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Re: [PrimeNumbers] Re: question on (un)boundedness

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  • James Merickel
    Empirically it appears there is for any n a finite number of values b (both b and n in {2,3,4,...}) such that the prime preceding b^n is not greater than
    Message 1 of 3 , Sep 26, 2013
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      Empirically it appears there is for any n a finite number of values b (both b and n in {2,3,4,...}) such that the prime preceding b^n is not greater than b^n-b, with for reasonably small n an easy-to-find value that makes a good conjecture for the count.  For n=2 it seems certain there are no such, for example, just empirically.  If this finitude is provable it would settle the question asked.
      JGM  

      From: Kermit Rose <kermit@...>
      To: primenumbers@yahoogroups.com
      Sent: Wednesday, September 25, 2013 2:30 PM
      Subject: [PrimeNumbers] Re: question on (un)boundedness
       
      On 9/25/2013 7:03 AM, primenumbers@yahoogroups.com wrote:
      > question on (un)boundedness
      > Posted by: "James Merickel"moralforce120@... moralforce120
      > Date: Tue Sep 24, 2013 11:17 am ((PDT))
      >
      > Dear group members:
      > Does anybody know of anything solid to point to indicating whether or not there is a bound on the smallest N(b) for which there is no prime between b^N(b)-b and b^N(b)? I imagine I may be able to address this heuristically myself without much bother (and the answer is there is no bound, but of course until I actually try I won't be too sure), but I wonder if there is anything stronger than heuristics on this or it's 'the usual for such number theory questions'.
      > Jim

      It is not an easy question to answer.

      (2^2-2,2^2) = (2,4)
      (2^3-2,2^3) = (6,8)
      (2^4-2,2^4) = (14,16) N(2) = 4

      (3^2-3,3^2) = (6,9)
      (3^3-3,3^3) = (24,27) N(3) = 3

      (4^2-4,4^2) = (12,16)
      (4^3-4,4^3) = (60,64)
      (4^4-4,4^4) = (252,256) N(4) = 4

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