From: gerry@mpce.mq.edu.au (Gerry Myerson)
Newsgroups: sci.math
Subject: Re: A Question about Prime Numbers
Date: Thu, 08 Oct 1998 16:03:01 +1100
In article <6vh9cl$dpl$1@news-2.news.gte.net>, "Robert \"Tralfaz\"
Armagost" wrote:
> I seem to recall that there is a proof that there exists a real number M
> such that the greatest integer in M^3n (If I remember the formula
> correctly) is prime.
That's M^(3^n).
W H Mills, A prime-representing function, Bull Amer Math Soc 53 (1947) 604,
MR 8, 567d.
Gerry Myerson (gerry@mpce.mq.edu.au)
==============================================================================
From: thomaso@best.com (Thomas Andrews)
Newsgroups: sci.crypt,sci.math
Subject: Re: science, vol. 279, Jan 2: refined asymptotic sieve
Date: 21 Jan 1998 10:47:55 -0800
This theorem is not hard to believe, either; all you need to
"believe" is the conjecture:
(1) There is always at least one prime between N^3 and (N+1)^3 for every N.
Once you believe that, you define your number as follows:
X(0)=2, A(0)=2.
X(1)=a prime between X(0)^3 and (X(0)+1)^3. A(1)=X(1)^(1/3).
...
X(n+1)=a prime between X(n)^3 and (X(n)+1)^3. A(n+1)=X(n+1)^(1/3^n).
Notice that A(n+1)^(3^n) is X(n), a prime, A(n+1)>A(n), and A(n)<2
for all n.
For example, we'd start with:
n X(n) A(n)
0 2 2
1 11 2.2239801... (cube root of 11)
2 1361 2.2294948... (ninth root of 1361)
Note that [A(2)^(3^0)]=2, [A(2)^(3^1)]=11, [A(2)^(3^2)]=1361.
If we let A be the limit, we are (mostly) done. It becomes necessary
to prove that A(n)^(3^m) ,
Yannick SAOUTER wrote:
>In article <34c61a31.1178892@news.prosurfr.com>, jsavard@nospam.ath (John Savard) writes:
>|> In fact, in a paper from the 1950s cited in "Introduction to Analytic
>|> Number Theory" by Apostol, there was a proof that for some number A,
>|> the integer part of A^(3^n) is always a prime number. (Of course, A
>|> can't be an integer, since this is a power of A.) However, the value
>|> of A is not known.
>
>1947, W.H. Mills "A prime-representing function." Bull. of the AMS,
>53 p. 604.
>
>It works also if you replace 3 by 4, 5 ... or even any real value
>greater than 3. In general it should (i.e. most probable but not
>proven) also work for any real value strictly greater than 1.
>
--
Thomas Andrews thomaso@best.com http://www.best.com/~thomaso/
"Show me somebody who is always smiling, always cheerful, always
optimistic, and I will show you somebody who hasn't the faintest
idea what the heck is really going on." - Mike Royko