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