Does this sequence of product of two primes always exist












2












$begingroup$


Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:




  1. Each $a_i$ is distinct.


  2. $m$ is the least integer so that $frac{m(m-1)}{2} ge n$


  3. Any three consecutive elements of sequence are coprime, but any two are not.



For example,




  1. for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$


  2. For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$


  3. For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$



The question is to show that such a sequence exists for any $n$.



If this sequence exists for all $n$, is there a simple pattern for $A$ ?



Edit



There was an error in example 3, no such sequence exists for $n=6$.






sequences-and-series algebra-precalculus discrete-mathematics prime-numbers







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  • 1




    $begingroup$
    There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
    $endgroup$
    – ajotatxe
    Jan 8 at 15:13










  • $begingroup$
    @ajotatxe sorry I overlooked this one!
    $endgroup$
    – jeea
    Jan 8 at 15:23
















2












$begingroup$


Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:




  1. Each $a_i$ is distinct.


  2. $m$ is the least integer so that $frac{m(m-1)}{2} ge n$


  3. Any three consecutive elements of sequence are coprime, but any two are not.



For example,




  1. for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$


  2. For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$


  3. For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$



The question is to show that such a sequence exists for any $n$.



If this sequence exists for all $n$, is there a simple pattern for $A$ ?



Edit



There was an error in example 3, no such sequence exists for $n=6$.






sequences-and-series algebra-precalculus discrete-mathematics prime-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
    $endgroup$
    – ajotatxe
    Jan 8 at 15:13










  • $begingroup$
    @ajotatxe sorry I overlooked this one!
    $endgroup$
    – jeea
    Jan 8 at 15:23














2












2








2





$begingroup$


Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:




  1. Each $a_i$ is distinct.


  2. $m$ is the least integer so that $frac{m(m-1)}{2} ge n$


  3. Any three consecutive elements of sequence are coprime, but any two are not.



For example,




  1. for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$


  2. For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$


  3. For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$



The question is to show that such a sequence exists for any $n$.



If this sequence exists for all $n$, is there a simple pattern for $A$ ?



Edit



There was an error in example 3, no such sequence exists for $n=6$.






sequences-and-series algebra-precalculus discrete-mathematics prime-numbers







share|cite|improve this question











$endgroup$




Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:




  1. Each $a_i$ is distinct.


  2. $m$ is the least integer so that $frac{m(m-1)}{2} ge n$


  3. Any three consecutive elements of sequence are coprime, but any two are not.



For example,




  1. for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$


  2. For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$


  3. For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$



The question is to show that such a sequence exists for any $n$.



If this sequence exists for all $n$, is there a simple pattern for $A$ ?



Edit



There was an error in example 3, no such sequence exists for $n=6$.






sequences-and-series algebra-precalculus discrete-mathematics prime-numbers




sequences-and-series algebra-precalculus discrete-mathematics prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 8 at 15:22







jeea

















asked Jan 8 at 15:01









jeeajeea

57314




57314








  • 1




    $begingroup$
    There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
    $endgroup$
    – ajotatxe
    Jan 8 at 15:13










  • $begingroup$
    @ajotatxe sorry I overlooked this one!
    $endgroup$
    – jeea
    Jan 8 at 15:23














  • 1




    $begingroup$
    There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
    $endgroup$
    – ajotatxe
    Jan 8 at 15:13










  • $begingroup$
    @ajotatxe sorry I overlooked this one!
    $endgroup$
    – jeea
    Jan 8 at 15:23








1




1




$begingroup$
There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
$endgroup$
– ajotatxe
Jan 8 at 15:13




$begingroup$
There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
$endgroup$
– ajotatxe
Jan 8 at 15:13












$begingroup$
@ajotatxe sorry I overlooked this one!
$endgroup$
– jeea
Jan 8 at 15:23




$begingroup$
@ajotatxe sorry I overlooked this one!
$endgroup$
– jeea
Jan 8 at 15:23










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$begingroup$

Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.



This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.






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    $begingroup$

    Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.



    This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.



      This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.



        This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.






        share|cite|improve this answer









        $endgroup$



        Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.



        This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 15:15









        Especially LimeEspecially Lime

        21.9k22858




        21.9k22858






























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