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ip +/- n = primenumbers

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  • FILIPPO GIORDANO
    - If p = prime numbers 3, 5, 7 or 13; (Se ip = numero primo 3, 5, 7 oppure 13); - if i = n^2 -1 (se i = n^2 -1)... for each n 1
    Message 1 of 2 , Apr 2, 2002
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      - If p = prime numbers 3, 5, 7 or 13;
      (Se ip = numero primo 3, 5, 7 oppure 13);
      - if i = n^2 -1
      (se i = n^2 -1)...
      for each n> 1 <p, ip - n = prime number,
      ip + n = prime number,
      (per ciascun n>1<p, ip -n= numero primo,
      ip+n=numero primo.
      -----------------------------------------------------------
      p = 3:
      n; i; ip; ip-n; ip+n
      2; 3; 9; 13; 17;
      -------------------------------------
      p = 5:
      n; i; ip; ip-n; ip+n
      2; 3; 15; 13; 17;
      3; 8; 40; 37; 43;
      4; 15; 75; 71; 79;
      --------------------------------------
      p = 7:
      n; i; ip; ip-n; ip+n
      2; 3; 21; 19; 23;
      3; 8; 56; 53; 59;
      4; 15; 105; 101; 109;
      5; 24; 168; 163; 173;
      6; 35; 245; 239; 251.
      ---------------------------------------
      p = 13:
      n; i; ip; ip-n; ip+n
      2; 3; 39; 37; 41;
      3; 8; 104; 101; 107;
      4; 15; 195; 191; 199;
      5; 24; 312; 307; 317;
      6; 35; 455; 449; 461;
      7; 48; 624; 617; 631;
      8; 63; 819; 811; 827;
      9; 80; 1040; 1031; 1049;
      10; 99; 1287; 1277; 1297;
      11; 120; 1560; 1549; 1571;
      12; 143; 1859; 1847; 1871.

      Regards.
      Filippo Giordano.



      [Non-text portions of this message have been removed]
    • Phil Carmody
      ... [Erk! webmail is misbehaving - unable to quote] Tables of i and n for each prime p, and all 1
      Message 2 of 2 , Apr 2, 2002
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        --- FILIPPO GIORDANO <fil.giordano@...> wrote:
        [Erk! webmail is misbehaving - unable to quote]

        Tables of i and n for each prime p, and all 1<n<p, all naturals, such
        that i is minimal and i.p+/-n are simultaniously prime.
        e.g. p=13, n=12 has i=143, as 1847 and 1871 (1859+/-12) are both
        prime.

        A little challenge...

        Find the i values such that i.p+/-n are _consecutive_ primes.

        No particular reason, just a whim, and as the number or results
        increases quadratically in the value of the prime chosen, I'm most
        interested in the n=p-1 results, as they require the largest
        composite run, and are thus heuristically hardest.

        Phil



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