Loading ...
Sorry, an error occurred while loading the content.

Products of k distinct primes that are concatenations of first k: Coincidence

Expand Messages
  • James Merickel
    1319732172923115 1713323729219115 2191317297323115 2291117192373135 2291772319133115 2322931117719135 2329131171719235 7112917313223195 7191317229231135
    Message 1 of 2 , Apr 3, 2011
    • 0 Attachment
      1319732172923115
      1713323729219115
      2191317297323115
      2291117192373135
      2291772319133115
      2322931117719135
      2329131171719235
      7112917313223195
      7191317229231135
      7191331129217235
       
      These are the only values through k=10 (k>1).  I have a very inefficient program and a much better one for k=11 and 12, respectively, running.  I expect a null result, but will report if something comes out (couple of days).  I think the problem is tractable with maximal resources and ideal programming through about k=16.  Looking for either a further example or a heuristic proof one likely doesn't exist.
       
      JGM

      [Non-text portions of this message have been removed]
    • James Merickel
      Sorry if I was a bit unclear.  The title was supposed to indicate what the meaning of the post was.  The ten numbers are concatenations of the first ten
      Message 2 of 2 , Apr 4, 2011
      • 0 Attachment
        Sorry if I was a bit unclear.  The title was supposed to indicate what the meaning of the post was.  The ten numbers are concatenations of the first ten primes that are also products of ten distinct primes.  I have run the more efficient version on the case k=11, so that's done.  For k=2 through 11, the only examples are the ones given.  I've also now determined that the way I have programmed this so far is backwards, and that k=16 as a limit is probably very pessimistic.  The initial program used 'if(isquarefree(n),if(numdiv(n)==2^k..', and my more efficient one basically factors up until a concatenation is no longer a plausible solution; but, of course, it would be a lot more sensible--and more difficult to optimize--to look for matches of products with possible initial and/or terminal digit strings.  I would now guess that the problem can be fully worked out perhaps as far as k=25 (optimally programmed using massive currently available
        resources).
        JGM 

        --- On Mon, 4/4/11, James Merickel <merk7777777@...> wrote:


        From: James Merickel <merk7777777@...>
        Subject: [PrimeNumbers] Products of k distinct primes that are concatenations of first k: Coincidence
        To: primenumbers@yahoogroups.com
        Date: Monday, April 4, 2011, 3:05 AM


         



        1319732172923115
        1713323729219115
        2191317297323115
        2291117192373135
        2291772319133115
        2322931117719135
        2329131171719235
        7112917313223195
        7191317229231135
        7191331129217235
         
        These are the only values through k=10 (k>1).  I have a very inefficient program and a much better one for k=11 and 12, respectively, running.  I expect a null result, but will report if something comes out (couple of days).  I think the problem is tractable with maximal resources and ideal programming through about k=16.  Looking for either a further example or a heuristic proof one likely doesn't exist.
         
        JGM

        [Non-text portions of this message have been removed]








        [Non-text portions of this message have been removed]
      Your message has been successfully submitted and would be delivered to recipients shortly.