generating function problem












0












$begingroup$


I need to solve this problem using generating functions:
What is the generating function to the number of ways of expressing n dollars as a 1,2 and 5 dollar coins.



I wasn't able to solve the $ (1+x^2+x^4)^n $ part.






combinatorics generating-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is this the full and exact problem statement? It's not very clear to me
    $endgroup$
    – Yuriy S
    Dec 29 '18 at 15:45










  • $begingroup$
    Yes, this is the question
    $endgroup$
    – Igor
    Dec 29 '18 at 15:46








  • 1




    $begingroup$
    Maybe you mean number of ways of expressing $n$ dollars as a collection of $1$, $2$ and $5$ dollar coins?
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:58












  • $begingroup$
    Yes, It looked clear enough for me. I will edit it,
    $endgroup$
    – Igor
    Dec 29 '18 at 16:02
















0












$begingroup$


I need to solve this problem using generating functions:
What is the generating function to the number of ways of expressing n dollars as a 1,2 and 5 dollar coins.



I wasn't able to solve the $ (1+x^2+x^4)^n $ part.






combinatorics generating-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is this the full and exact problem statement? It's not very clear to me
    $endgroup$
    – Yuriy S
    Dec 29 '18 at 15:45










  • $begingroup$
    Yes, this is the question
    $endgroup$
    – Igor
    Dec 29 '18 at 15:46








  • 1




    $begingroup$
    Maybe you mean number of ways of expressing $n$ dollars as a collection of $1$, $2$ and $5$ dollar coins?
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:58












  • $begingroup$
    Yes, It looked clear enough for me. I will edit it,
    $endgroup$
    – Igor
    Dec 29 '18 at 16:02














0












0








0


0



$begingroup$


I need to solve this problem using generating functions:
What is the generating function to the number of ways of expressing n dollars as a 1,2 and 5 dollar coins.



I wasn't able to solve the $ (1+x^2+x^4)^n $ part.






combinatorics generating-functions







share|cite|improve this question











$endgroup$




I need to solve this problem using generating functions:
What is the generating function to the number of ways of expressing n dollars as a 1,2 and 5 dollar coins.



I wasn't able to solve the $ (1+x^2+x^4)^n $ part.






combinatorics generating-functions




combinatorics generating-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 16:03







Igor

















asked Dec 29 '18 at 15:35









IgorIgor

315




315












  • $begingroup$
    Is this the full and exact problem statement? It's not very clear to me
    $endgroup$
    – Yuriy S
    Dec 29 '18 at 15:45










  • $begingroup$
    Yes, this is the question
    $endgroup$
    – Igor
    Dec 29 '18 at 15:46








  • 1




    $begingroup$
    Maybe you mean number of ways of expressing $n$ dollars as a collection of $1$, $2$ and $5$ dollar coins?
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:58












  • $begingroup$
    Yes, It looked clear enough for me. I will edit it,
    $endgroup$
    – Igor
    Dec 29 '18 at 16:02


















  • $begingroup$
    Is this the full and exact problem statement? It's not very clear to me
    $endgroup$
    – Yuriy S
    Dec 29 '18 at 15:45










  • $begingroup$
    Yes, this is the question
    $endgroup$
    – Igor
    Dec 29 '18 at 15:46








  • 1




    $begingroup$
    Maybe you mean number of ways of expressing $n$ dollars as a collection of $1$, $2$ and $5$ dollar coins?
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:58












  • $begingroup$
    Yes, It looked clear enough for me. I will edit it,
    $endgroup$
    – Igor
    Dec 29 '18 at 16:02
















$begingroup$
Is this the full and exact problem statement? It's not very clear to me
$endgroup$
– Yuriy S
Dec 29 '18 at 15:45




$begingroup$
Is this the full and exact problem statement? It's not very clear to me
$endgroup$
– Yuriy S
Dec 29 '18 at 15:45












$begingroup$
Yes, this is the question
$endgroup$
– Igor
Dec 29 '18 at 15:46






$begingroup$
Yes, this is the question
$endgroup$
– Igor
Dec 29 '18 at 15:46






1




1




$begingroup$
Maybe you mean number of ways of expressing $n$ dollars as a collection of $1$, $2$ and $5$ dollar coins?
$endgroup$
– Jakobian
Dec 29 '18 at 15:58






$begingroup$
Maybe you mean number of ways of expressing $n$ dollars as a collection of $1$, $2$ and $5$ dollar coins?
$endgroup$
– Jakobian
Dec 29 '18 at 15:58














$begingroup$
Yes, It looked clear enough for me. I will edit it,
$endgroup$
– Igor
Dec 29 '18 at 16:02




$begingroup$
Yes, It looked clear enough for me. I will edit it,
$endgroup$
– Igor
Dec 29 '18 at 16:02










2 Answers
2






active

oldest

votes


















0












$begingroup$

The generating function for coins of value $k$ is
$dfrac1{1-x^k}$.



For different value coins, multiply their generating functions.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We're looking for
    $$F(x) = sum_{n=0}^infty a_nx^n $$
    where $a_n$ is the number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins.
    $$a_n = sum_{k_1+k_2+k_3 = n} m_{k_1, 1}m_{k_2, 2}m_{k_3, 5} $$
    where $m_{k_1, 1}$ is the number of ways to express $k_1$ dollars in terms of $1$ dollar coins, $m_{k_2, 2}$ is the number of ways of expressing $k_2$ dollars in terms of $2$ dollar coins, and $m_{k_3, 5}$ is the number of ways of expressing $k_3$ dollars in terms of $5$ dollar coins. From this:
    $$F(x) = left(sum_{n=0}^infty m_{n, 1}x^nright)left(sum_{n=0}^infty m_{n, 2}x^nright)left(sum_{n=0}^infty m_{n, 5}x^nright) = left(sum_{n=0}^infty x^nright)left(sum_{n=0}^infty x^{2n}right)left(sum_{n=0}^infty x^{5n}right)=\ = frac{1}{1-x}cdot frac{1}{1-x^2}cdot frac{1}{1-x^5} $$
    If the order of the coins is important, then we can derive a recurrence:
    $$b_{n+5} = b_{n+4}+b_{n+3}+b_n, ngeq 0 \ b_0 = 0, b_1 = 1, b_2 = 2, b_3 = 3, b_4 = 5$$
    It's because there is $3$ possible cases, we have $1, 2$ or $5$ dollar coin at the end, and we take as much dollars from the whole if that's the case. Then we have
    $$G(x) = sumlimits_{n=0}^infty b_nx^n = x+2x^2+3x^3+5x^4+sumlimits_{n=0}^infty b_{n+5}x^{n+5} = \ = p(x)+sum_{n=0}^infty (b_{n+4}+b_{n+3}+b_n)x^{n+5} = p(x)+sum_{n=0}^infty b_nx^{n+5} + sum_{n=0}^infty b_{n+3}x^{n+5} + sum_{n=0}^infty b_{n+4}x^{n+5} $$
    where $p(x) = x+2x^2+3x^3+5x^4$. Then we can simplify the $3$ sums
    $$sum_{n=0}^infty b_nx^{n+5} = x^5sum_{n=0}^infty b_nx^n = x^5G(x) $$
    $$sum_{n=0}^infty b_{n+3}x^{n+5} = x^2sum_{n=0}^infty b_{n+3}x^{n+3} = x^2sum_{n=3}^infty b_nx^n = x^2(sum_{n=0}^infty b_nx^n - sum_{n=0}^2 b_nx^n)= \ = x^2(G(x)-2x^2-x) = x^2G(x) - 2x^4-x^3 $$
    $$sum_{n=0}^infty b_{n+4}x^{n+5} = xsum_{n=0}^infty b_{n+4}x^{n+4} = xsum_{n=4}^infty b_nx^n = x(sum_{n=0}^infty b_nx^n - sum_{n=0}^3 b_nx^n) = \ = x(G(x)-3x^3-2x^2-x) = xG(x)-3x^4-2x^3-x^2 $$
    So we can now calulate what $G(x)$ is
    $$G(x) = p(x)+(x^5+x^2+x)G(x)-5x^4-3x^3-x^2 = (x^5+x^2+x)G(x)+x+x^2 $$
    $$G(x) = frac{x+x^2}{1-(x+x^2+x^5)} $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Follow up question - What if I the order of the coins is important. How will it change?
      $endgroup$
      – Igor
      Dec 30 '18 at 11:33








    • 1




      $begingroup$
      @Igor I've edited my answer
      $endgroup$
      – Jakobian
      Dec 30 '18 at 12:37












    • $begingroup$
      I didn't quiet understand the transition. Why is G(X) defined this way and what is x+2x^2+3x^3+5x^4? I lost you there..
      $endgroup$
      – Igor
      Dec 30 '18 at 15:14












    • $begingroup$
      @Igor If $b_n$ is number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins, so that the order of coins matters, then from definition, it's generating function is $sum_{n=0}^infty b_nx^n$. This is just the definition. $x+2x^2+3x^3+5x^4 = sum_{n=0}^4 b_nx^n$, we split the sum in two parts.
      $endgroup$
      – Jakobian
      Dec 30 '18 at 17:08












    • $begingroup$
      Why can't you say this is the answer? $$ frac{1}{(1-x^2-x^3-x^5)} $$ I am just summing all the possibilities for choosing coins.
      $endgroup$
      – Igor
      Jan 1 at 17:36











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The generating function for coins of value $k$ is
    $dfrac1{1-x^k}$.



    For different value coins, multiply their generating functions.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The generating function for coins of value $k$ is
      $dfrac1{1-x^k}$.



      For different value coins, multiply their generating functions.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The generating function for coins of value $k$ is
        $dfrac1{1-x^k}$.



        For different value coins, multiply their generating functions.






        share|cite|improve this answer









        $endgroup$



        The generating function for coins of value $k$ is
        $dfrac1{1-x^k}$.



        For different value coins, multiply their generating functions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 16:14









        marty cohenmarty cohen

        72.9k549128




        72.9k549128























            0












            $begingroup$

            We're looking for
            $$F(x) = sum_{n=0}^infty a_nx^n $$
            where $a_n$ is the number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins.
            $$a_n = sum_{k_1+k_2+k_3 = n} m_{k_1, 1}m_{k_2, 2}m_{k_3, 5} $$
            where $m_{k_1, 1}$ is the number of ways to express $k_1$ dollars in terms of $1$ dollar coins, $m_{k_2, 2}$ is the number of ways of expressing $k_2$ dollars in terms of $2$ dollar coins, and $m_{k_3, 5}$ is the number of ways of expressing $k_3$ dollars in terms of $5$ dollar coins. From this:
            $$F(x) = left(sum_{n=0}^infty m_{n, 1}x^nright)left(sum_{n=0}^infty m_{n, 2}x^nright)left(sum_{n=0}^infty m_{n, 5}x^nright) = left(sum_{n=0}^infty x^nright)left(sum_{n=0}^infty x^{2n}right)left(sum_{n=0}^infty x^{5n}right)=\ = frac{1}{1-x}cdot frac{1}{1-x^2}cdot frac{1}{1-x^5} $$
            If the order of the coins is important, then we can derive a recurrence:
            $$b_{n+5} = b_{n+4}+b_{n+3}+b_n, ngeq 0 \ b_0 = 0, b_1 = 1, b_2 = 2, b_3 = 3, b_4 = 5$$
            It's because there is $3$ possible cases, we have $1, 2$ or $5$ dollar coin at the end, and we take as much dollars from the whole if that's the case. Then we have
            $$G(x) = sumlimits_{n=0}^infty b_nx^n = x+2x^2+3x^3+5x^4+sumlimits_{n=0}^infty b_{n+5}x^{n+5} = \ = p(x)+sum_{n=0}^infty (b_{n+4}+b_{n+3}+b_n)x^{n+5} = p(x)+sum_{n=0}^infty b_nx^{n+5} + sum_{n=0}^infty b_{n+3}x^{n+5} + sum_{n=0}^infty b_{n+4}x^{n+5} $$
            where $p(x) = x+2x^2+3x^3+5x^4$. Then we can simplify the $3$ sums
            $$sum_{n=0}^infty b_nx^{n+5} = x^5sum_{n=0}^infty b_nx^n = x^5G(x) $$
            $$sum_{n=0}^infty b_{n+3}x^{n+5} = x^2sum_{n=0}^infty b_{n+3}x^{n+3} = x^2sum_{n=3}^infty b_nx^n = x^2(sum_{n=0}^infty b_nx^n - sum_{n=0}^2 b_nx^n)= \ = x^2(G(x)-2x^2-x) = x^2G(x) - 2x^4-x^3 $$
            $$sum_{n=0}^infty b_{n+4}x^{n+5} = xsum_{n=0}^infty b_{n+4}x^{n+4} = xsum_{n=4}^infty b_nx^n = x(sum_{n=0}^infty b_nx^n - sum_{n=0}^3 b_nx^n) = \ = x(G(x)-3x^3-2x^2-x) = xG(x)-3x^4-2x^3-x^2 $$
            So we can now calulate what $G(x)$ is
            $$G(x) = p(x)+(x^5+x^2+x)G(x)-5x^4-3x^3-x^2 = (x^5+x^2+x)G(x)+x+x^2 $$
            $$G(x) = frac{x+x^2}{1-(x+x^2+x^5)} $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Follow up question - What if I the order of the coins is important. How will it change?
              $endgroup$
              – Igor
              Dec 30 '18 at 11:33








            • 1




              $begingroup$
              @Igor I've edited my answer
              $endgroup$
              – Jakobian
              Dec 30 '18 at 12:37












            • $begingroup$
              I didn't quiet understand the transition. Why is G(X) defined this way and what is x+2x^2+3x^3+5x^4? I lost you there..
              $endgroup$
              – Igor
              Dec 30 '18 at 15:14












            • $begingroup$
              @Igor If $b_n$ is number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins, so that the order of coins matters, then from definition, it's generating function is $sum_{n=0}^infty b_nx^n$. This is just the definition. $x+2x^2+3x^3+5x^4 = sum_{n=0}^4 b_nx^n$, we split the sum in two parts.
              $endgroup$
              – Jakobian
              Dec 30 '18 at 17:08












            • $begingroup$
              Why can't you say this is the answer? $$ frac{1}{(1-x^2-x^3-x^5)} $$ I am just summing all the possibilities for choosing coins.
              $endgroup$
              – Igor
              Jan 1 at 17:36
















            0












            $begingroup$

            We're looking for
            $$F(x) = sum_{n=0}^infty a_nx^n $$
            where $a_n$ is the number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins.
            $$a_n = sum_{k_1+k_2+k_3 = n} m_{k_1, 1}m_{k_2, 2}m_{k_3, 5} $$
            where $m_{k_1, 1}$ is the number of ways to express $k_1$ dollars in terms of $1$ dollar coins, $m_{k_2, 2}$ is the number of ways of expressing $k_2$ dollars in terms of $2$ dollar coins, and $m_{k_3, 5}$ is the number of ways of expressing $k_3$ dollars in terms of $5$ dollar coins. From this:
            $$F(x) = left(sum_{n=0}^infty m_{n, 1}x^nright)left(sum_{n=0}^infty m_{n, 2}x^nright)left(sum_{n=0}^infty m_{n, 5}x^nright) = left(sum_{n=0}^infty x^nright)left(sum_{n=0}^infty x^{2n}right)left(sum_{n=0}^infty x^{5n}right)=\ = frac{1}{1-x}cdot frac{1}{1-x^2}cdot frac{1}{1-x^5} $$
            If the order of the coins is important, then we can derive a recurrence:
            $$b_{n+5} = b_{n+4}+b_{n+3}+b_n, ngeq 0 \ b_0 = 0, b_1 = 1, b_2 = 2, b_3 = 3, b_4 = 5$$
            It's because there is $3$ possible cases, we have $1, 2$ or $5$ dollar coin at the end, and we take as much dollars from the whole if that's the case. Then we have
            $$G(x) = sumlimits_{n=0}^infty b_nx^n = x+2x^2+3x^3+5x^4+sumlimits_{n=0}^infty b_{n+5}x^{n+5} = \ = p(x)+sum_{n=0}^infty (b_{n+4}+b_{n+3}+b_n)x^{n+5} = p(x)+sum_{n=0}^infty b_nx^{n+5} + sum_{n=0}^infty b_{n+3}x^{n+5} + sum_{n=0}^infty b_{n+4}x^{n+5} $$
            where $p(x) = x+2x^2+3x^3+5x^4$. Then we can simplify the $3$ sums
            $$sum_{n=0}^infty b_nx^{n+5} = x^5sum_{n=0}^infty b_nx^n = x^5G(x) $$
            $$sum_{n=0}^infty b_{n+3}x^{n+5} = x^2sum_{n=0}^infty b_{n+3}x^{n+3} = x^2sum_{n=3}^infty b_nx^n = x^2(sum_{n=0}^infty b_nx^n - sum_{n=0}^2 b_nx^n)= \ = x^2(G(x)-2x^2-x) = x^2G(x) - 2x^4-x^3 $$
            $$sum_{n=0}^infty b_{n+4}x^{n+5} = xsum_{n=0}^infty b_{n+4}x^{n+4} = xsum_{n=4}^infty b_nx^n = x(sum_{n=0}^infty b_nx^n - sum_{n=0}^3 b_nx^n) = \ = x(G(x)-3x^3-2x^2-x) = xG(x)-3x^4-2x^3-x^2 $$
            So we can now calulate what $G(x)$ is
            $$G(x) = p(x)+(x^5+x^2+x)G(x)-5x^4-3x^3-x^2 = (x^5+x^2+x)G(x)+x+x^2 $$
            $$G(x) = frac{x+x^2}{1-(x+x^2+x^5)} $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Follow up question - What if I the order of the coins is important. How will it change?
              $endgroup$
              – Igor
              Dec 30 '18 at 11:33








            • 1




              $begingroup$
              @Igor I've edited my answer
              $endgroup$
              – Jakobian
              Dec 30 '18 at 12:37












            • $begingroup$
              I didn't quiet understand the transition. Why is G(X) defined this way and what is x+2x^2+3x^3+5x^4? I lost you there..
              $endgroup$
              – Igor
              Dec 30 '18 at 15:14












            • $begingroup$
              @Igor If $b_n$ is number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins, so that the order of coins matters, then from definition, it's generating function is $sum_{n=0}^infty b_nx^n$. This is just the definition. $x+2x^2+3x^3+5x^4 = sum_{n=0}^4 b_nx^n$, we split the sum in two parts.
              $endgroup$
              – Jakobian
              Dec 30 '18 at 17:08












            • $begingroup$
              Why can't you say this is the answer? $$ frac{1}{(1-x^2-x^3-x^5)} $$ I am just summing all the possibilities for choosing coins.
              $endgroup$
              – Igor
              Jan 1 at 17:36














            0












            0








            0





            $begingroup$

            We're looking for
            $$F(x) = sum_{n=0}^infty a_nx^n $$
            where $a_n$ is the number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins.
            $$a_n = sum_{k_1+k_2+k_3 = n} m_{k_1, 1}m_{k_2, 2}m_{k_3, 5} $$
            where $m_{k_1, 1}$ is the number of ways to express $k_1$ dollars in terms of $1$ dollar coins, $m_{k_2, 2}$ is the number of ways of expressing $k_2$ dollars in terms of $2$ dollar coins, and $m_{k_3, 5}$ is the number of ways of expressing $k_3$ dollars in terms of $5$ dollar coins. From this:
            $$F(x) = left(sum_{n=0}^infty m_{n, 1}x^nright)left(sum_{n=0}^infty m_{n, 2}x^nright)left(sum_{n=0}^infty m_{n, 5}x^nright) = left(sum_{n=0}^infty x^nright)left(sum_{n=0}^infty x^{2n}right)left(sum_{n=0}^infty x^{5n}right)=\ = frac{1}{1-x}cdot frac{1}{1-x^2}cdot frac{1}{1-x^5} $$
            If the order of the coins is important, then we can derive a recurrence:
            $$b_{n+5} = b_{n+4}+b_{n+3}+b_n, ngeq 0 \ b_0 = 0, b_1 = 1, b_2 = 2, b_3 = 3, b_4 = 5$$
            It's because there is $3$ possible cases, we have $1, 2$ or $5$ dollar coin at the end, and we take as much dollars from the whole if that's the case. Then we have
            $$G(x) = sumlimits_{n=0}^infty b_nx^n = x+2x^2+3x^3+5x^4+sumlimits_{n=0}^infty b_{n+5}x^{n+5} = \ = p(x)+sum_{n=0}^infty (b_{n+4}+b_{n+3}+b_n)x^{n+5} = p(x)+sum_{n=0}^infty b_nx^{n+5} + sum_{n=0}^infty b_{n+3}x^{n+5} + sum_{n=0}^infty b_{n+4}x^{n+5} $$
            where $p(x) = x+2x^2+3x^3+5x^4$. Then we can simplify the $3$ sums
            $$sum_{n=0}^infty b_nx^{n+5} = x^5sum_{n=0}^infty b_nx^n = x^5G(x) $$
            $$sum_{n=0}^infty b_{n+3}x^{n+5} = x^2sum_{n=0}^infty b_{n+3}x^{n+3} = x^2sum_{n=3}^infty b_nx^n = x^2(sum_{n=0}^infty b_nx^n - sum_{n=0}^2 b_nx^n)= \ = x^2(G(x)-2x^2-x) = x^2G(x) - 2x^4-x^3 $$
            $$sum_{n=0}^infty b_{n+4}x^{n+5} = xsum_{n=0}^infty b_{n+4}x^{n+4} = xsum_{n=4}^infty b_nx^n = x(sum_{n=0}^infty b_nx^n - sum_{n=0}^3 b_nx^n) = \ = x(G(x)-3x^3-2x^2-x) = xG(x)-3x^4-2x^3-x^2 $$
            So we can now calulate what $G(x)$ is
            $$G(x) = p(x)+(x^5+x^2+x)G(x)-5x^4-3x^3-x^2 = (x^5+x^2+x)G(x)+x+x^2 $$
            $$G(x) = frac{x+x^2}{1-(x+x^2+x^5)} $$






            share|cite|improve this answer











            $endgroup$



            We're looking for
            $$F(x) = sum_{n=0}^infty a_nx^n $$
            where $a_n$ is the number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins.
            $$a_n = sum_{k_1+k_2+k_3 = n} m_{k_1, 1}m_{k_2, 2}m_{k_3, 5} $$
            where $m_{k_1, 1}$ is the number of ways to express $k_1$ dollars in terms of $1$ dollar coins, $m_{k_2, 2}$ is the number of ways of expressing $k_2$ dollars in terms of $2$ dollar coins, and $m_{k_3, 5}$ is the number of ways of expressing $k_3$ dollars in terms of $5$ dollar coins. From this:
            $$F(x) = left(sum_{n=0}^infty m_{n, 1}x^nright)left(sum_{n=0}^infty m_{n, 2}x^nright)left(sum_{n=0}^infty m_{n, 5}x^nright) = left(sum_{n=0}^infty x^nright)left(sum_{n=0}^infty x^{2n}right)left(sum_{n=0}^infty x^{5n}right)=\ = frac{1}{1-x}cdot frac{1}{1-x^2}cdot frac{1}{1-x^5} $$
            If the order of the coins is important, then we can derive a recurrence:
            $$b_{n+5} = b_{n+4}+b_{n+3}+b_n, ngeq 0 \ b_0 = 0, b_1 = 1, b_2 = 2, b_3 = 3, b_4 = 5$$
            It's because there is $3$ possible cases, we have $1, 2$ or $5$ dollar coin at the end, and we take as much dollars from the whole if that's the case. Then we have
            $$G(x) = sumlimits_{n=0}^infty b_nx^n = x+2x^2+3x^3+5x^4+sumlimits_{n=0}^infty b_{n+5}x^{n+5} = \ = p(x)+sum_{n=0}^infty (b_{n+4}+b_{n+3}+b_n)x^{n+5} = p(x)+sum_{n=0}^infty b_nx^{n+5} + sum_{n=0}^infty b_{n+3}x^{n+5} + sum_{n=0}^infty b_{n+4}x^{n+5} $$
            where $p(x) = x+2x^2+3x^3+5x^4$. Then we can simplify the $3$ sums
            $$sum_{n=0}^infty b_nx^{n+5} = x^5sum_{n=0}^infty b_nx^n = x^5G(x) $$
            $$sum_{n=0}^infty b_{n+3}x^{n+5} = x^2sum_{n=0}^infty b_{n+3}x^{n+3} = x^2sum_{n=3}^infty b_nx^n = x^2(sum_{n=0}^infty b_nx^n - sum_{n=0}^2 b_nx^n)= \ = x^2(G(x)-2x^2-x) = x^2G(x) - 2x^4-x^3 $$
            $$sum_{n=0}^infty b_{n+4}x^{n+5} = xsum_{n=0}^infty b_{n+4}x^{n+4} = xsum_{n=4}^infty b_nx^n = x(sum_{n=0}^infty b_nx^n - sum_{n=0}^3 b_nx^n) = \ = x(G(x)-3x^3-2x^2-x) = xG(x)-3x^4-2x^3-x^2 $$
            So we can now calulate what $G(x)$ is
            $$G(x) = p(x)+(x^5+x^2+x)G(x)-5x^4-3x^3-x^2 = (x^5+x^2+x)G(x)+x+x^2 $$
            $$G(x) = frac{x+x^2}{1-(x+x^2+x^5)} $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 30 '18 at 12:36

























            answered Dec 29 '18 at 16:11









            JakobianJakobian

            2,493720




            2,493720












            • $begingroup$
              Follow up question - What if I the order of the coins is important. How will it change?
              $endgroup$
              – Igor
              Dec 30 '18 at 11:33








            • 1




              $begingroup$
              @Igor I've edited my answer
              $endgroup$
              – Jakobian
              Dec 30 '18 at 12:37












            • $begingroup$
              I didn't quiet understand the transition. Why is G(X) defined this way and what is x+2x^2+3x^3+5x^4? I lost you there..
              $endgroup$
              – Igor
              Dec 30 '18 at 15:14












            • $begingroup$
              @Igor If $b_n$ is number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins, so that the order of coins matters, then from definition, it's generating function is $sum_{n=0}^infty b_nx^n$. This is just the definition. $x+2x^2+3x^3+5x^4 = sum_{n=0}^4 b_nx^n$, we split the sum in two parts.
              $endgroup$
              – Jakobian
              Dec 30 '18 at 17:08












            • $begingroup$
              Why can't you say this is the answer? $$ frac{1}{(1-x^2-x^3-x^5)} $$ I am just summing all the possibilities for choosing coins.
              $endgroup$
              – Igor
              Jan 1 at 17:36


















            • $begingroup$
              Follow up question - What if I the order of the coins is important. How will it change?
              $endgroup$
              – Igor
              Dec 30 '18 at 11:33








            • 1




              $begingroup$
              @Igor I've edited my answer
              $endgroup$
              – Jakobian
              Dec 30 '18 at 12:37












            • $begingroup$
              I didn't quiet understand the transition. Why is G(X) defined this way and what is x+2x^2+3x^3+5x^4? I lost you there..
              $endgroup$
              – Igor
              Dec 30 '18 at 15:14












            • $begingroup$
              @Igor If $b_n$ is number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins, so that the order of coins matters, then from definition, it's generating function is $sum_{n=0}^infty b_nx^n$. This is just the definition. $x+2x^2+3x^3+5x^4 = sum_{n=0}^4 b_nx^n$, we split the sum in two parts.
              $endgroup$
              – Jakobian
              Dec 30 '18 at 17:08












            • $begingroup$
              Why can't you say this is the answer? $$ frac{1}{(1-x^2-x^3-x^5)} $$ I am just summing all the possibilities for choosing coins.
              $endgroup$
              – Igor
              Jan 1 at 17:36
















            $begingroup$
            Follow up question - What if I the order of the coins is important. How will it change?
            $endgroup$
            – Igor
            Dec 30 '18 at 11:33






            $begingroup$
            Follow up question - What if I the order of the coins is important. How will it change?
            $endgroup$
            – Igor
            Dec 30 '18 at 11:33






            1




            1




            $begingroup$
            @Igor I've edited my answer
            $endgroup$
            – Jakobian
            Dec 30 '18 at 12:37






            $begingroup$
            @Igor I've edited my answer
            $endgroup$
            – Jakobian
            Dec 30 '18 at 12:37














            $begingroup$
            I didn't quiet understand the transition. Why is G(X) defined this way and what is x+2x^2+3x^3+5x^4? I lost you there..
            $endgroup$
            – Igor
            Dec 30 '18 at 15:14






            $begingroup$
            I didn't quiet understand the transition. Why is G(X) defined this way and what is x+2x^2+3x^3+5x^4? I lost you there..
            $endgroup$
            – Igor
            Dec 30 '18 at 15:14














            $begingroup$
            @Igor If $b_n$ is number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins, so that the order of coins matters, then from definition, it's generating function is $sum_{n=0}^infty b_nx^n$. This is just the definition. $x+2x^2+3x^3+5x^4 = sum_{n=0}^4 b_nx^n$, we split the sum in two parts.
            $endgroup$
            – Jakobian
            Dec 30 '18 at 17:08






            $begingroup$
            @Igor If $b_n$ is number of ways of expressing $n$ dollars as a collection of $1, 2, 5$ dollar coins, so that the order of coins matters, then from definition, it's generating function is $sum_{n=0}^infty b_nx^n$. This is just the definition. $x+2x^2+3x^3+5x^4 = sum_{n=0}^4 b_nx^n$, we split the sum in two parts.
            $endgroup$
            – Jakobian
            Dec 30 '18 at 17:08














            $begingroup$
            Why can't you say this is the answer? $$ frac{1}{(1-x^2-x^3-x^5)} $$ I am just summing all the possibilities for choosing coins.
            $endgroup$
            – Igor
            Jan 1 at 17:36




            $begingroup$
            Why can't you say this is the answer? $$ frac{1}{(1-x^2-x^3-x^5)} $$ I am just summing all the possibilities for choosing coins.
            $endgroup$
            – Igor
            Jan 1 at 17:36


















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