theory on Mersenne primes ?
[QUOTE=science_man_88]I know that according to resource on the divisors of Mersenne numbers of prime index that aren't prime are +1/1 mod 8 and of the form 2kp+1 which limits possible k values depending on the exponent mod 8 . I've look at all divisors of the exceptions to 2^371 so far it seems if p mod 6 =5 or 1 then 2 of the factors of 2^p1 seem to be also the same modulo 6 is this verifiable if so could this be used to further reduce the k values needed to be checked ?[/QUOTE]
this is from a pm I sent (bet if ever pm I sent was deleted from people's inbox's the server would run faster lol) 
Note that all primes above 3 are either 1 or 5 mod 6.
[QUOTE=science_man_88]I know that according to resource on the divisors of Mersenne numbers of prime index that aren't prime are +1/1 mod 8 and of the form 2kp+1 which limits possible k values depending on the exponent mod 8 . I've look at all divisors of the exceptions to 2^371 so far it seems if p mod 6 =5 or 1 then 2 of the factors of 2^p1 seem to be also the same modulo 6 is this verifiable if so could this be used to further reduce the k values needed to be checked ?[/QUOTE] Another way to write 5 mod 6 is 1 mod 6. When p is odd, 2^p1 is 1 mod 6. If N is 1 mod 6, then there must be an even number of factors that are 1 mod 6 (because the factors mod 6 have to multiply to the number mod 6). As far as I can tell, this can't be used to make it easier to find factors. 
[QUOTE=MiniGeek;235461]Note that all primes above 3 are either 1 or 5 mod 6.
Another way to write 5 mod 6 is 1 mod 6. When p is odd, 2^p1 is 1 mod 6. If N is 1 mod 6, then there must be an even number of factors that are 1 mod 6 (because the factors mod 6 have to multiply to the number mod 6). As far as I can tell, this can't be used to make it easier to find factors.[/QUOTE] well for the 1/1 mod 8 as well if p=3 mod 8 as I've listed before k = 1,5,9,etc. for 7(1) mod 8 and k=0,4,8,12,16 etc. for 1 mod 8, if p=5 mod 6 to get 1 mod 6 use k=0,3,6,9,etc. ? and for 5 mod 6 k= 1,4,7,etc. ? is so when do k match up for the given mod 8 and mod 6 such that they can equal a common thing number that can be a factor. 
[QUOTE=science_man_88;235466]well for the 1/1 mod 8 as well if p=3 mod 8 as I've listed before k = 1,5,9,etc. for 7(1) mod 8 and k=0,4,8,12,16 etc. for 1 mod 8, if p=5 mod 6 to get 1 mod 6 use k=0,3,6,9,etc. ? and for 5 mod 6 k= 1,4,7,etc. ? is so when do k match up for the given mod 8 and mod 6 such that they can equal a common thing number that can be a factor.[/QUOTE]
Since only primes need to be considered for potential factors, all factors that are not 1 or 5 mod 6 are ignored anyway (all primes over 3 are 1 or 5 mod 6 because all other values mod 6 have 2 and/or 3 as factors). Factors that are 1 mod 6 still have to be tested, as do factors that are 5 mod 6. In the event that it'd be better to test 1 mod 6 and 5 mod 6 factors separately, it might be useful to find out which k's produce which sort of factor and work with that. IIRC, Prime95 does factors within a bit level according to their value mod 120 for efficiency, in which case the value mod 6 isn't at all helpful except in simple implementations. I don't see any way this can be an improvement on current methods. 
My understanding of sm88's idea: if you're factoring a number that is 5 mod 6, at least one of its prime divisors must be 5 mod 6. Can this be used to speed trial division? (This was stated only in the case of Mersenne numbers, but it seems to be more general.)
Generally, the answer seems to be "no". You could search only for primes that are 5 mod 6, but it's quite possible that all such primes are large  greater than sqrt(n). In essence, you're trading a factor of 2 for a factor of sqrt(n) which is a losing proposition. Example: Suppose you're factoring 15419076477348026044248723582269. There are about 1.12e14 primes below the square root of this number, so trial division will take at most this many divisions to factor the number. (In fact it will take 1.3926475881e10 divisions, since the smaller factor is 355894230031.) But the smallest factor that is 5 mod 6 is 43324884688366413299. Now only about half the primes up to that number need be tested, but this is 4.9e17 which is not only greater than 1.4e10 but greater than 1.1e14. 
want an example of what I mean ? too bad I'll give you one anyways lol.
for p=11 p=3 mod 8 for factors 7 mod 8 k=1,5,9,etc. (4n+1) for 1 mod 8 k=0,4,8,12 since they can be 6n+1 or 6n1 since 11= 5 mod 6 6n+1 k=0,3,6,9, (3x+0) 6n1 k = 1,4,7 (3x+1) if we say we need at least 1 factor of the same type mod 6 then we only have to check when 4n+1 or 4n match up with 3x+1 and we should find possible factors k= 1,13,etc. because 4n+1 meets 3x+1 every fourth value and every 3rd value greater than index =2 for 4n this limits it to at most if we can figure on thing out i could narrow it down to finding factors on one or the other k list for mod 8 but anyways. 
[QUOTE=science_man_88;235473]if we say we need at least 1 factor of the same type mod 6 then we only have to check when 4n+1 or 4n match up with 3x+1 and we should find possible factors k= 1,13,etc. because 4n+1 meets 3x+1 every fourth value and every 3rd value greater than index =2 for 4n this limits it to at most if we can figure on thing out i could narrow it down to finding factors on one or the other k list for mod 8 but anyways.[/QUOTE]
We don't always need at least 1 factor of the same type mod 6, and on large numbers we have no clue how many other factors there are. Consider a number that is 1 mod 6. It must have an even number of factors that are 1 mod 6, but that includes 0, 2, 4, etc. And it can have any number of factors that are 1 mod 6. It really doesn't help. It could have two factors that are 1 mod 6 and none that are 1 mod 6. Or 0 factors that are 1 mod 6 and 1 that is 1 mod 6 (a.k.a. it's prime), or anything else. Yes, excluding k's that make 2kp+1 not 1 or 5 mod 6 narrows down the list. It removes a great deal of candidates, but it does so by removing all composites with 2 or 3 as a factor. You can find the same thing by just checking each 2kp+1 to see if it has 3 as a factor (since the form is 2kp+1, they're all odd, 2 can never be a factor). The question, then, is: Is it better to figure out which k's don't have some small numbers as factors and only consider those, or to first consider all k's and then remove them if they have the small factors? I'm pretty sure Prime95 and other modern implementations would have already asked this and have chosen the most efficient way. 
[QUOTE=MiniGeek;235475]Yes, excluding k's that make 2kp+1 not 1 or 5 mod 6 narrows down the list. It removes a great deal of candidates, but it does so by removing all composites with 2 or 3 as a factor. You can find the same thing by just checking each 2kp+1 to see if it has 3 as a factor (since the form is 2kp+1, they're all odd, 2 can never be a factor). The question, then, is: Is it better to figure out which k's don't have some small numbers as factors and only consider those, or to first consider all k's and then remove them if they have the small factors? I'm pretty sure Prime95 and other modern implementations would have already asked this and have chosen the most efficient way.[/QUOTE]
In other words: the idea works and is probably already in use, unless a yet better way has been found by the Prime95 crew. 
You want to combine those factors (2kp+1) = +/ 1 (mod 8) with those that can be prime (2kp+1)= +/ 1 (mod 6).
Prime95 already does this, but instead of using +/ 1 mod 6, it uses those numbers mod 120 which can be prime, i.e. those that are coprime with 120: 1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59,61,67,71,73,77,79,83,89,91,97,101,103,107,109,113,119 Combining the above list with the requirement +/ 1 mod 8 gives 16 residue classes mod 120, which is those 16 Prime95 uses: 1,7,17,23,31,41,47,49,71,73,79,89,97,103,113,119 Your idea gives 20 residue classes mod 120: 1,7,17,23,25,31,41,47,49,55,65,71,73,79,89,95,97,103,113,119 so what Prime95 already uses is more efficient. 
[QUOTE=ATH;235565]You want to combine those factors (2kp+1) = +/ 1 (mod 8) with those that can be prime (2kp+1)= +/ 1 (mod 6).
Prime95 already does this, but instead of using +/ 1 mod 6, it uses those numbers mod 120 which can be prime, i.e. those that are coprime with 120: 1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59,61,67,71,73,77,79,83,89,91,97,101,103,107,109,113,119 Combining the above list with the requirement +/ 1 mod 8 gives 16 residue classes mod 120, which is those 16 Prime95 uses: 1,7,17,23,31,41,47,49,71,73,79,89,97,103,113,119 Your idea gives 20 residue classes mod 120: 1,7,17,23,25,31,41,47,49,55,65,71,73,79,89,95,97,103,113,119 so what Prime95 already uses is more efficient.[/QUOTE] okay well can we alter it all to make it even more efficient ? I know now that it eliminates the multiples of 5 I still must try and figure this out lol. doh dah multiples of 5 still god can i get any dumber lol. 
another theory
I have another theory could the p for mersenne primes be the minimum prime p such that x^p[SUB]n[/SUB]1  x^p[SUB]n1[/SUB]1 is divisible by 6 for some set of x values? it seems to work for x=2 and x=3 so far that I've done. I forgot that p=3 and 2 don't work for x=2 lol but does it work for some set for p>3

[CODE](19:15) gp > for(x=2,20,for(i=2,30,print1(((x^mersenne[i]1)(x^mersenne[i1]1))%6","));print())
4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,[/CODE] anyone else see a pattern ? lol never mind that was disproved easy lol. 
I think it's disproved anyway even if you allow exceptions with this pattern lol.

[QUOTE=science_man_88;235604]I think it's disproved anyway even if you allow exceptions with this pattern lol.[/QUOTE]
is my new theory provable ? / proven ? 
[QUOTE=science_man_88;235933]is my new theory provable ? / proven ?[/QUOTE]
I don't understand what the new conjecture is. 
[QUOTE=CRGreathouse;235942]I don't understand what the new conjecture is.[/QUOTE]
in the code above of : [CODE]for(x=2,20,for(i=2,30,print1(((x^mersenne[i]1)(x^mersenne[i1]1))%6","));print()) 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,[/CODE] there's a pattern if we can prove this stays true could we use it to figure out the p values ? 
[QUOTE=science_man_88;235950]there's a pattern if we can prove this stays true could we use it to figure out the p values ?[/QUOTE]
What pattern are you claiming, precisely? 
[QUOTE=CRGreathouse;235952]What pattern are you claiming, precisely?[/QUOTE]
[CODE]4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,[/CODE] repeats lol 
You're still not being specific enough. You seem to be saying that
[code](x^mersenne[i]1)(x^mersenne[i1]1)[/code] is 4 mod 6 when x is 2 mod 3 and i = 2, but is 0 mod 6 otherwise. But what is mersenne[i]? A000043? A000668? Something else? You really can't expect other people to do the work for you when you're asking them to check your conjectures! 
[QUOTE=CRGreathouse;235960]You're still not being specific enough. You seem to be saying that
[code](x^mersenne[i]1)(x^mersenne[i1]1)[/code] is 4 mod 6 when x is 2 mod 3 and i = 2, but is 0 mod 6 otherwise. But what is mersenne[i]? A000043? A000668? Something else? You really can't expect other people to do the work for you when you're asking them to check your conjectures![/QUOTE] A000043 
now that pattern looks stupid to help.
even less likely as it seems prime works in that same pattern as well as mersenne 
I have stuff to add I think to this theory the hard part is how to sum it up.

[CODE](12:16) gp > for(x=2,30,for(i=2,20,print1(((mersenne[i]^x1)(mersenne[i1]^x1))%6","));print())
5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,0,4,2,0,0,4,0,2,4,2,0,0,0,0,4,2, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, (12:17) gp > for(x=2,30,for(i=2,20,print1(((prime(i)^x1)(prime(i1)^x1))%6","));print()) 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 1,2,2,4,2,4,2,4,0,2,0,4,2,4,0,0,2,0,4, 5,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, (12:18) gp > for(x=2,30,for(i=2,20,print1(((prime(x)^i1)(prime(x1)^i1))%6","));print()) 5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5, 4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, (12:19) gp > for(x=2,30,for(i=2,20,print1(((mersene[x]^i1)(mersenne[x1]^i1))%6","));print()) *** for: _[_]: not a vector. (12:19) gp > for(x=2,30,for(i=2,20,print1(((mersenne[x]^i1)(mersenne[x1]^i1))%6","));print()) 5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5, 4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0, 0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0,2,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,[/CODE] this help for information I've been figuring out? 
I may have found something in this data involving:
[url]http://www.research.att.com/~njas/sequences/A176260[/url] as it's one of the sequences as far as I can see and maybe I we can find a way to like the Mersenne prime exponents to Fibonacci numbers through> [url]http://www.research.att.com/~njas/sequences/A047264[/url] just a guess anyone else want to see if it's useful ? 
What the hell are you going on about? :huh: X 2

[QUOTE=3.14159;236349]What the hell are you going on about? :huh: X 2[/QUOTE]
do you really want to know ? okay maybe it's not as useful as i thought as I don't know if or when it will repeat lol. but if can relate the others likewise to sequences maybe we can sort of which x work out lol 
[QUOTE=science_man_88;236348][url]http://www.research.att.com/~njas/sequences/A176260[/url] as it's one of the sequences as far as I can see and maybe I we can find a way to like the Mersenne prime exponents to Fibonacci numbers through> [url]http://www.research.att.com/~njas/sequences/A047264[/url]
just a guess anyone else want to see if it's useful ?[/QUOTE] Not useful. If you found something that made it useful, that something would still be useful without A176260. 
[QUOTE=science_man_88;236353]okay maybe it's not as useful as i thought as I don't know if or when it will repeat lol. but if can relate the others likewise to sequences maybe we can sort of which x work out lol[/QUOTE]
There's an amusing site for ideas like this: [url]http://www.halfbakery.com/[/url] Even if you don't submit anything you should try reading some of the entries. Some are crazy, others are (naturally!) halfbaked, while some almost seem plausible... 
[QUOTE=CRGreathouse;236384]There's an amusing site for ideas like this:
[url]http://www.halfbakery.com/[/url] Even if you don't submit anything you should try reading some of the entries. Some are crazy, others are (naturally!) halfbaked, while some almost seem plausible...[/QUOTE] then why is this mersenne forums ? 
[QUOTE=science_man_88;236483]then why is this mersenne forums ?[/QUOTE]
I don't understand the question. 
[QUOTE=CRGreathouse;236488]I don't understand the question.[/QUOTE]
I figured as much. nobody understands me. 
[QUOTE=science_man_88;236510]I figured as much. nobody understands me.[/QUOTE]I understood the question to mean: if this is the Mersenne forum, why is only a small minority of the material posted here anything to do with Mersenne and/or the eponymous numbers.
Of course, I could be wrong. Paul 
[QUOTE=xilman;236515]I understood the question to mean: if this is the Mersenne forum, why is only a small minority of the material posted here anything to do with Mersenne and/or the eponymous numbers.[/QUOTE]
If that's the case then I understand the question now but can't answer it. 
[QUOTE=science_man_88;236510]I figured as much. nobody understands me.[/QUOTE]
Our problems are mathematically dual: You are rarely understood by anyone and I rarely understand anyone. 
[QUOTE=CRGreathouse;236518]Our problems are mathematically dual: You are rarely understood by anyone and I rarely understand anyone.[/QUOTE]
maybe if i can figure out that think I thought of we can get both solved lol 
[QUOTE=CRGreathouse;236518]Our problems are mathematically dual: You are rarely understood by anyone and I rarely understand anyone.[/QUOTE]
Well, even if you make a serious attempt to understand what he posts, you might understand about 10% of what he posts. 
[QUOTE=3.14159;236587]Well, even if you make a serious attempt to understand what he posts, you might understand about 10% of what he posts.[/QUOTE]
I'm amused by the ambiguity in the post.:smile: 
[QUOTE=CRGreathouse;236606]I'm amused by the ambiguity in the post.:smile:[/QUOTE]
Ambiguity? Where? 
[QUOTE=3.14159;236607]Ambiguity? Where?[/QUOTE]
The antecedents, of course  you could easily reverse them. 
could we add sumdigits(sumdigits(x)+sundigits(y)) = sumdigits(x*y) into the first theory ? I think it works can anyone help prove it ?.

What are x and y? (Remember: when you're asking for help you need to be specific!)

for example 2047 = 23*89
2+4+7=4 mod 9 2+3+8+9 = 4 mod 9 5+8 = 4 mod 9 so the sumdigits for each multiplier 2kp+1 summed together gives the same as the sumdigits for 2^p+1. 
So... what are x and y?

[QUOTE=CRGreathouse;236835]So... what are x and y?[/QUOTE]
23 and 89 in this case. 
[QUOTE=science_man_88;236836]23 and 89 in this case.[/QUOTE]
If you want us to comment on more than just this case, you'll have to give us more information. If not, that's cool too. 
[QUOTE=CRGreathouse;236837]If you want us to comment on more than just this case, you'll have to give us more information. If not, that's cool too.[/QUOTE]
[CODE] (11:39) gp > sumdigits(47)+sumdigits(178481) %69 = 4 (16:19) gp > sumdigits(47*178481) %70 = 4[/CODE] in this case it works without the outer sumdigits in the first expression because both sumdigits are below 5 so it can't add up above 9. 
okay found a bad exception 2^291 because it uses 3 but if we can reduce it to 2 maybe it still works this is confirmed as a good change lol.

okay verified exception = 2^371 so how to compensate for exceptions if not to many.0
maybe it works for sumdigits(2^p1) = 4 the hard part is adapting to the case when it's 1. 
[QUOTE=science_man_88;236838][CODE] (11:39) gp > sumdigits(47)[B][COLOR="Red"]*[/COLOR][/B]sumdigits(178481)
%69 = 4 (16:19) gp > sumdigits(47*178481) %70 = 4[/CODE] in this case it works without the outer sumdigits in the first expression because both sumdigits are below 5 so it can't add up above 9.[/QUOTE] See the highlighted part. That is the correct relation. This follows from the fact that sumdigits(x) is equivalent to x%9. 
[QUOTE=axn;236845]See the highlighted part. That is the correct relation. This follows from the fact that sumdigits(x) is equivalent to x%9.[/QUOTE]
it works for addition for sumdigits(x*y)=4 lol 
[QUOTE=science_man_88;236846]it works for addition for sumdigits(x*y)=4 lol[/QUOTE]
not really. try the trivial case 1*4. 
[QUOTE=axn;236848]not really. try the trivial case 1*4.[/QUOTE]
i mean't the mersenne prime factors and so far I've got it working for the first 3 of sumdigits(2^p1)=4 
okay i found the exception at 101 point noted. I'll see if either of the idea's work if one does maybe we can limit even more.

[CODE][COLOR="red"]12[/COLOR]3[COLOR="Red"]45[/COLOR]6[COLOR="Red"]78[/COLOR]9
[COLOR="red"]24[/COLOR]6[COLOR="Red"]8[/COLOR][COLOR="red"]1[/COLOR]3[COLOR="Red"]5[/COLOR][COLOR="red"]7[/COLOR]9 369369369 [COLOR="red"]48[/COLOR]3[COLOR="red"]72[/COLOR]6[COLOR="red"]15[/COLOR]9 [COLOR="red"]51[/COLOR]6[COLOR="Red"]27[/COLOR]3[COLOR="red"]84[/COLOR]9 639639639 [COLOR="red"]75[/COLOR]3[COLOR="red"]18[/COLOR]6[COLOR="red"]42[/COLOR]9 [COLOR="Red"]87[/COLOR]6[COLOR="Red"]54[/COLOR]3[COLOR="red"]21[/COLOR]9 999999999[/CODE] this is the 2 way multiplication for modulo if I have it correct, the red are ones that can if multiplied by something else or (for some) left alone can be brought to 4 or 1 as sumdigits(x) 
I think this is a longwinded circular argument again.. Hopefully it doesn't take 25 pages of posts to realize this.

[QUOTE=3.14159;236897]I think this is a longwinded circular argument again.. Hopefully it doesn't take 25 pages of posts to realize this.[/QUOTE]
It may be. Can you prove it? Alternately, can you phrase his request unambiguously? If so I may be able to help. As it stands I have a few guesses as to where he might be going, but no firm ideas and certainly not a complete understanding. 
Unambiguously? I think that may be an oxymoron.

I was saying that numbers of a specific sumdigits like 2 can be found to be possible in one case of 5mod9*8mod9 = 4mod9 so 2*2mod9 =4mod9 + 1 =5 so k=1 is an answer for a 5mod9 if we look for 8mod9 8*2mod9+1= 8mod9 so k=4 is a possibility in this case it works for 11 which is sumdigits(x)=2

[QUOTE=3.14159;236909]Unambiguously? I think that may be an oxymoron.[/QUOTE]
You're every bit as lucid as sm. 
[QUOTE=CRGreathouse;236948]You're every bit as lucid as sm.[/QUOTE]
Were you intending to insult me there? 
[QUOTE=3.14159;236956]Were you intending to insult me there?[/QUOTE]
unless you think I'm a creative genius I don't know how else you'd take that lol 
[QUOTE=3.14159;236956]Were you intending to insult me there?[/QUOTE]
I was asking for clarification on what you found oxymoronic in my request for unambiguous phrasing. 
[QUOTE=CRGreathouse;236967]I was asking for clarification on what you found oxymoronic in my request for unambiguous phrasing.[/QUOTE]
Figure it out on your own. 
[QUOTE=3.14159;236969]Figure it out on your own.[/QUOTE]
Based on the the content of my post  no thanks. :smile: 
does anyone else care what I'm talking about ?

[QUOTE=science_man_88;236985]does anyone else care what I'm talking about ?[/QUOTE]
I can't tell if I care, because I can't understand you. The above flurry of posts was an attempt to get someone, perhaps Pi, to translate. If you were more clear and explicit this wouldn't be needed, but that's life. You may be unable to write that way, or you may be unwilling to put in the effort. In the former case, that's a pity; you may have an interesting idea that no one will be able to get at. In the latter, don't expect others to do the work of extracting the information for you! 
Now, back to the main focus of the thread..
Some obscure theory on primes of the form 2[sup]p[/sup]1. What about them? Want to find one? P95 is your friend. Why don't you try 2[sup]64354649[/sup]  1? By the way; What is the status of 2[sup]64354649[/sup]  1? (A number of about 19.37 million digits.) If it's uncharted territory, perhaps sm88 can pursue [B]actually looking for one[/B], as opposed to the usual business of seeing patterns where none exist, and babbling nonsense no one understands. Try these values for p, for 2[sup]p[/sup]1, if that one is composite. These are all probably uncharted territory anyway, and I'm a tad too lazy to go looking around for each exponent's status, although I'm sure that at least one has already been meddled with at least once. [code]60687901 38938813 45615587 60254287 73008427 80300791 62029711 50742259 78463417 86100557 70978823 30219743 42805117 97802563 54000169 45921541 67627421 50714737 39618451 49718023 80976547 71843129 82795997 37144193 32807759 52988981 85507519 81177307 29024783 94450157 49717169 84537977 52294097 53169451 70975439 62525147[/code] 
[QUOTE=3.14159;237008]Now, back to the main focus of the thread..
Some obscure theory on primes of the form 2[sup]p[/sup]1. What about them? Want to find one? P95 is your friend. Why don't you try 2[sup]64354649[/sup]  1? By the way; What is the status of 2[sup]64354649[/sup]  1? (A number of about 19.37 million digits.) If it's uncharted territory, perhaps sm88 can pursue [B]actually looking for one[/B], as opposed to the usual business of seeing patterns where none exist, and babbling nonsense no one understands. Try these values for p, for 2[sup]p[/sup]1, if that one is composite. These are all probably uncharted territory anyway, and I'm a tad too lazy to go looking around for each exponent's status, although I'm sure that at least one has already been meddled with at least once. [code]60687901 38938813 45615587 60254287 73008427 80300791 62029711 50742259 78463417 86100557 70978823 30219743 42805117 97802563 54000169 45921541 67627421 50714737 39618451 49718023 80976547 71843129 82795997 37144193 32807759 52988981 85507519 81177307 29024783 94450157 49717169 84537977 52294097 53169451 70975439 62525147[/code][/QUOTE] get me the worlds fastest computer and I'll care if i can't get pari to print the biggest prime known in under 30 minutes then I'm not worth it to help anyone find one. so unless I retest and get faster than I did before by 25% I'm not worth it to be searching for them. apparently that was checking if it was prime because it took 31 ms to calculate this time. 
my computer isn't worth breaking to be in the record books, plain and simple

[QUOTE=science_man_88;237053]my computer isn't worth breaking to be in the record books, plain and simple[/QUOTE]
P95 is your friend.. A random c160, completely factored: 1613 * 16685818316410231 * 15694342916407420598282061127 * 21443356185490234674701067126317 * 11753380988799343801639257420578787695318692321305544085583187497380249619645889. 
[QUOTE=3.14159;237057].. P95 is your friend..[/QUOTE]
all that means to me is a prime of 95 digits. 
[QUOTE=science_man_88;237058]all that means to me is a prime of 95 digits.[/QUOTE]
[URL="http://www.mersenne.org/freesoft/"]Hey, look! P95![/URL] And, an actual p95; 82449289611552559574990925283007152491333479550118068579078519555198400142339056437465979206437. 
[QUOTE=3.14159;237062][URL="http://www.mersenne.org/freesoft/"]Hey, look! P95![/URL][/QUOTE]
that helps me none 
[QUOTE=science_man_88;237067]that helps me none[/QUOTE]
Prime95 is (Mersenne prime)testing software. Pi is apparently suggesting that, rather than think about the problem, you apply some cycles to it. 
[QUOTE=CRGreathouse;237072]Prime95 is (Mersenne prime)testing software. Pi is apparently suggesting that, rather than think about the problem, you apply some cycles to it.[/QUOTE]
I have before and it would take me over 50 days to test one exponent pointless to me. that and it teaches me nothing about them. 
[QUOTE=science_man_88;237073]I have before and it would take me over 50 days to test one exponent pointless to me.
that and it teaches me nothing about them.[/QUOTE] I don't see why 50 days/test makes it bad. But I agree that it doesn't teach you anything. 
[QUOTE=CRGreathouse;237072]Prime95 is (Mersenne prime)testing software. Pi is apparently suggesting that, rather than think about the problem, you apply some cycles to it.[/QUOTE]
Think about [B]what[/B] problem? No one's even made it clear what the supposed "problem" is, besides looking for some Mersenne number. It's really simple; P95 is your friend. 
[QUOTE=CRGreathouse;237076]I don't see why 50 days/test makes it bad. But I agree that it doesn't teach you anything.[/QUOTE]
All there is to know about them is; Mersenne numbers are numbers of the form 2[sup]p[/sup]  1, where p is any prime number; There may be a chance that 2[sup]p[/sup]  1 is prime, and any factors of such a number are of the form 2kp + 1, where p is the exponent, and are probably the easiest to test, with a convenient little test for specifically those types of primes. Other than that, there is not much more to learn about them. 
[QUOTE=3.14159;237082]Other than that, there is not much more to learn about them.[/QUOTE]
:rofl: 
[QUOTE=CRGreathouse;237084]:rofl:[/QUOTE]
:orly owl: Show me something else that is not useless trivia. 
[QUOTE=3.14159;237085]:orly owl: Show me something else that is not useless trivia.[/QUOTE]
They're pairwise coprime. 
[QUOTE=3.14159;237085]:orly owl: Show me something else that is not useless trivia.[/QUOTE]
you missed that the 2kp+1 must be =1/1 mod 8 which is what started this thread lol. 
[QUOTE=3.14159;237085]:orly owl: Show me something else that is not useless trivia.[/QUOTE]
They'e used in the generation of highquality pseudorandom numbers. 
[QUOTE=CRGreathouse;237089]They'e used in the generation of highquality pseudorandom numbers.[/QUOTE]
Fine.. You win.. 
[QUOTE=science_man_88;237088]you missed that the 2kp+1 must be =1/1 mod 8 which is what started this thread lol.[/QUOTE]
Thanks, Captain Obvious, but it wasn't necessary. 5. 
The best known scheme for private information retrieval and locallydecodable codes, a vast improvement on the previous method, uses Mersenne primes.

[QUOTE=CRGreathouse;237092]The best known scheme for private information retrieval and locallydecodable codes, a vast improvement on the previous method, uses Mersenne primes.[/QUOTE]
I wish to know from which website you are pasting these from. 
[QUOTE=3.14159;237093]I wish to know from which website you are pasting these from.[/QUOTE]
[url]http://oeis.org/[/url] heres the oeis for you you can come up with other facts. 
They're used in efficient algorithms (because modular reduction can be done with shifts and subtraction), especially in cryptology.
[QUOTE=3.14159;237093]I wish to know from which website you are pasting these from.[/QUOTE] My own knowledge. I don't know of a site collecting all this information together. As far as I know, this thread is the only place on the Internet that does that. :smile: 
Don't you mean, these forums in general?

They can be used to calculate the convolution or correlation of 2D images in parallel.
[QUOTE=3.14159;237109]Don't you mean, these forums in general?[/QUOTE] No, I meant this particular thread. If you know of other threads giving this kind of information, let me know! I'd be happy to pool my information with others. 
[QUOTE=CRGreathouse;237111]They can be used to calculate the convolution or correlation of 2D images in parallel.
No, I meant this particular thread. If you know of other threads giving this kind of information, let me know! I'd be happy to pool my information with others.[/QUOTE] Displaying 110 of 347 results found. for A000043 Displaying 110 of 368 results found. for mersenne prime A000043 Displaying 110 of 234 results found. for "mersenne prime" A000043 
A053648 has a possible term missing 31 when listing of from 131071
oh it's already listed under 31 dah lol. 
The factorization of small Mersenne numbers resulted in a significant speedup for Jan Feitsma's calculation of the pseudoprimes.
[QUOTE=science_man_88;237132]Displaying 110 of 347 results found. for A000043 Displaying 110 of 368 results found. for mersenne prime A000043 Displaying 110 of 234 results found. for "mersenne prime" A000043[/QUOTE] I'm pretty familiar with the OEIS, but much/most of the information I've listed so far isn't there as far as I know. 
I have a challenge for you;
Why don't you set off and find me a composite number which passes the BPSW test? Good luck! 
If m is a Mersenne prime, then [TEX]\sigma(m+1)\sigma(m)=m.[/TEX]
[QUOTE=3.14159;237157]I have a challenge for you; Why don't you set off and find me a composite number which passes the BPSW test? Good luck![/QUOTE] There's a $620 bounty for that. What's more, there is a (finite) collection of numbers chosen in such a fashion that it is believed that a product of certain terms is a BPSWpseudoprime. So that's something of a head start. :smile: Also, the "cheap third author" as I believe Pomerance referred to himself, has considered raising his share of the prize from $20, so you might even get more out of it than $620. 
[QUOTE=science_man_88;236942]I was saying that numbers of a specific sumdigits like 2 can be found to be possible in one case of 5mod9*8mod9 = 4mod9 so 2*2mod9 =4mod9 + 1 =5 so k=1 is an answer for a 5mod9 if we look for 8mod9 8*2mod9+1= 8mod9 so k=4 is a possibility in this case it works for 11 which is sumdigits(x)=2[/QUOTE]
anyone actually figure me out yet ? since this was a while ago i figured your minds need refreshing. 
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