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n, 2n-1, 2n+1 all prime or prime-power (maybe n-2 also)

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  • WarrenS
    If we demand that n, 2n-1, and 2n+1 all simultaneously be prime or prime-power, then initial examples found by hand are n, 2n-1, 2n+1 2, 3, 5 3, 5, 7 5, 9*, 11
    Message 1 of 6 , Oct 1, 2012
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      If we demand that n, 2n-1, and 2n+1 all simultaneously be prime or prime-power,
      then initial examples found by hand are

      n, 2n-1, 2n+1
      2, 3, 5
      3, 5, 7
      5, 9*, 11
      9*, 17, 19
      13, 25*, 27*
      41, 81*, 83
      121*,241,243*
      (next one over 10^12)

      where * for prime powers.
      You can easily see that is is impossible for all three to be prime (except in the
      first two lines) since at least one member of the troika must be divisible by 3. Hence the only way to accomplish it, is to make that one be a power of 3.
      So then we necessarily have an exponentially-sparse set of primes, in that sense resembling the famous "Mersenne primes" 2^p-1 as well as "Proth primes" like 3*2^n+1.
      Another kind of resemblance is the fact that if Q=3^k then
      Q+2 will be easy to test for primality if we know (Q+1)/2 is prime [since then it
      would be a "safeprime"]. In the other direction, (Q-1)/2 might also be easy to test for primality...

      Perhaps the 13,25,27 and 121,241,243 lines are the only ones featuring a
      a double (or triple) *.

      It seems to me that this is an interesting class of primes.
      Perhaps even more interesting are the "pack of four" cases where n-2 is ALSO prime or prime-power.
    • Jack Brennen
      First, you missed an easy one: 4, 7, 9 Second, the next one seems to be: (3^541-1)/2, 3^541-2, 3^541 As far as the conjecture about the small examples with
      Message 2 of 6 , Oct 1, 2012
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        First, you missed an easy one:

        4, 7, 9

        Second, the next one seems to be:

        (3^541-1)/2, 3^541-2, 3^541

        As far as the conjecture about the small examples with double
        powers being the only ones, that would seem to be related to
        the ABC Conjecture.



        On 10/1/2012 9:33 AM, WarrenS wrote:
        > If we demand that n, 2n-1, and 2n+1 all simultaneously be prime or prime-power,
        > then initial examples found by hand are
        >
        > n, 2n-1, 2n+1
        > 2, 3, 5
        > 3, 5, 7
        > 5, 9*, 11
        > 9*, 17, 19
        > 13, 25*, 27*
        > 41, 81*, 83
        > 121*,241,243*
        > (next one over 10^12)
        >
        > where * for prime powers.
        > You can easily see that is is impossible for all three to be prime (except in the
        > first two lines) since at least one member of the troika must be divisible by 3. Hence the only way to accomplish it, is to make that one be a power of 3.
        > So then we necessarily have an exponentially-sparse set of primes, in that sense resembling the famous "Mersenne primes" 2^p-1 as well as "Proth primes" like 3*2^n+1.
        > Another kind of resemblance is the fact that if Q=3^k then
        > Q+2 will be easy to test for primality if we know (Q+1)/2 is prime [since then it
        > would be a "safeprime"]. In the other direction, (Q-1)/2 might also be easy to test for primality...
        >
        > Perhaps the 13,25,27 and 121,241,243 lines are the only ones featuring a
        > a double (or triple) *.
        >
        > It seems to me that this is an interesting class of primes.
        > Perhaps even more interesting are the "pack of four" cases where n-2 is ALSO prime or prime-power.
        >
        >
        >
        >
        >
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      • WarrenS
        ... --thanks, this confirms my own (now computerized) results +example, 1; 2, 3, 5 -example, 2; 4, 7, 9 & +example, 2; 5, 9, 11 & -example, 3; 13, 25, 27 &
        Message 3 of 6 , Oct 1, 2012
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          --- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
          >
          > First, you missed an easy one:
          >
          > 4, 7, 9
          >
          > Second, the next one seems to be:
          >
          > (3^541-1)/2, 3^541-2, 3^541
          >
          > As far as the conjecture about the small examples with double
          > powers being the only ones, that would seem to be related to
          > the ABC Conjecture.

          --thanks, this confirms my own (now computerized) results
          +example, 1; 2, 3, 5
          -example, 2; 4, 7, 9 &
          +example, 2; 5, 9, 11 &
          -example, 3; 13, 25, 27 &
          +example, 4; 41, 81, 83
          -example, 5; 121, 241, 243
          -example, 541; (large)

          where the lines are of the form

          +example, n; (1+3^n)/2, 3^n, 2+3^n
          or
          -example, n; (-1+3^n)/2, -2+3^n, 3^n
          all three at the end of the lines being prime-power.
          I also have awarded a "&" iff "pack of four."
          It claims there are no further examples
          for n<=4096.

          You are correct that the ABC conjecture looks related to my conjecture that
          there are only a finite set of such examples involving TWO or more prime powers.
          Weaker conjectures going in same direction are "Pillai's conjecture"
          and -- which actually now is a theorem by Mihailescu --
          Catalan's conjecture. The success on Catalan suggests to me
          that my conjecture might be within reach, although
          the proof would, if so, require a lot of effort.
        • WarrenS
          Also, at least one heuristic argument (involving 1/lnX probability that X is prime) suggests the conjecture that the set of n with n, 2n-1, 2n+1 all
          Message 4 of 6 , Oct 1, 2012
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            Also, at least one heuristic argument
            (involving 1/lnX "probability" that X is prime)
            suggests the conjecture that the set
            of n with n, 2n-1, 2n+1 all simultaneously prime or prime power, is a FINITE set.

            [On the other hand, I can also dream up a different heuristic argument (involving
            sieving the exponent of 3) which suggests it is an INFINITE set! You can place
            your bets on which heuristic to believe...]

            In the former case, it seems reasonably likely that
            Brennan & I have actually already found every example.

            It would be very interesting if anybody could prove this or any similar
            nontrivial finiteness theorem.

            I wondered if such a theorem could be proven under the assumption of the Riemann
            hypothesis & Montgomery pair correlation conjectures, and whatever other standard conjectures about nature of Riemann zeta zeros.
            I made a quick try to produce such a proof, but my attempt failed.
          • Phil Carmody
            ... I can t see an infinite set coming from any reasonable heuristic. You re summing 1/n^2. ... On its own, RH just helps shore up the finite heuristic, as it
            Message 5 of 6 , Oct 1, 2012
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              --- On Mon, 10/1/12, WarrenS <warren.wds@...> wrote:
              > Also, at least one heuristic argument
              > (involving 1/lnX "probability" that X is prime)
              > suggests the conjecture that the set
              > of n with n, 2n-1, 2n+1 all simultaneously prime or prime
              > power, is a FINITE set.
              >
              > [On the other hand, I can also dream up a different
              > heuristic argument (involving
              > sieving the exponent of 3) which suggests it is an INFINITE
              > set!   You can place
              > your bets on which heuristic to believe...]

              I can't see an infinite set coming from any reasonable heuristic. You're summing 1/n^2.

              > In the former case, it seems reasonably likely that
              > Brennan & I have actually already found every example.
              >
              > It would be very interesting if anybody could prove this or
              > any similar nontrivial finiteness theorem.
              >
              > I wondered if such a theorem could be proven under the
              > assumption of the Riemann
              > hypothesis & Montgomery pair correlation conjectures,
              > and whatever other standard conjectures about nature of
              > Riemann zeta zeros.
              > I made a quick try to produce such a proof, but my attempt
              > failed.

              On its own, RH just helps shore up the finite heuristic, as it makes the probabilities better justified.

              Phil
            • WarrenS
              ... --well, I basically agree with you. I can think of a heuristic that says infinite, but I don t like that heuristic :) Meanwhile I happened to notice
              Message 6 of 6 , Oct 1, 2012
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                > I can't see an infinite set coming from any reasonable heuristic. You're summing 1/n^2.

                --well, I basically agree with you. I can think of a heuristic that says infinite, but I don't like that heuristic :) Meanwhile I happened to notice these summaries of immense computations:

                http://oeis.org/A014224
                http://oeis.org/A028491

                http://oeis.org/A051783
                http://oeis.org/A171381 (this last one surprised me!)

                which show that Brennan & my examples are all there are,
                up to 3^195430 at least (wow!) PROVIDED we only allow
                ONE prime-power, the other two need to be genuine primes.
                It would not be hard to use these to genuinely deal with prime powers too,
                but I haven't.

                In the proofs of things like the Catalan conjecture, usually they prove it for very large
                numbers, then deal with the small numbers by computer.
                The fact that these computations have been so immense may make it
                feasible to prove there are no more examples with TWO nonprime prime-powers.
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