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-   -   theory on Mersenne primes ? (https://www.mersenneforum.org/showthread.php?t=14151)

 science_man_88 2010-11-03 15:38

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 2^37-1 so far it seems if p mod 6 =5 or 1 then 2 of the factors of 2^p-1 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)

 Mini-Geek 2010-11-03 16:12

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 2^37-1 so far it seems if p mod 6 =5 or 1 then 2 of the factors of 2^p-1 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^p-1 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.

 science_man_88 2010-11-03 16:30

[QUOTE=Mini-Geek;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^p-1 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.

 Mini-Geek 2010-11-03 16:43

[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.

 CRGreathouse 2010-11-03 17:10

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.

 science_man_88 2010-11-03 17:17

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 6n-1

since 11= 5 mod 6

6n+1 k=0,3,6,9, (3x+0)
6n-1 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.

 Mini-Geek 2010-11-03 17:25

[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.

 CRGreathouse 2010-11-03 17:39

[QUOTE=Mini-Geek;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.

 ATH 2010-11-04 16:37

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 co-prime 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.

 science_man_88 2010-11-04 17:03

[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 co-prime 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.

 science_man_88 2010-11-04 22:07

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]n-1[/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

 science_man_88 2010-11-04 22:16

[CODE](19:15) gp > for(x=2,20,for(i=2,30,print1(((x^mersenne[i]-1)-(x^mersenne[i-1]-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.

 science_man_88 2010-11-04 23:46

I think it's disproved anyway even if you allow exceptions with this pattern lol.

 science_man_88 2010-11-07 16:44

[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 ?

 CRGreathouse 2010-11-07 18:24

[QUOTE=science_man_88;235933]is my new theory provable ? / proven ?[/QUOTE]

I don't understand what the new conjecture is.

 science_man_88 2010-11-07 18:57

[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[i-1]-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 ?

 CRGreathouse 2010-11-07 19:33

[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?

 science_man_88 2010-11-07 20:32

[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

 CRGreathouse 2010-11-07 20:37

You're still not being specific enough. You seem to be saying that

[code](x^mersenne[i]-1)-(x^mersenne[i-1]-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!

 science_man_88 2010-11-07 20:38

[QUOTE=CRGreathouse;235960]You're still not being specific enough. You seem to be saying that

[code](x^mersenne[i]-1)-(x^mersenne[i-1]-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

 science_man_88 2010-11-07 21:05

now that pattern looks stupid to help.

even less likely as it seems prime works in that same pattern as well as mersenne

 science_man_88 2010-11-09 23:09

I have stuff to add I think to this theory the hard part is how to sum it up.

 science_man_88 2010-11-09 23:10

[CODE](12:16) gp > for(x=2,30,for(i=2,20,print1(((mersenne[i]^x-1)-(mersenne[i-1]^x-1))%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)^x-1)-(prime(i-1)^x-1))%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)^i-1)-(prime(x-1)^i-1))%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]^i-1)-(mersenne[x-1]^i-1))%6","));print())
*** for: _[_]: not a vector.
(12:19) gp > for(x=2,30,for(i=2,20,print1(((mersenne[x]^i-1)-(mersenne[x-1]^i-1))%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?

 science_man_88 2010-11-10 00:52

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 ?

 3.14159 2010-11-10 00:52

What the hell are you going on about? :huh: X 2

 science_man_88 2010-11-10 00:54

[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

 CRGreathouse 2010-11-10 05:00

[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.

 CRGreathouse 2010-11-10 05:01

[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!) half-baked, while some almost seem plausible...

 science_man_88 2010-11-10 16:38

[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!) half-baked, while some almost seem plausible...[/QUOTE]

then why is this mersenne forums ?

 CRGreathouse 2010-11-10 17:18

[QUOTE=science_man_88;236483]then why is this mersenne forums ?[/QUOTE]

I don't understand the question.

 science_man_88 2010-11-10 19:06

[QUOTE=CRGreathouse;236488]I don't understand the question.[/QUOTE]

I figured as much. nobody understands me.

 xilman 2010-11-10 19:25

[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

 CRGreathouse 2010-11-10 19:29

[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.

 CRGreathouse 2010-11-10 19:30

[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.

 science_man_88 2010-11-10 20:26

[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

 3.14159 2010-11-11 01:32

[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.

 CRGreathouse 2010-11-11 04:22

[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:

 3.14159 2010-11-11 04:42

[QUOTE=CRGreathouse;236606]I'm amused by the ambiguity in the post.:smile:[/QUOTE]

Ambiguity? Where?

 CRGreathouse 2010-11-11 04:55

[QUOTE=3.14159;236607]Ambiguity? Where?[/QUOTE]

The antecedents, of course -- you could easily reverse them.

 science_man_88 2010-11-12 15:44

could we add sumdigits(sumdigits(x)+sundigits(y)) = sumdigits(x*y) into the first theory ? I think it works can anyone help prove it ?.

 CRGreathouse 2010-11-12 19:43

 science_man_88 2010-11-12 20:01

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.

 CRGreathouse 2010-11-12 20:03

So... what are x and y?

 science_man_88 2010-11-12 20:03

[QUOTE=CRGreathouse;236835]So... what are x and y?[/QUOTE]

23 and 89 in this case.

 CRGreathouse 2010-11-12 20:18

[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.

 science_man_88 2010-11-12 20:22

[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.

 science_man_88 2010-11-12 20:24

okay found a bad exception 2^29-1 because it uses 3 but if we can reduce it to 2 maybe it still works this is confirmed as a good change lol.

 science_man_88 2010-11-12 20:33

okay verified exception = 2^37-1 so how to compensate for exceptions if not to many.0

maybe it works for sumdigits(2^p-1) = 4 the hard part is adapting to the case when it's 1.

 axn 2010-11-12 20:41

[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.

 science_man_88 2010-11-12 20:44

[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

 axn 2010-11-12 20:46

[QUOTE=science_man_88;236846]it works for addition for sumdigits(x*y)=4 lol[/QUOTE]

not really. try the trivial case 1*4.

 science_man_88 2010-11-12 20:52

[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^p-1)=4

 science_man_88 2010-11-12 21:40

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.

 science_man_88 2010-11-12 23:32

[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)

 3.14159 2010-11-13 02:17

I think this is a long-winded circular argument again.. Hopefully it doesn't take 25 pages of posts to realize this.

 CRGreathouse 2010-11-13 04:41

[QUOTE=3.14159;236897]I think this is a long-winded 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.

 3.14159 2010-11-13 05:05

Unambiguously? I think that may be an oxymoron.

 science_man_88 2010-11-13 13:27

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

 CRGreathouse 2010-11-13 14:10

[QUOTE=3.14159;236909]Unambiguously? I think that may be an oxymoron.[/QUOTE]

You're every bit as lucid as sm.

 3.14159 2010-11-13 15:01

[QUOTE=CRGreathouse;236948]You're every bit as lucid as sm.[/QUOTE]

Were you intending to insult me there?

 science_man_88 2010-11-13 15:09

[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

 CRGreathouse 2010-11-13 18:15

[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.

 3.14159 2010-11-13 18:21

[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.

 CRGreathouse 2010-11-13 21:31

[QUOTE=3.14159;236969]Figure it out on your own.[/QUOTE]

Based on the the content of my post -- no thanks. :smile:

 science_man_88 2010-11-13 22:49

does anyone else care what I'm talking about ?

 CRGreathouse 2010-11-14 03:53

[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!

 3.14159 2010-11-14 04:00

Now, back to the main focus of the thread..

Some obscure theory on primes of the form 2[sup]p[/sup]-1.

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]

 science_man_88 2010-11-14 13:31

[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.

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.

 science_man_88 2010-11-14 15:11

my computer isn't worth breaking to be in the record books, plain and simple

 3.14159 2010-11-14 15:20

[QUOTE=science_man_88;237053]my computer isn't worth breaking to be in the record books, plain and simple[/QUOTE]

A random c160, completely factored: 1613 * 16685818316410231 * 15694342916407420598282061127 * 21443356185490234674701067126317 * 11753380988799343801639257420578787695318692321305544085583187497380249619645889.

 science_man_88 2010-11-14 15:20

all that means to me is a prime of 95 digits.

 3.14159 2010-11-14 15:37

[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.

 science_man_88 2010-11-14 15:44

[QUOTE=3.14159;237062][URL="http://www.mersenne.org/freesoft/"]Hey, look! P95![/URL][/QUOTE]

that helps me none

 CRGreathouse 2010-11-14 16:19

[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.

 science_man_88 2010-11-14 16:21

[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.

 CRGreathouse 2010-11-14 16:42

[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.

 3.14159 2010-11-14 17:12

[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]

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.

 3.14159 2010-11-14 17:20

[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.

 CRGreathouse 2010-11-14 17:22

[QUOTE=3.14159;237082]Other than that, there is not much more to learn about them.[/QUOTE]

:rofl:

 3.14159 2010-11-14 17:29

[QUOTE=CRGreathouse;237084]:rofl:[/QUOTE]

:orly owl: Show me something else that is not useless trivia.

 CRGreathouse 2010-11-14 17:36

[QUOTE=3.14159;237085]:orly owl: Show me something else that is not useless trivia.[/QUOTE]

They're pairwise coprime.

 science_man_88 2010-11-14 17:49

[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.

 CRGreathouse 2010-11-14 17:55

[QUOTE=3.14159;237085]:orly owl: Show me something else that is not useless trivia.[/QUOTE]

They'e used in the generation of high-quality pseudorandom numbers.

 3.14159 2010-11-14 18:09

[QUOTE=CRGreathouse;237089]They'e used in the generation of high-quality pseudorandom numbers.[/QUOTE]

Fine.. You win..

 3.14159 2010-11-14 18:11

[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.

 CRGreathouse 2010-11-14 18:11

The best known scheme for private information retrieval and locally-decodable codes, a vast improvement on the previous method, uses Mersenne primes.

 3.14159 2010-11-14 18:12

[QUOTE=CRGreathouse;237092]The best known scheme for private information retrieval and locally-decodable codes, a vast improvement on the previous method, uses Mersenne primes.[/QUOTE]

I wish to know from which website you are pasting these from.

 science_man_88 2010-11-14 18:15

[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.

 CRGreathouse 2010-11-14 18:19

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:

 3.14159 2010-11-14 19:01

Don't you mean, these forums in general?

 CRGreathouse 2010-11-14 19:03

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.

 science_man_88 2010-11-14 22:29

[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 1-10 of 347 results found. for A000043
Displaying 1-10 of 368 results found. for mersenne prime -A000043
Displaying 1-10 of 234 results found. for "mersenne prime" -A000043

 science_man_88 2010-11-14 22:34

A053648 has a possible term missing 31 when listing of from 131071

oh it's already listed under 31 dah lol.

 CRGreathouse 2010-11-15 01:39

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 1-10 of 347 results found. for A000043
Displaying 1-10 of 368 results found. for mersenne prime -A000043
Displaying 1-10 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.

 3.14159 2010-11-15 01:51

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!

 CRGreathouse 2010-11-15 04:54

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 BPSW-pseudoprime. 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.

 science_man_88 2010-11-16 17:29

[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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