How to find a recurrence relation for counting the number of solutions?












2












$begingroup$


Consider the diophantine equation
$$x_1+3x_2+5x_3 = n$$
where $x_igeq 0$ and $ngeq 1.$
Let $P_n(1,3,5)$ denote the number of solutions to this equation. I want to express $P_n(1,3,5)$ in terms of $P_{1}(1,3,5), P_{2}(1,3,5),cdots, P_{n-1}(1,3,5).$



Here is what I observed:



If $(x_1,x_2,x_3)$ is a solution to
$$x_1+3x_2+5x_3 = k$$
then $(x_1+1,x_2,x_3)$ is a solution to
$$x_1+3x_2+5x_3 = k+1$$
$(x_1,x_2+1,x_3)$ is a solution to
$$x_1+3x_2+5x_3 = k+3$$
and $(x_1,x_2,x_3+1)$ is a solution to
$$x_1+3x_2+5x_3 = k+5.$$
But I don't know know how to combine this to get the desired relation. Any ideas will be much appreciated.



Edit: Based on the answer given below, we observe that a solution (x_1,x_2,x_3) to
$$x_1+3x_2+5x_3 = n$$
must have $x_1>0$ or $x_2>0$ or $x_3>0.$ If $x_1>0$ then (x_1-1,x_2,x_3) is a solution
$$x_1+3x_2+5x_3 = n-1$$
and there are $P_{n-1}(1,3,5).$ Proceeding in a similar manner for $x_2$ and $x_3$ and applying the inclusion-exclusion principle we get:
$$P_{n}(1,3,5) = P_{n-1}(1,3,5)+P_{n-3}(1,3,5)+P_{n-5}(1,3,5)-P_{n-4}(1,3,5)-P_{n-8}(1,3,5)-P_{n-6}(1,3,5)+P_{n-9}(1,3,5).$$






number-theory elementary-set-theory diophantine-equations linear-diophantine-equations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Consider the diophantine equation
    $$x_1+3x_2+5x_3 = n$$
    where $x_igeq 0$ and $ngeq 1.$
    Let $P_n(1,3,5)$ denote the number of solutions to this equation. I want to express $P_n(1,3,5)$ in terms of $P_{1}(1,3,5), P_{2}(1,3,5),cdots, P_{n-1}(1,3,5).$



    Here is what I observed:



    If $(x_1,x_2,x_3)$ is a solution to
    $$x_1+3x_2+5x_3 = k$$
    then $(x_1+1,x_2,x_3)$ is a solution to
    $$x_1+3x_2+5x_3 = k+1$$
    $(x_1,x_2+1,x_3)$ is a solution to
    $$x_1+3x_2+5x_3 = k+3$$
    and $(x_1,x_2,x_3+1)$ is a solution to
    $$x_1+3x_2+5x_3 = k+5.$$
    But I don't know know how to combine this to get the desired relation. Any ideas will be much appreciated.



    Edit: Based on the answer given below, we observe that a solution (x_1,x_2,x_3) to
    $$x_1+3x_2+5x_3 = n$$
    must have $x_1>0$ or $x_2>0$ or $x_3>0.$ If $x_1>0$ then (x_1-1,x_2,x_3) is a solution
    $$x_1+3x_2+5x_3 = n-1$$
    and there are $P_{n-1}(1,3,5).$ Proceeding in a similar manner for $x_2$ and $x_3$ and applying the inclusion-exclusion principle we get:
    $$P_{n}(1,3,5) = P_{n-1}(1,3,5)+P_{n-3}(1,3,5)+P_{n-5}(1,3,5)-P_{n-4}(1,3,5)-P_{n-8}(1,3,5)-P_{n-6}(1,3,5)+P_{n-9}(1,3,5).$$






    number-theory elementary-set-theory diophantine-equations linear-diophantine-equations







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Consider the diophantine equation
      $$x_1+3x_2+5x_3 = n$$
      where $x_igeq 0$ and $ngeq 1.$
      Let $P_n(1,3,5)$ denote the number of solutions to this equation. I want to express $P_n(1,3,5)$ in terms of $P_{1}(1,3,5), P_{2}(1,3,5),cdots, P_{n-1}(1,3,5).$



      Here is what I observed:



      If $(x_1,x_2,x_3)$ is a solution to
      $$x_1+3x_2+5x_3 = k$$
      then $(x_1+1,x_2,x_3)$ is a solution to
      $$x_1+3x_2+5x_3 = k+1$$
      $(x_1,x_2+1,x_3)$ is a solution to
      $$x_1+3x_2+5x_3 = k+3$$
      and $(x_1,x_2,x_3+1)$ is a solution to
      $$x_1+3x_2+5x_3 = k+5.$$
      But I don't know know how to combine this to get the desired relation. Any ideas will be much appreciated.



      Edit: Based on the answer given below, we observe that a solution (x_1,x_2,x_3) to
      $$x_1+3x_2+5x_3 = n$$
      must have $x_1>0$ or $x_2>0$ or $x_3>0.$ If $x_1>0$ then (x_1-1,x_2,x_3) is a solution
      $$x_1+3x_2+5x_3 = n-1$$
      and there are $P_{n-1}(1,3,5).$ Proceeding in a similar manner for $x_2$ and $x_3$ and applying the inclusion-exclusion principle we get:
      $$P_{n}(1,3,5) = P_{n-1}(1,3,5)+P_{n-3}(1,3,5)+P_{n-5}(1,3,5)-P_{n-4}(1,3,5)-P_{n-8}(1,3,5)-P_{n-6}(1,3,5)+P_{n-9}(1,3,5).$$






      number-theory elementary-set-theory diophantine-equations linear-diophantine-equations







      share|cite|improve this question











      $endgroup$




      Consider the diophantine equation
      $$x_1+3x_2+5x_3 = n$$
      where $x_igeq 0$ and $ngeq 1.$
      Let $P_n(1,3,5)$ denote the number of solutions to this equation. I want to express $P_n(1,3,5)$ in terms of $P_{1}(1,3,5), P_{2}(1,3,5),cdots, P_{n-1}(1,3,5).$



      Here is what I observed:



      If $(x_1,x_2,x_3)$ is a solution to
      $$x_1+3x_2+5x_3 = k$$
      then $(x_1+1,x_2,x_3)$ is a solution to
      $$x_1+3x_2+5x_3 = k+1$$
      $(x_1,x_2+1,x_3)$ is a solution to
      $$x_1+3x_2+5x_3 = k+3$$
      and $(x_1,x_2,x_3+1)$ is a solution to
      $$x_1+3x_2+5x_3 = k+5.$$
      But I don't know know how to combine this to get the desired relation. Any ideas will be much appreciated.



      Edit: Based on the answer given below, we observe that a solution (x_1,x_2,x_3) to
      $$x_1+3x_2+5x_3 = n$$
      must have $x_1>0$ or $x_2>0$ or $x_3>0.$ If $x_1>0$ then (x_1-1,x_2,x_3) is a solution
      $$x_1+3x_2+5x_3 = n-1$$
      and there are $P_{n-1}(1,3,5).$ Proceeding in a similar manner for $x_2$ and $x_3$ and applying the inclusion-exclusion principle we get:
      $$P_{n}(1,3,5) = P_{n-1}(1,3,5)+P_{n-3}(1,3,5)+P_{n-5}(1,3,5)-P_{n-4}(1,3,5)-P_{n-8}(1,3,5)-P_{n-6}(1,3,5)+P_{n-9}(1,3,5).$$






      number-theory elementary-set-theory diophantine-equations linear-diophantine-equations




      number-theory elementary-set-theory diophantine-equations linear-diophantine-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 11 at 5:52







      Hello_World

















      asked Jan 11 at 5:20









      Hello_WorldHello_World

      4,12621731




      4,12621731






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Let's do a simpler example: $P_n(2,3)$, the number of solutions to $2x+3y=n$
          in nonnegative integers. For $n>0$ a solution must have either $x>0$ or $y>0$.
          A solution with $x>0$ means that $(x-1,y)$ is a solution to $2X+3Y=n-2$
          so there are $P_{n-2}(2,3)$ of these. Likewise there are $P_{n-3}(2,3)$
          solutions with $y>0$. But some solutions have $x>0$ and $y>0$, and there
          are $P_{n-5}(2,3)$ of these. By the inclusion/exclusion principle,
          $$P_n(2,3)=P_{n-2}(2,3)+P_{n-3}(2,3)-P_{n-5}(2,3).$$



          In your example, you'll have to do three-fold inclusion/exclusion...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I have two questions: when $x>0$ and $y>0$ then if $(x-1,y-1)$ is a solution to $2x+3y=n-5$ and that is why there are $P_{n-5}(2,3)$ of them, right? And second, I made an edit. Do you think that the recurrence is correct?
            $endgroup$
            – Hello_World
            Jan 11 at 5:54










          • $begingroup$
            Yes, and yes. Now, a better way to do all this is to use generating functions.... @Hello_World
            $endgroup$
            – Lord Shark the Unknown
            Jan 11 at 5:56










          • $begingroup$
            Yeah, I know for $P_n(2,3)$ we want the n'th coefficient of $$frac{1}{(1-x^2)(1-x^3)}.$$ But it's not clear what the power series looks like.
            $endgroup$
            – Hello_World
            Jan 11 at 6:00











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          Let's do a simpler example: $P_n(2,3)$, the number of solutions to $2x+3y=n$
          in nonnegative integers. For $n>0$ a solution must have either $x>0$ or $y>0$.
          A solution with $x>0$ means that $(x-1,y)$ is a solution to $2X+3Y=n-2$
          so there are $P_{n-2}(2,3)$ of these. Likewise there are $P_{n-3}(2,3)$
          solutions with $y>0$. But some solutions have $x>0$ and $y>0$, and there
          are $P_{n-5}(2,3)$ of these. By the inclusion/exclusion principle,
          $$P_n(2,3)=P_{n-2}(2,3)+P_{n-3}(2,3)-P_{n-5}(2,3).$$



          In your example, you'll have to do three-fold inclusion/exclusion...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I have two questions: when $x>0$ and $y>0$ then if $(x-1,y-1)$ is a solution to $2x+3y=n-5$ and that is why there are $P_{n-5}(2,3)$ of them, right? And second, I made an edit. Do you think that the recurrence is correct?
            $endgroup$
            – Hello_World
            Jan 11 at 5:54










          • $begingroup$
            Yes, and yes. Now, a better way to do all this is to use generating functions.... @Hello_World
            $endgroup$
            – Lord Shark the Unknown
            Jan 11 at 5:56










          • $begingroup$
            Yeah, I know for $P_n(2,3)$ we want the n'th coefficient of $$frac{1}{(1-x^2)(1-x^3)}.$$ But it's not clear what the power series looks like.
            $endgroup$
            – Hello_World
            Jan 11 at 6:00
















          1












          $begingroup$

          Let's do a simpler example: $P_n(2,3)$, the number of solutions to $2x+3y=n$
          in nonnegative integers. For $n>0$ a solution must have either $x>0$ or $y>0$.
          A solution with $x>0$ means that $(x-1,y)$ is a solution to $2X+3Y=n-2$
          so there are $P_{n-2}(2,3)$ of these. Likewise there are $P_{n-3}(2,3)$
          solutions with $y>0$. But some solutions have $x>0$ and $y>0$, and there
          are $P_{n-5}(2,3)$ of these. By the inclusion/exclusion principle,
          $$P_n(2,3)=P_{n-2}(2,3)+P_{n-3}(2,3)-P_{n-5}(2,3).$$



          In your example, you'll have to do three-fold inclusion/exclusion...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I have two questions: when $x>0$ and $y>0$ then if $(x-1,y-1)$ is a solution to $2x+3y=n-5$ and that is why there are $P_{n-5}(2,3)$ of them, right? And second, I made an edit. Do you think that the recurrence is correct?
            $endgroup$
            – Hello_World
            Jan 11 at 5:54










          • $begingroup$
            Yes, and yes. Now, a better way to do all this is to use generating functions.... @Hello_World
            $endgroup$
            – Lord Shark the Unknown
            Jan 11 at 5:56










          • $begingroup$
            Yeah, I know for $P_n(2,3)$ we want the n'th coefficient of $$frac{1}{(1-x^2)(1-x^3)}.$$ But it's not clear what the power series looks like.
            $endgroup$
            – Hello_World
            Jan 11 at 6:00














          1












          1








          1





          $begingroup$

          Let's do a simpler example: $P_n(2,3)$, the number of solutions to $2x+3y=n$
          in nonnegative integers. For $n>0$ a solution must have either $x>0$ or $y>0$.
          A solution with $x>0$ means that $(x-1,y)$ is a solution to $2X+3Y=n-2$
          so there are $P_{n-2}(2,3)$ of these. Likewise there are $P_{n-3}(2,3)$
          solutions with $y>0$. But some solutions have $x>0$ and $y>0$, and there
          are $P_{n-5}(2,3)$ of these. By the inclusion/exclusion principle,
          $$P_n(2,3)=P_{n-2}(2,3)+P_{n-3}(2,3)-P_{n-5}(2,3).$$



          In your example, you'll have to do three-fold inclusion/exclusion...






          share|cite|improve this answer









          $endgroup$



          Let's do a simpler example: $P_n(2,3)$, the number of solutions to $2x+3y=n$
          in nonnegative integers. For $n>0$ a solution must have either $x>0$ or $y>0$.
          A solution with $x>0$ means that $(x-1,y)$ is a solution to $2X+3Y=n-2$
          so there are $P_{n-2}(2,3)$ of these. Likewise there are $P_{n-3}(2,3)$
          solutions with $y>0$. But some solutions have $x>0$ and $y>0$, and there
          are $P_{n-5}(2,3)$ of these. By the inclusion/exclusion principle,
          $$P_n(2,3)=P_{n-2}(2,3)+P_{n-3}(2,3)-P_{n-5}(2,3).$$



          In your example, you'll have to do three-fold inclusion/exclusion...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 5:31









          Lord Shark the UnknownLord Shark the Unknown

          103k1160132




          103k1160132












          • $begingroup$
            I have two questions: when $x>0$ and $y>0$ then if $(x-1,y-1)$ is a solution to $2x+3y=n-5$ and that is why there are $P_{n-5}(2,3)$ of them, right? And second, I made an edit. Do you think that the recurrence is correct?
            $endgroup$
            – Hello_World
            Jan 11 at 5:54










          • $begingroup$
            Yes, and yes. Now, a better way to do all this is to use generating functions.... @Hello_World
            $endgroup$
            – Lord Shark the Unknown
            Jan 11 at 5:56










          • $begingroup$
            Yeah, I know for $P_n(2,3)$ we want the n'th coefficient of $$frac{1}{(1-x^2)(1-x^3)}.$$ But it's not clear what the power series looks like.
            $endgroup$
            – Hello_World
            Jan 11 at 6:00


















          • $begingroup$
            I have two questions: when $x>0$ and $y>0$ then if $(x-1,y-1)$ is a solution to $2x+3y=n-5$ and that is why there are $P_{n-5}(2,3)$ of them, right? And second, I made an edit. Do you think that the recurrence is correct?
            $endgroup$
            – Hello_World
            Jan 11 at 5:54










          • $begingroup$
            Yes, and yes. Now, a better way to do all this is to use generating functions.... @Hello_World
            $endgroup$
            – Lord Shark the Unknown
            Jan 11 at 5:56










          • $begingroup$
            Yeah, I know for $P_n(2,3)$ we want the n'th coefficient of $$frac{1}{(1-x^2)(1-x^3)}.$$ But it's not clear what the power series looks like.
            $endgroup$
            – Hello_World
            Jan 11 at 6:00
















          $begingroup$
          I have two questions: when $x>0$ and $y>0$ then if $(x-1,y-1)$ is a solution to $2x+3y=n-5$ and that is why there are $P_{n-5}(2,3)$ of them, right? And second, I made an edit. Do you think that the recurrence is correct?
          $endgroup$
          – Hello_World
          Jan 11 at 5:54




          $begingroup$
          I have two questions: when $x>0$ and $y>0$ then if $(x-1,y-1)$ is a solution to $2x+3y=n-5$ and that is why there are $P_{n-5}(2,3)$ of them, right? And second, I made an edit. Do you think that the recurrence is correct?
          $endgroup$
          – Hello_World
          Jan 11 at 5:54












          $begingroup$
          Yes, and yes. Now, a better way to do all this is to use generating functions.... @Hello_World
          $endgroup$
          – Lord Shark the Unknown
          Jan 11 at 5:56




          $begingroup$
          Yes, and yes. Now, a better way to do all this is to use generating functions.... @Hello_World
          $endgroup$
          – Lord Shark the Unknown
          Jan 11 at 5:56












          $begingroup$
          Yeah, I know for $P_n(2,3)$ we want the n'th coefficient of $$frac{1}{(1-x^2)(1-x^3)}.$$ But it's not clear what the power series looks like.
          $endgroup$
          – Hello_World
          Jan 11 at 6:00




          $begingroup$
          Yeah, I know for $P_n(2,3)$ we want the n'th coefficient of $$frac{1}{(1-x^2)(1-x^3)}.$$ But it's not clear what the power series looks like.
          $endgroup$
          – Hello_World
          Jan 11 at 6:00


















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