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 2005-11-11, 00:05 #1 pacionet     Oct 2005 Italy 3·113 Posts General formula Is it possible that a general formula exist for all Mersenne primes ?
 2005-11-11, 00:55 #2 ewmayer ∂2ω=0     Sep 2002 República de California 5·2,339 Posts If you're talking about all that are known *and* those yet to be found, then it's exceedingly improbable. Finding a formula that yields all that are already known (say a smooth function f(x) such that f(n) gives the exponent of the (n)th Mersenne prime) is of course a (rather silly) exercise in curve fitting.
 2005-11-11, 09:43 #3 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 12018 Posts of course there is. M(n) where M(n) is the n-th Mersenne prime. if you don't mind filtering through results, you can even write M'(n) where M'(n) = 2^n-1. I guarantee you that every mersenne prime will be listed, you just have to disregard the composite mersenne numbers
2005-11-11, 13:57   #4
pacionet

Oct 2005
Italy

15316 Posts

Quote:
 Originally Posted by TravisT of course there is. M(n) where M(n) is the n-th Mersenne prime. if you don't mind filtering through results, you can even write M'(n) where M'(n) = 2^n-1. I guarantee you that every mersenne prime will be listed, you just have to disregard the composite mersenne numbers

I mean , of course, a formula f(n) where , when n varies , f(n) returns a Mersenne prime number (for all n values)

2005-11-11, 14:23   #5
R.D. Silverman

Nov 2003

746010 Posts

Quote:
 Originally Posted by pacionet I mean , of course, a formula f(n) where , when n varies , f(n) returns a Mersenne prime number (for all n values)
Yes. Such a formula exists. But it is *useless* for computational
purposes.

I will give a hint how to construct such a formula.

Let the n'th Mersenne prime be given as M(n) = 2^(p_n) - 1
where p_n is the exponent of the n'th prime.

Consider the constant:

alpha = sum(i=1 to oo) of 10^(-2i) p_i. This is a well defined
real number. The sum clearly converges.

Now, given this constant, one can compute M(n) by multiplying
alpha by 10^(2n), extracting the fractional part
of 10^(2n) alpha, subtracting then truncating the part of the fraction after the trailing 0's that follow p_n. One can do this with a suitable
combination of floor functions and simple multiplications that I am too
lazy to work out at the moment. A similar formula may be found in
Hardy and Wright except it gives the n'th prime, instead of the n'th
Mersenne prime.

It is useless, because we have no way of computing alpha to find
as yet unknown primes.

But the formula does indeed *exist* because alpha exists.

 2005-12-04, 17:31 #6 sghodeif   Sep 2005 2×32 Posts we find this of course there is.2^F-1 where F one of Fermates prime numbers(in special form) .
 2005-12-04, 18:11 #7 akruppa     "Nancy" Aug 2002 Alexandria 2,467 Posts sghodeif, what exactly are you referring to, what precisely do you mean by "special form" and how do Fermat numbers enter the picture? Alex
2005-12-04, 22:05   #8
ewmayer
2ω=0

Sep 2002
República de California

5×2,339 Posts

Quote:
 Originally Posted by akruppa sghodeif, what exactly are you referring to, what precisely do you mean by "special form" and how do Fermat numbers enter the picture?
sghodeif is simply trying very hard to get this thread moved to the "miscellaneous math threads forum," Alex.

 2005-12-04, 22:52 #9 Citrix     Jun 2003 3·232 Posts I think he is telling us that he needs coffee, as the icon says. Citrix
 2005-12-05, 08:02 #10 akruppa     "Nancy" Aug 2002 Alexandria 246710 Posts Well, lets not dismiss it without at least giving him a chance to explain himself. Alex
 2005-12-05, 08:13 #11 Citrix     Jun 2003 3·232 Posts He might be saying that 2^(2^2^x)+1)-1 is always prime, for prime fermat's. So 2^3-1 is prime 2^5-1 is prime 2^17-1 is prime but 2^257-1 is not prime? So still not sure.

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