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Re: additive combinations all prime?

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  • Mark Underwood
    Well surprise of surprises: The six numbers (5,30,33,42,60,63) yield primes for all 32 additive combinations of the six numbers.(!) I thought it would have
    Message 1 of 10 , May 6, 2006
      Well surprise of surprises:

      The six numbers (5,30,33,42,60,63) yield primes for all 32 additive
      combinations of the six numbers.(!) I thought it would have been
      waaay higher.

      So we have as first cases:

      (1,4) has all 2 additive combinations yielding 2 primes.
      (1,4,8) has all 4 additive combos yielding 4 primes.
      (3,5,8,13) has all 8 additive combos yielding 8 primes.
      (3,10,12,15,27) has all 16 additive combos yielding 16 primes.
      (5,30,33,42,60,63) has all 32 additive combos yielding 32 primes.

      Interesting that thus far the greatest number in each set does not
      quite exceed 2^n. Surely that can't hold up for long...or can it?

      Wonders never cease!
      Mark
    • Patrick Capelle
      ... Something interesting with the product of the numbers ? (1,4)-- 4 = 2^2 (1,4,8)-- 32 = 2^5 (3,5,8,13)-- 1560 = 2^3 * 3 * 5 * 13 (3,10,12,15,27)--
      Message 2 of 10 , May 7, 2006
        --- "Mark Underwood" <mark.underwood@...> wrote:
        >
        > Well surprise of surprises:
        >
        > The six numbers (5,30,33,42,60,63) yield primes for all 32 additive
        > combinations of the six numbers.(!) I thought it would have been
        > waaay higher.
        >
        > So we have as first cases:
        >
        > (1,4) has all 2 additive combinations yielding 2 primes.
        > (1,4,8) has all 4 additive combos yielding 4 primes.
        > (3,5,8,13) has all 8 additive combos yielding 8 primes.
        > (3,10,12,15,27) has all 16 additive combos yielding 16 primes.
        > (5,30,33,42,60,63) has all 32 additive combos yielding 32 primes.
        >
        > Interesting that thus far the greatest number in each set does not
        > quite exceed 2^n. Surely that can't hold up for long...or can it?
        >
        > Wonders never cease!
        > Mark
        ----------------------------------------------------------------------

        Something interesting with the product of the numbers ?

        (1,4)--> 4 = 2^2
        (1,4,8)--> 32 = 2^5
        (3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
        (3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
        (5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11

        In each case the product of the n numbers gives a number whose number
        of different prime factors (in the factorization) is smaller or equal
        to n.
        Is it always the case ?

        Patrick Capelle.
      • Mark Underwood
        ... number ... equal ... Perhaps. Anyways, the numbers are relatively factor rich. A prime guru on very quick order came up with 7 more examples where all 32
        Message 3 of 10 , May 8, 2006
          --- In primenumbers@yahoogroups.com, "Patrick Capelle"
          <patrick.capelle@...> wrote:
          >
          > Something interesting with the product of the numbers ?
          >
          > (1,4)--> 4 = 2^2
          > (1,4,8)--> 32 = 2^5
          > (3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
          > (3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
          > (5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11
          >
          > In each case the product of the n numbers gives a number whose
          number
          > of different prime factors (in the factorization) is smaller or
          equal
          > to n.
          > Is it always the case ?
          >

          Perhaps. Anyways, the numbers are relatively factor rich.

          A prime guru on very quick order came up with 7 more examples where
          all 32 additive combinations yield primes (!):

          (30, 33, 35, 60, 63, 72)
          (30, 33, 47, 60, 72, 75)
          (30, 42, 47, 60, 63, 75)
          (15, 42, 48, 57, 70, 75)
          (30, 33, 42, 60, 75, 77)
          (15, 33, 42, 57, 70, 90)
          (15, 35, 42, 57, 72, 90)

          What is curious to me is that each set has three numbers with factors
          of two. Also each set has three numbers with factors of five.
          But it is easy to see why each set has five numbers with factors of
          three.

          So far, no luck finding 7 numbers with all 64 additive combinations
          yielding primes.

          Mark
        • Patrick Capelle
          ... Something interesting with the product of the numbers ? (1,4)-- 4 = 2^2 (1,4,8)-- 32 = 2^5 (3,5,8,13)-- 1560 = 2^3 * 3 * 5 * 13 (3,10,12,15,27)--
          Message 4 of 10 , May 8, 2006
            --- "Patrick Capelle" <patrick.capelle@> wrote:
            Something interesting with the product of the numbers ?
            (1,4)--> 4 = 2^2
            (1,4,8)--> 32 = 2^5
            (3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
            (3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
            (5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11

            In each case the product of the n numbers gives a number whose
            number of different prime factors (in the factorization) is smaller
            or equal to n. Is it always the case ?
            --------------------------------------------------------------------
            --- "Mark Underwood" <mark.underwood@...> wrote:
            Perhaps. Anyways, the numbers are relatively factor rich.

            A prime guru on very quick order came up with 7 more examples where
            all 32 additive combinations yield primes (!)
            ...
            --------------------------------------------------------------------

            It holds for n = 6 :
            ( 5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11
            (30,33,35,60,63,72)--> 9430344000 = 2^6 * 3^7 * 5^3 * 7^2 * 11
            (30,33,47,60,72,75)--> 15075720000 = 2^6 * 3^6 * 5^4 * 11^ * 47
            (30,42,47,60,63,75)--> 16788870000 = 2^4 * 3^6 * 5^4 * 7^2 * 47
            (15,42,48,57,70,75)--> 9049320000 = 2^6 * 3^5 * 5^4 * 7^2 * 19
            (30,33,42,60,75,77)--> 14407470000 = 2^4 * 3^5 * 5^4 * 7^2 * 11^2
            (15,33,42,57,70,90)--> 7465689000 = 2^3 * 3^6 * 5^3 * 7^2 * 11 * 19
            (15,35,42,57,72,90)--> 8144388000 = 2^5 * 3^7 * 5^3 * 7^2 * 19

            Patrick Capelle.
          • Patrick Capelle
            ... Something interesting with the product of the numbers ? (1,4)-- 4 = 2^2 (1,4,8)-- 32 = 2^5 (3,5,8,13)-- 1560 = 2^3 * 3 * 5 * 13 (3,10,12,15,27)--
            Message 5 of 10 , May 9, 2006
              --- "Patrick Capelle" <patrick.capelle@...> wrote:
              Something interesting with the product of the numbers ?
              (1,4)--> 4 = 2^2
              (1,4,8)--> 32 = 2^5
              (3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
              (3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
              (5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11

              In each case the product of the n numbers gives a number whose
              number of different prime factors (in the factorization) is smaller
              or equal to n. Is it always the case ?
              ---------------------------------------------------------------------
              --- "Patrick Capelle" <patrick.capelle@...> wrote:
              It holds for n = 6 :
              ( 5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11
              (30,33,35,60,63,72)--> 9430344000 = 2^6 * 3^7 * 5^3 * 7^2 * 11
              (30,33,47,60,72,75)--> 15075720000 = 2^6 * 3^6 * 5^4 * 11^ * 47
              (30,42,47,60,63,75)--> 16788870000 = 2^4 * 3^6 * 5^4 * 7^2 * 47
              (15,42,48,57,70,75)--> 9049320000 = 2^6 * 3^5 * 5^4 * 7^2 * 19
              (30,33,42,60,75,77)--> 14407470000 = 2^4 * 3^5 * 5^4 * 7^2 * 11^2
              (15,33,42,57,70,90)--> 7465689000 = 2^3 * 3^6 * 5^3 * 7^2 * 11 * 19
              (15,35,42,57,72,90)--> 8144388000 = 2^5 * 3^7 * 5^3 * 7^2 * 19
              ---------------------------------------------------------------------
              --- "Mark Underwood" <mark.underwood@...> wrote:
              I assumed you meant it would hold for first cases. Afterall it
              certainly doesn't hold for some cases after the first case. For
              instance

              14 + 9 = 23
              14 - 9 = 5

              And 14*9 has three different prime factors.

              I'm going to try to look for solutions for n=7 and all 64 additive
              combinations yielding primes. My computer is so slow however it does
              not look promising.

              kind regards,
              Mark
              ---------------------------------------------------------------------

              Important precision.
              Thank you Mark.
              I only started with the examples that you gave, without thinking to
              other cases.
              When we look at each n it is possible that it holds only for the
              smallest set(s).
              Or for all the sets when n is not too small ?

              Best regards,
              Patrick Capelle.
            • Mark Underwood
              ... Took a nice sabbatical from numbers, namely the primes. The things drive me crazy. But in a momement of weakness/boredom, perhaps abit of seasonal
              Message 6 of 10 , Feb 5, 2008
                --- In primenumbers@yahoogroups.com, "Mark Underwood"
                <mark.underwood@...> wrote:
                >
                > Well surprise of surprises:
                >
                > The six numbers (5,30,33,42,60,63) yield primes for all 32 additive
                > combinations of the six numbers.(!) I thought it would have been
                > waaay higher.
                >
                > So we have as first cases:
                >
                > (1,4) has all 2 additive combinations yielding 2 primes.
                > (1,4,8) has all 4 additive combos yielding 4 primes.
                > (3,5,8,13) has all 8 additive combos yielding 8 primes.
                > (3,10,12,15,27) has all 16 additive combos yielding 16 primes.
                > (5,30,33,42,60,63) has all 32 additive combos yielding 32 primes.
                >
                > Interesting that thus far the greatest number in each set does not
                > quite exceed 2^n. Surely that can't hold up for long...or can it?
                >
                > Wonders never cease!
                > Mark
                >


                Took a nice sabbatical from numbers, namely the primes. The things
                drive me crazy. But in a momement of weakness/boredom, perhaps abit of
                seasonal affective disorder, I looked back over some of my
                investigations.


                Very shortly after this post from about a year and a half ago, I found
                a way to greatly speed up the search for additive combinations which
                yeild all primes. For instance: for six numbers, each not exceeding
                500, I found on the order of about 70 different solutions, each
                yeilding (in their 32 additive combinations) all primes. But I noted
                only one which yielded 32 *distinct* primes:


                42,60,75,77,105,108

                in its 32 additive combinations yields the 32 unique primes

                13,17,19,23,41,43,47,53,79,97,
                101,103,107,109,113,131,137,163,167,173,193,197,
                229,233,251,257,263,313,317,347,383,467.


                I tried a similar technique to get 7 numbers with all 64 additive
                combinations yielding only primes, but with no success. I don't think
                there is a solution if all the numbers are below 1000, but it wasn't
                quite an exhaustive search.

                Just throwing this out in case anyone is interested, before I entirely
                forgot about it.

                Mark
              • Mark Underwood
                ... [snip] ... [snip] I meant each number not exceeding 150 (not 500). Mark
                Message 7 of 10 , Feb 5, 2008
                  --- In primenumbers@yahoogroups.com, "Mark Underwood" <mark.underwood@...> wrote:

                  [snip]
                  > For instance: for six numbers, each not exceeding
                  > 500, I found on the order of about 70 different solutions, each
                  > yeilding (in their 32 additive combinations) all primes.
                  [snip]

                  I meant each number not exceeding 150 (not 500).

                  Mark
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