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Re: [PrimeNumbers] adjacents and primes

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  • Jens Kruse Andersen
    ... 7 divides a^3 + (a+1)^4 for a==5 (mod 7). 6 consecutive primes is then the best possible for that formula. It first occurs starting at a = 841369395. a^5 +
    Message 1 of 6 , Jan 3 5:21 PM
      December 17 Mark Underwood wrote:

      > Have been looking at a^r + b^s = prime,
      > where a,b are adjacent integers as are r,s.
      >
      > Occassionally there is a glimmer of order, like for instance
      >
      > 20^3 + 21^4 = prime
      > 21^3 + 22^4 = prime
      > 22^3 + 23^4 = prime
      > 23^3 + 24^4 = prime
      > 24^3 + 25^4 = prime.
      >
      > Is there any longer? (I don't know, haven't looked into it.)

      7 divides a^3 + (a+1)^4 for a==5 (mod 7).
      6 consecutive primes is then the best possible for that formula.
      It first occurs starting at a = 841369395.

      a^5 + (a+1)^4 never has a factor below 17.
      17 divides for both a==1 and a==4 (mod 17).
      The formula can avoid small factors for 13 consecutive a.
      The first case of 6 primes starts at a = 1469217.
      The first 7 is:

      188934537^5 + 188934538^4
      188934538^5 + 188934539^4
      188934539^5 + 188934540^4
      188934540^5 + 188934541^4
      188934541^5 + 188934542^4
      188934542^5 + 188934543^4
      188934543^5 + 188934544^4

      > Anyways I wonder, given any integer a >1 is there always at least one
      > adjacent integer b such that a^r + b^s is prime, where r,s are
      > adjacent integers no more than a?
      >
      > For instance:
      >
      > 26: 26^2 + 27^3 = prime
      > 29: 29^10 + 28^9 = prime
      > 35: 35^2 + 34^3 = prime

      I don't allow exponent 0 (a^0=1).

      If exponent 1 is allowed:
      The smallest a>1 for which a^r + b^s is never prime for
      b = a+/-1, s = r+/-1, and r,s <= a, is a = 13361.
      This assumes prp's for smaller a are really primes.

      The cases where prp's above 3000 digits were needed:
      1252^1155 + 1253^1154 (3578 digits)
      3319^992 + 3318^991 (3493 digits)
      9818^765 + 9819^764 (3054 digits)
      9819^764 + 9818^765 (3054 digits, same prp as for a=9818)
      10127^888 + 10126^887 (3557 digits)
      11051^1176 + 11050^1177 (4760 digits)

      PrimeForm/GW made prp tests. 13361^13361 has 55126 digits.
      I stopped there but guess there are primes for larger exponents.

      If exponent 1 is not allowed then the prime 3743^1 + 3744^2 is
      skipped and the smallest a>2 is a = 3743.
      For a=2, there are no allowed r,s values at all.

      --
      Jens Kruse Andersen
    • Mark Underwood
      Jens I don t know how you can go up that high, good lord! Amazing. Thank you for finding (at over 55,000 digits!) what I could never have hoped to. kind
      Message 2 of 6 , Jan 3 10:34 PM
        Jens I don't know how you can go up that high, good lord! Amazing.
        Thank you for finding (at over 55,000 digits!) what I could never
        have hoped to.

        kind regards,
        Mark




        --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
        <jens.k.a@g...> wrote:
        >
        (snip)

        > The smallest a>1 for which a^r + b^s is never prime for
        > b = a+/-1, s = r+/-1, and r,s <= a, is a = 13361.
        > This assumes prp's for smaller a are really primes.
        >
        > The cases where prp's above 3000 digits were needed:
        > 1252^1155 + 1253^1154 (3578 digits)
        > 3319^992 + 3318^991 (3493 digits)
        > 9818^765 + 9819^764 (3054 digits)
        > 9819^764 + 9818^765 (3054 digits, same prp as for a=9818)
        > 10127^888 + 10126^887 (3557 digits)
        > 11051^1176 + 11050^1177 (4760 digits)
        >
        > PrimeForm/GW made prp tests. 13361^13361 has 55126 digits.
        > I stopped there but guess there are primes for larger exponents.
        >
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