Assorted formulas for exponents of Mersenne primes

[ewmayer]

Originally Posted by CRGreathouse

In fact, if like science_man_88 and myself, the OP lives in a Berne signatory state then just by writing it a copyright is created. But if taken to court you'd need to prove copyright, so your choices would be to either make it public somewhere that records dates (say, on the mersenne forums) or, better but more expensive, register a copyright. In my country, this can be as cheap as $35.

I notice that 2^127-1 is 39 digits in length. I hope I don't get into any hot water with Swiss interzone crank-patent enforcement authorities for pointing out that "coincidence".

p.s.: What exactly is a "10 digit integer float"? Has it been patented?

[CRGreathouse]

Originally Posted by ewmayer

I notice that 2^127-1 is 39 digits in length. I hope I don't get into any hot water with Swiss interzone crank-patent enforcement authorities for pointing out that "coincidence".

Facts can't be copyrighted, just tangible expressions of facts. If I write a proof that 1 + 1 = 2 you can't copy that particular proof (or perform it publicly, etc.) but you can still use the fact itself.

Originally Posted by ewmayer

p.s.: What exactly is a "10 digit integer float"? Has it been patented?

To give more credit than is due, a floating-point number with significand large enough that the number represented is an integer.

[xilman]

Originally Posted by CRGreathouse

Facts can't be copyrighted, just tangible expressions of facts. If I write a proof that 1 + 1 = 2 you can't copy that particular proof (or perform it publicly, etc.) but you can still use the fact itself..

Ordinarily I wouldn't bother but you introduced the law into this thread.

I can, in fact, copy that particular proof of 1+1=2 if you give me permission to do so. If your proof has been written before and is out of copyright, I can copy your proof.

For you to be able to prevent me from your proof, it not only has to be written by you, it also has to have an element of novelty. (Arguing whether something is novel could turn out to be expensive.)

A simple example: I have just now written (typed, actually) "To be or not to be, that is the question.". I can not prevent you from copying that sentence within the quote characters because it has no novelty. It was written 400 years ago and I have merely copied it, not created it.

Paul

[CRGreathouse]

Originally Posted by xilman

I can, in fact, copy that particular proof of 1+1=2 if you give me permission to do so.

Certainly, and there are § 107 limitations and many other things beside. I was giving the general overview, not every contour.

Originally Posted by xilman

If your proof has been written before and is out of copyright, I can copy your proof.

This is actually not entirely clear, cases have gone both ways AFAIK. You can certainly copy from the old source, but whether you can copy from the new source might depend on whether it is derivative of the old or not. I agree with your interpretation, for what it's worth, but would be careful applying it.

Originally Posted by xilman

For you to be able to prevent me from your proof, it not only has to be written by you, it also has to have an element of novelty. (Arguing whether something is novel could turn out to be expensive.)

Patent law is where that is most critical and most expensive. In copyright law, a person with a registered copyright has the presumption of ownership -- the other side must present C&C evidence to the contrary to prevail.

Of course we all know Feist v Rural, but that's a very low bar: even a shred of originality will suffice.

[Lee Yiyuan]

I am not a programmer, so i cant really test it.

the prime is 2^(2^127 - 1) -1

Would really appreciate it if you guys could help.

Thanks in advance.

[firejuggler]

ooooooooook....
lol.. even with state of the art computer, you can't prove it is prime (yet) unless you own a quantic computer , a farm of teragrid, or revolutionarize the prime proving algorythm.

[science_man_88]

Originally Posted by Lee Yiyuan

I am not a programmer, so i cant really test it.

the prime is 2^(2^127 - 1) -1

Would really appreciate it if you guys could help.

Thanks in advance.

once again to program it I'd need the prime generating function which you don't want to release, without that even though I too have thought this we can't prove it.

[CRGreathouse]

Originally Posted by Lee Yiyuan

I am not a programmer, so i cant really test it.

the prime is 2^(2^127 - 1) -1

What makes you think it's prime? It's a huge number, 170141183460469231731687303715884105727 bits long, and the chance that a random number that size is prime is tiny. This one is better than most -- it has no prime factors under 2^128 -- but if you pick a random number that size with no prime factors under 2^200 it has only about a 0.0000000000000000000000000000000002% chance to be prime. (I don't know exactly how far the number has been checked for small factors, but probably not that far.)

[ewmayer]

Originally Posted by CRGreathouse

What makes you think it's prime?

Others have conjectured that this, the 5th (and smallest of unknown status) Catalan-Mersenne Number, a.k.a. MM127, is prime. It's reminiscent of Fermat's conjecture about primality of the number sequence that now bears his name.

AFAIK no one has ever produced any number-theoretic evidence that MM127 is more likely to be prime than one would expect for any other M-number with a prime exponent in that size range. Empirically, MM127 has been TFed to a factor index (that's what I call the k in the standard form-of-Mersenne-factors formula q = 2kp+1) of around 50 bits, with no factors found, so MM127 is perhaps as much as 2x more likely to be prime than a randomly-chosen M(p) with p of 127 bits. But 2x a very tiny number is still a very tiny number.

[Lee Yiyuan]

(P.S. This is the formula for Catalan-Mersenne Numbers, just realized it after reading ewmayer's post)



The recursive function f(x) = 2^f(x-1) -1 where f(1) = 3
f(x) will always be a Mersenne prime for all positive integers x > 1


E.G. f(3) = 2^f(2) - 1
= 2^(2^f(1) - 1) -1
= 2^(2^3 - 1) - 1
= 2^(7) - 1
= 128 - 1
= 127 (Prime)


f(1) = M3,
f(2) = M7,
f(3) = M127,
f(4) = M170141183460469231731687303715884105727,
f(5) = M(2^170141183460469231731687303715884105727 -1)

[Lee Yiyuan]

Originally Posted by CRGreathouse

What makes you think it's prime? It's a huge number, 170141183460469231731687303715884105727 bits long, and the chance that a random number that size is prime is tiny. This one is better than most -- it has no prime factors under 2^128 -- but if you pick a random number that size with no prime factors under 2^200 it has only about a 0.0000000000000000000000000000000002% chance to be prime. (I don't know exactly how far the number has been checked for small factors, but probably not that far.)

An exponent that ends with a 7 or 1 makes it more probable for the calculated value of 2^170141183460469231731687303715884105727 - 1 to be prime. The chances of a randomly picked number lesser than 2^200 being a prime is 1/ln(2^200) = 0.007213475 ( 7 significant figures)