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Consecutive integers: 1*prime, 2*prime, 3*prime, 4*prime,5*prime,...

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  • Mark
    Andrey s line of questioning brought this (much simpler!) question to mind. We want to find consecutive integers: 1*prime,2*prime, 3*prime,
    Message 1 of 6 , Feb 25, 2012
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      Andrey's line of questioning brought this (much simpler!) question to mind.

      We want to find consecutive integers: 1*prime,2*prime, 3*prime, 4*prime,5*prime,...

      How far can you go? I got to three using an advanced carbon based, parallel bioprocessor running at about 15 Hz:

      1*13, 2*7, 3*5.

      Mark
    • Jens Kruse Andersen
      ... Somebody once posted this problem here. Guess who! http://tech.groups.yahoo.com/group/primenumbers/message/16364 It s also at
      Message 2 of 6 , Feb 25, 2012
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        Mark wrote:
        > We want to find consecutive integers: 1*prime,2*prime, 3*prime,
        > 4*prime,5*prime,...

        Somebody once posted this problem here. Guess who!
        http://tech.groups.yahoo.com/group/primenumbers/message/16364

        It's also at http://www.primepuzzles.net/puzzles/puzz_181.htm
        and http://oeis.org/A074200

        --
        Jens Kruse Andersen
      • Mark
        Good lord! It s rather disconcerting that I have absolutely no memory of that. Jens, I admire your power of recall. I think I ve regressed, hehe! Mark
        Message 3 of 6 , Feb 25, 2012
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          Good lord! It's rather disconcerting that I have absolutely no memory of that. Jens, I admire your power of recall. I think I've regressed, hehe!

          Mark

          --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...> wrote:
          >
          > Mark wrote:
          > > We want to find consecutive integers: 1*prime,2*prime, 3*prime,
          > > 4*prime,5*prime,...
          >
          > Somebody once posted this problem here. Guess who!
          > http://tech.groups.yahoo.com/group/primenumbers/message/16364
          >
          > It's also at http://www.primepuzzles.net/puzzles/puzz_181.htm
          > and http://oeis.org/A074200
          >
          > --
          > Jens Kruse Andersen
          >
        • WarrenS
          ... --The number just before such an n-term sequence must be divisible by n!, according to my advanced bioprocessor. Say it is n!*k. Then n!*k+1=prime,
          Message 4 of 6 , Feb 28, 2012
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            --- In primenumbers@yahoogroups.com, "Mark" <mark.underwood@...> wrote:
            >
            > Andrey's line of questioning brought this (much simpler!) question to mind.
            >
            > We want to find consecutive integers: 1*prime,2*prime, 3*prime, 4*prime,5*prime,...
            >
            > How far can you go? I got to three using an advanced carbon based, parallel bioprocessor running at about 15 Hz:
            >
            > 1*13, 2*7, 3*5.
            >
            > Mark

            --The number just before such an n-term sequence must be divisible by n!,
            according to my advanced bioprocessor. Say it is n!*k.
            Then n!*k+1=prime, n!*k/2+1=prime, n!*k/3+1=prime, n!*k/4+1=prime, etc.

            I think this is a pretty good problem, actually. I can't see why it could not happen for arbitrarily large n, and furthermore, whenever it does happen, it'll be pretty easy to prove primality for the primes p, since p-1 is easily factored.
          • Peter Kosinar
            ... Not exactly so; it only needs to be divisible by least common multiple of numbers [1..n]. For example, for n = 6, the first occurrence is V = 5516280 =
            Message 5 of 6 , Feb 28, 2012
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              > --The number just before such an n-term sequence must be divisible by n!,
              > according to my advanced bioprocessor. Say it is n!*k.
              > Then n!*k+1=prime, n!*k/2+1=prime, n!*k/3+1=prime, n!*k/4+1=prime, etc.

              Not exactly so; it only needs to be divisible by least common multiple of
              numbers [1..n]. For example, for n = 6, the first occurrence is

              V = 5516280 = 60*91938 = 720*7661.5
              V+1 = 1*5516281
              V+2 = 2*2758141
              V+3 = 3*1838761
              V+4 = 4*1379071
              V+5 = 5*1103257
              V+6 = 6*919381

              Peter
            • Warren Smith
              right, good correction.
              Message 6 of 6 , Feb 28, 2012
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                right, good correction.

                >Not exactly so; it only needs to be divisible by least common multiple of numbers [1..n].
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