Pair up ${1..2n}$ that the sums of each pair are different primes.












4












$begingroup$


Pair up ${1..2n}$ that the sums of each pair are different primes.
I found 9 examples:



${(1,2)},$



${(1,2),(3,4)},$



${(1,2),(3,4),(5,6)},$



${(1,4),(2,5),(3,8),(6,7)},{(2,3),(1,6),(4,7),(5,8)},$



${(1,4),(2,5),(3,8),(6,7),(9,10)},{(2,3),(1,6),(4,7),(5,8),(9,10)},$



${(1,4),(2,5),(3,8),(6,7),(9,10),(11,12)},{(2,3),(1,6),(4,7),(5,8),(9,10),(11,12)}.$



Is there another example like this?






combinatorics prime-numbers







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    4












    $begingroup$


    Pair up ${1..2n}$ that the sums of each pair are different primes.
    I found 9 examples:



    ${(1,2)},$



    ${(1,2),(3,4)},$



    ${(1,2),(3,4),(5,6)},$



    ${(1,4),(2,5),(3,8),(6,7)},{(2,3),(1,6),(4,7),(5,8)},$



    ${(1,4),(2,5),(3,8),(6,7),(9,10)},{(2,3),(1,6),(4,7),(5,8),(9,10)},$



    ${(1,4),(2,5),(3,8),(6,7),(9,10),(11,12)},{(2,3),(1,6),(4,7),(5,8),(9,10),(11,12)}.$



    Is there another example like this?






    combinatorics prime-numbers







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      3



      $begingroup$


      Pair up ${1..2n}$ that the sums of each pair are different primes.
      I found 9 examples:



      ${(1,2)},$



      ${(1,2),(3,4)},$



      ${(1,2),(3,4),(5,6)},$



      ${(1,4),(2,5),(3,8),(6,7)},{(2,3),(1,6),(4,7),(5,8)},$



      ${(1,4),(2,5),(3,8),(6,7),(9,10)},{(2,3),(1,6),(4,7),(5,8),(9,10)},$



      ${(1,4),(2,5),(3,8),(6,7),(9,10),(11,12)},{(2,3),(1,6),(4,7),(5,8),(9,10),(11,12)}.$



      Is there another example like this?






      combinatorics prime-numbers







      share|cite|improve this question











      $endgroup$




      Pair up ${1..2n}$ that the sums of each pair are different primes.
      I found 9 examples:



      ${(1,2)},$



      ${(1,2),(3,4)},$



      ${(1,2),(3,4),(5,6)},$



      ${(1,4),(2,5),(3,8),(6,7)},{(2,3),(1,6),(4,7),(5,8)},$



      ${(1,4),(2,5),(3,8),(6,7),(9,10)},{(2,3),(1,6),(4,7),(5,8),(9,10)},$



      ${(1,4),(2,5),(3,8),(6,7),(9,10),(11,12)},{(2,3),(1,6),(4,7),(5,8),(9,10),(11,12)}.$



      Is there another example like this?






      combinatorics prime-numbers




      combinatorics prime-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 14:49







      Manyama

















      asked Dec 8 '18 at 14:16









      ManyamaManyama

      4971315




      4971315






















          1 Answer
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          $begingroup$

          There are no more. There are none for $n=7$. We would need the sums to be seven distinct odd primes below $27$. The most the sum of these sums can be is $5+7+11+13+17+19+23=95$ but the sum of all the numbers up to $14$ is $105$.



          For $n=8$ we would need the sum of eight primes below $31$ to be $120$. We can only do this with $3,5,11,13,17,19,23,29$. We need $1+2$ to get $3$ but cannot get $5$.



          For $n=9$ we need all the primes up to $31$ except $2,5$, but then $1+2=3,3+4=7,5+5=11$ and we are stuck for $13$.



          For $n=10$ we need all the primes up to $37$ except $2,5$ and the $n=9$ proof works.



          For $n=11$ we need all the primes up to $43$ except $2,7,11$ or $2,5,13$. We can do $21+22=43$ but cannot get $41$.



          For $n=12$ the sum of all the primes up to $43$ is $271$ while the sum of the numbers up to $24$ is $276$



          Now going from $n$ to $n+1$ the sum of the numbers increases by $4n+3$ while the sum of primes increases by $4n+1, 4n+3,$ or $8n+4$. As less than $frac {2cdot 3 cdot 5}{3 cdot 5 cdot 7} lt frac 12$ are primes, the sum of primes will never catch up. $n=13$ doesn't add any new primes because the odd numbers added are $49,51$ and again we miss at $n=19$. Each intervening $n$ has only added one prime.






          share|cite|improve this answer











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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            There are no more. There are none for $n=7$. We would need the sums to be seven distinct odd primes below $27$. The most the sum of these sums can be is $5+7+11+13+17+19+23=95$ but the sum of all the numbers up to $14$ is $105$.



            For $n=8$ we would need the sum of eight primes below $31$ to be $120$. We can only do this with $3,5,11,13,17,19,23,29$. We need $1+2$ to get $3$ but cannot get $5$.



            For $n=9$ we need all the primes up to $31$ except $2,5$, but then $1+2=3,3+4=7,5+5=11$ and we are stuck for $13$.



            For $n=10$ we need all the primes up to $37$ except $2,5$ and the $n=9$ proof works.



            For $n=11$ we need all the primes up to $43$ except $2,7,11$ or $2,5,13$. We can do $21+22=43$ but cannot get $41$.



            For $n=12$ the sum of all the primes up to $43$ is $271$ while the sum of the numbers up to $24$ is $276$



            Now going from $n$ to $n+1$ the sum of the numbers increases by $4n+3$ while the sum of primes increases by $4n+1, 4n+3,$ or $8n+4$. As less than $frac {2cdot 3 cdot 5}{3 cdot 5 cdot 7} lt frac 12$ are primes, the sum of primes will never catch up. $n=13$ doesn't add any new primes because the odd numbers added are $49,51$ and again we miss at $n=19$. Each intervening $n$ has only added one prime.






            share|cite|improve this answer











            $endgroup$


















              5












              $begingroup$

              There are no more. There are none for $n=7$. We would need the sums to be seven distinct odd primes below $27$. The most the sum of these sums can be is $5+7+11+13+17+19+23=95$ but the sum of all the numbers up to $14$ is $105$.



              For $n=8$ we would need the sum of eight primes below $31$ to be $120$. We can only do this with $3,5,11,13,17,19,23,29$. We need $1+2$ to get $3$ but cannot get $5$.



              For $n=9$ we need all the primes up to $31$ except $2,5$, but then $1+2=3,3+4=7,5+5=11$ and we are stuck for $13$.



              For $n=10$ we need all the primes up to $37$ except $2,5$ and the $n=9$ proof works.



              For $n=11$ we need all the primes up to $43$ except $2,7,11$ or $2,5,13$. We can do $21+22=43$ but cannot get $41$.



              For $n=12$ the sum of all the primes up to $43$ is $271$ while the sum of the numbers up to $24$ is $276$



              Now going from $n$ to $n+1$ the sum of the numbers increases by $4n+3$ while the sum of primes increases by $4n+1, 4n+3,$ or $8n+4$. As less than $frac {2cdot 3 cdot 5}{3 cdot 5 cdot 7} lt frac 12$ are primes, the sum of primes will never catch up. $n=13$ doesn't add any new primes because the odd numbers added are $49,51$ and again we miss at $n=19$. Each intervening $n$ has only added one prime.






              share|cite|improve this answer











              $endgroup$
















                5












                5








                5





                $begingroup$

                There are no more. There are none for $n=7$. We would need the sums to be seven distinct odd primes below $27$. The most the sum of these sums can be is $5+7+11+13+17+19+23=95$ but the sum of all the numbers up to $14$ is $105$.



                For $n=8$ we would need the sum of eight primes below $31$ to be $120$. We can only do this with $3,5,11,13,17,19,23,29$. We need $1+2$ to get $3$ but cannot get $5$.



                For $n=9$ we need all the primes up to $31$ except $2,5$, but then $1+2=3,3+4=7,5+5=11$ and we are stuck for $13$.



                For $n=10$ we need all the primes up to $37$ except $2,5$ and the $n=9$ proof works.



                For $n=11$ we need all the primes up to $43$ except $2,7,11$ or $2,5,13$. We can do $21+22=43$ but cannot get $41$.



                For $n=12$ the sum of all the primes up to $43$ is $271$ while the sum of the numbers up to $24$ is $276$



                Now going from $n$ to $n+1$ the sum of the numbers increases by $4n+3$ while the sum of primes increases by $4n+1, 4n+3,$ or $8n+4$. As less than $frac {2cdot 3 cdot 5}{3 cdot 5 cdot 7} lt frac 12$ are primes, the sum of primes will never catch up. $n=13$ doesn't add any new primes because the odd numbers added are $49,51$ and again we miss at $n=19$. Each intervening $n$ has only added one prime.






                share|cite|improve this answer











                $endgroup$



                There are no more. There are none for $n=7$. We would need the sums to be seven distinct odd primes below $27$. The most the sum of these sums can be is $5+7+11+13+17+19+23=95$ but the sum of all the numbers up to $14$ is $105$.



                For $n=8$ we would need the sum of eight primes below $31$ to be $120$. We can only do this with $3,5,11,13,17,19,23,29$. We need $1+2$ to get $3$ but cannot get $5$.



                For $n=9$ we need all the primes up to $31$ except $2,5$, but then $1+2=3,3+4=7,5+5=11$ and we are stuck for $13$.



                For $n=10$ we need all the primes up to $37$ except $2,5$ and the $n=9$ proof works.



                For $n=11$ we need all the primes up to $43$ except $2,7,11$ or $2,5,13$. We can do $21+22=43$ but cannot get $41$.



                For $n=12$ the sum of all the primes up to $43$ is $271$ while the sum of the numbers up to $24$ is $276$



                Now going from $n$ to $n+1$ the sum of the numbers increases by $4n+3$ while the sum of primes increases by $4n+1, 4n+3,$ or $8n+4$. As less than $frac {2cdot 3 cdot 5}{3 cdot 5 cdot 7} lt frac 12$ are primes, the sum of primes will never catch up. $n=13$ doesn't add any new primes because the odd numbers added are $49,51$ and again we miss at $n=19$. Each intervening $n$ has only added one prime.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 9 '18 at 22:57

























                answered Dec 9 '18 at 19:40









                Ross MillikanRoss Millikan

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