (who am I take a guess lol).

[URL="http://www.research.att.com/~njas/sequences/A146768"]A146768[/URL]

adding 2 to every term leads to a formula for primes.

[URL="http://www.research.att.com/~njas/sequences/A138576"]A138576[/URL]