20051111, 00:05  #1 
Oct 2005
Italy
339_{10} Posts 
General formula
Is it possible that a general formula exist for all Mersenne primes ?

20051111, 00:55  #2 
∂^{2}ω=0
Sep 2002
República de California
26734_{8} 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.

20051111, 09:43  #3 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
of course there is. M(n) where M(n) is the nth Mersenne prime. if you don't mind filtering through results, you can even write M'(n) where M'(n) = 2^n1. I guarantee you that every mersenne prime will be listed, you just have to disregard the composite mersenne numbers

20051111, 13:57  #4  
Oct 2005
Italy
3·113 Posts 
Quote:
I mean , of course, a formula f(n) where , when n varies , f(n) returns a Mersenne prime number (for all n values) 

20051111, 14:23  #5  
"Bob Silverman"
Nov 2003
North of Boston
5×1,493 Posts 
Quote:
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. 

20051204, 17:31  #6 
Sep 2005
12_{16} Posts 
we find this
of course there is.2^F1 where F one of Fermates prime numbers(in special form) .

20051204, 18:11  #7 
"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 
20051204, 22:05  #8  
∂^{2}ω=0
Sep 2002
República de California
10110111011100_{2} Posts 
Quote:


20051204, 22:52  #9 
Jun 2003
2^{2}×397 Posts 
I think he is telling us that he needs coffee, as the icon says.
Citrix 
20051205, 08:02  #10 
"Nancy"
Aug 2002
Alexandria
100110100011_{2} Posts 
Well, lets not dismiss it without at least giving him a chance to explain himself.
Alex 
20051205, 08:13  #11 
Jun 2003
2^{2}×397 Posts 
He might be saying that
2^(2^2^x)+1)1 is always prime, for prime fermat's. So 2^31 is prime 2^51 is prime 2^171 is prime but 2^2571 is not prime? So still not sure. 
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