Loading ...
Sorry, an error occurred while loading the content.

Re: [PrimeNumbers] additive combinations all prime?

Expand Messages
  • Phil Carmody
    ... -d+c+b-a and d-c-b+a are the same prime -d+c-b+a and d-c+b-a are the same prime -d-c+b+a and d+c-b-a are the same prime ... Just line up the
    Message 1 of 10 , May 5, 2006
    • 0 Attachment
      --- Mark Underwood <mark.underwood@...> wrote:
      > Note for below: negatives included as primes.
      >
      > Given positive a,b, a<b, we want b+a and b-a both prime. First
      > solution: (a,b): (1,4)
      >
      > Given positive a,b,c, a<b<c, we want c+b+a, c+b-a, c-b+a,-c+b+a all
      > prime. First solution: (a,b,c) :(1,4,8)
      >
      > Given positive a,b,c,d, a<b<c<d we want d+c+b+a, d+c+b-a, d+c-b+a, d-
      > c+b+a, -d+c+b+a, d+c-b-a, d-c+b-a,-d+c+b-a, d-c-b+a, -d+c-b+a, -d-c+b+a
      > all prime. (11 combinations.) First solution: (a,b,c,d): (3,5,8,13)
      > (No "fibbing"!)

      -d+c+b-a and d-c-b+a are the same prime
      -d+c-b+a and d-c+b-a are the same prime
      -d-c+b+a and d+c-b-a are the same prime

      > Given positive a,b,c,d,e, a<b<c<d<e, we want all 16 combinations to be
      > prime. First solution: (a,b,c,d,e): (3,10,12,15,27)
      >
      > What about for 6 terms? That would be 1 + 6 + 15 + 20 = 42
      > combinations. All prime? Eeek! If it is solved it won't be on my
      > computer....

      Just line up the sums/differences, so that many combinations evaluate to the
      same thing.

      I think your count of combinations is incorrect. There are 2^(n-1) ways to
      chose the signs. See above examples of miscounting.

      Phil

      () ASCII ribbon campaign () Hopeless ribbon campaign
      /\ against HTML mail /\ against gratuitous bloodshed

      [stolen with permission from Daniel B. Cristofani]

      __________________________________________________
      Do You Yahoo!?
      Tired of spam? Yahoo! Mail has the best spam protection around
      http://mail.yahoo.com
    • Mark Underwood
      ... all ... b+a, d- ... c+b+a ... (3,5,8,13) ... to be ... evaluate to the ... ways to ... You re right Phil! As it turns out my method of counting works only
      Message 2 of 10 , May 6, 2006
      • 0 Attachment
        --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...>
        wrote:
        >
        > --- Mark Underwood <mark.underwood@...> wrote:
        > > Note for below: negatives included as primes.
        > >
        > > Given positive a,b, a<b, we want b+a and b-a both prime. First
        > > solution: (a,b): (1,4)
        > >
        > > Given positive a,b,c, a<b<c, we want c+b+a, c+b-a, c-b+a,-c+b+a
        all
        > > prime. First solution: (a,b,c) :(1,4,8)
        > >
        > > Given positive a,b,c,d, a<b<c<d we want d+c+b+a, d+c+b-a, d+c-
        b+a, d-
        > > c+b+a, -d+c+b+a, d+c-b-a, d-c+b-a,-d+c+b-a, d-c-b+a, -d+c-b+a, -d-
        c+b+a
        > > all prime. (11 combinations.) First solution: (a,b,c,d):
        (3,5,8,13)
        > > (No "fibbing"!)
        >
        > -d+c+b-a and d-c-b+a are the same prime
        > -d+c-b+a and d-c+b-a are the same prime
        > -d-c+b+a and d+c-b-a are the same prime
        >
        > > Given positive a,b,c,d,e, a<b<c<d<e, we want all 16 combinations
        to be
        > > prime. First solution: (a,b,c,d,e): (3,10,12,15,27)
        > >
        > > What about for 6 terms? That would be 1 + 6 + 15 + 20 = 42
        > > combinations. All prime? Eeek! If it is solved it won't be on my
        > > computer....
        >
        > Just line up the sums/differences, so that many combinations
        evaluate to the
        > same thing.
        >
        > I think your count of combinations is incorrect. There are 2^(n-1)
        ways to
        > chose the signs. See above examples of miscounting.
        >

        You're right Phil! As it turns out my method of counting works only
        when the number of terms is odd. If there are five terms:

        1 + 5 + (5*4)/2 = 16 = 2^(5-1) additive combinations.

        If there are seven terms:

        1 + 7 + (7*6)/2 + (7*6*5)/(3*2) = 64 = 2^(7-1) additive combinations.

        Mark
      • 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 3 of 10 , May 6, 2006
        • 0 Attachment
          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 4 of 10 , May 7, 2006
          • 0 Attachment
            --- "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 5 of 10 , May 8, 2006
            • 0 Attachment
              --- 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 6 of 10 , May 8, 2006
              • 0 Attachment
                --- "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 7 of 10 , May 9, 2006
                • 0 Attachment
                  --- "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 8 of 10 , Feb 5, 2008
                  • 0 Attachment
                    --- 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 9 of 10 , Feb 5, 2008
                    • 0 Attachment
                      --- 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
                    Your message has been successfully submitted and would be delivered to recipients shortly.