Solve for the generating function with $x^1, x^5, x^{10}, x^{20}, x^{50}, x^{100}, x^{500}$












0












$begingroup$


How would you find the coefficient of $x^{2000}$ in



$ = (x^0 + x + x^2 +...)(x^0 + x^5 + x^{10} +...)(x^0 + x^{10} + x^{20}+...)(x^0 + x^{20} + x^{40} + ...)(x^0 + \
x^{50} + x^{100} +...)(x^0 + x^{100} + x^{200} +...)(x^0 + x^{500} + x^{1000} +...)$



I've been trying to use Mathematica, but it's not giving me a solution (very new at using it).






combinatorics generating-functions







share|cite|improve this question









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  • 4




    $begingroup$
    You need to find non-negative integer solutions to $$a+5b+10c+20d+50e+100f+500g = 2000$$
    $endgroup$
    – Fly by Night
    Dec 18 '18 at 21:35












  • $begingroup$
    @FlybyNight could you explain how to do that?
    $endgroup$
    – Math Newbie
    Dec 18 '18 at 21:39












  • $begingroup$
    You can ask WA to compute the number using command SeriesCoefficient[ 1/((1-x)*(1-x^5)*(1-x^10)*(1-x^20)*(1-x^50)*(1-x^100)*(1-x^500)),{x,0,2000}] (same command should work on Mathematica), the number you want is 86950230.
    $endgroup$
    – achille hui
    Dec 18 '18 at 23:52












  • $begingroup$
    @MathNewbie No, sorry. This is related to "partitions" and that is a very difficult combinatorial problem. en.wikipedia.org/wiki/Partition_(number_theory)
    $endgroup$
    – Fly by Night
    Dec 19 '18 at 0:46


















0












$begingroup$


How would you find the coefficient of $x^{2000}$ in



$ = (x^0 + x + x^2 +...)(x^0 + x^5 + x^{10} +...)(x^0 + x^{10} + x^{20}+...)(x^0 + x^{20} + x^{40} + ...)(x^0 + \
x^{50} + x^{100} +...)(x^0 + x^{100} + x^{200} +...)(x^0 + x^{500} + x^{1000} +...)$



I've been trying to use Mathematica, but it's not giving me a solution (very new at using it).






combinatorics generating-functions







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    You need to find non-negative integer solutions to $$a+5b+10c+20d+50e+100f+500g = 2000$$
    $endgroup$
    – Fly by Night
    Dec 18 '18 at 21:35












  • $begingroup$
    @FlybyNight could you explain how to do that?
    $endgroup$
    – Math Newbie
    Dec 18 '18 at 21:39












  • $begingroup$
    You can ask WA to compute the number using command SeriesCoefficient[ 1/((1-x)*(1-x^5)*(1-x^10)*(1-x^20)*(1-x^50)*(1-x^100)*(1-x^500)),{x,0,2000}] (same command should work on Mathematica), the number you want is 86950230.
    $endgroup$
    – achille hui
    Dec 18 '18 at 23:52












  • $begingroup$
    @MathNewbie No, sorry. This is related to "partitions" and that is a very difficult combinatorial problem. en.wikipedia.org/wiki/Partition_(number_theory)
    $endgroup$
    – Fly by Night
    Dec 19 '18 at 0:46
















0












0








0


0



$begingroup$


How would you find the coefficient of $x^{2000}$ in



$ = (x^0 + x + x^2 +...)(x^0 + x^5 + x^{10} +...)(x^0 + x^{10} + x^{20}+...)(x^0 + x^{20} + x^{40} + ...)(x^0 + \
x^{50} + x^{100} +...)(x^0 + x^{100} + x^{200} +...)(x^0 + x^{500} + x^{1000} +...)$



I've been trying to use Mathematica, but it's not giving me a solution (very new at using it).






combinatorics generating-functions







share|cite|improve this question









$endgroup$




How would you find the coefficient of $x^{2000}$ in



$ = (x^0 + x + x^2 +...)(x^0 + x^5 + x^{10} +...)(x^0 + x^{10} + x^{20}+...)(x^0 + x^{20} + x^{40} + ...)(x^0 + \
x^{50} + x^{100} +...)(x^0 + x^{100} + x^{200} +...)(x^0 + x^{500} + x^{1000} +...)$



I've been trying to use Mathematica, but it's not giving me a solution (very new at using it).






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










asked Dec 18 '18 at 21:30









Math NewbieMath Newbie

428




428








  • 4




    $begingroup$
    You need to find non-negative integer solutions to $$a+5b+10c+20d+50e+100f+500g = 2000$$
    $endgroup$
    – Fly by Night
    Dec 18 '18 at 21:35












  • $begingroup$
    @FlybyNight could you explain how to do that?
    $endgroup$
    – Math Newbie
    Dec 18 '18 at 21:39












  • $begingroup$
    You can ask WA to compute the number using command SeriesCoefficient[ 1/((1-x)*(1-x^5)*(1-x^10)*(1-x^20)*(1-x^50)*(1-x^100)*(1-x^500)),{x,0,2000}] (same command should work on Mathematica), the number you want is 86950230.
    $endgroup$
    – achille hui
    Dec 18 '18 at 23:52












  • $begingroup$
    @MathNewbie No, sorry. This is related to "partitions" and that is a very difficult combinatorial problem. en.wikipedia.org/wiki/Partition_(number_theory)
    $endgroup$
    – Fly by Night
    Dec 19 '18 at 0:46
















  • 4




    $begingroup$
    You need to find non-negative integer solutions to $$a+5b+10c+20d+50e+100f+500g = 2000$$
    $endgroup$
    – Fly by Night
    Dec 18 '18 at 21:35












  • $begingroup$
    @FlybyNight could you explain how to do that?
    $endgroup$
    – Math Newbie
    Dec 18 '18 at 21:39












  • $begingroup$
    You can ask WA to compute the number using command SeriesCoefficient[ 1/((1-x)*(1-x^5)*(1-x^10)*(1-x^20)*(1-x^50)*(1-x^100)*(1-x^500)),{x,0,2000}] (same command should work on Mathematica), the number you want is 86950230.
    $endgroup$
    – achille hui
    Dec 18 '18 at 23:52












  • $begingroup$
    @MathNewbie No, sorry. This is related to "partitions" and that is a very difficult combinatorial problem. en.wikipedia.org/wiki/Partition_(number_theory)
    $endgroup$
    – Fly by Night
    Dec 19 '18 at 0:46










4




4




$begingroup$
You need to find non-negative integer solutions to $$a+5b+10c+20d+50e+100f+500g = 2000$$
$endgroup$
– Fly by Night
Dec 18 '18 at 21:35






$begingroup$
You need to find non-negative integer solutions to $$a+5b+10c+20d+50e+100f+500g = 2000$$
$endgroup$
– Fly by Night
Dec 18 '18 at 21:35














$begingroup$
@FlybyNight could you explain how to do that?
$endgroup$
– Math Newbie
Dec 18 '18 at 21:39






$begingroup$
@FlybyNight could you explain how to do that?
$endgroup$
– Math Newbie
Dec 18 '18 at 21:39














$begingroup$
You can ask WA to compute the number using command SeriesCoefficient[ 1/((1-x)*(1-x^5)*(1-x^10)*(1-x^20)*(1-x^50)*(1-x^100)*(1-x^500)),{x,0,2000}] (same command should work on Mathematica), the number you want is 86950230.
$endgroup$
– achille hui
Dec 18 '18 at 23:52






$begingroup$
You can ask WA to compute the number using command SeriesCoefficient[ 1/((1-x)*(1-x^5)*(1-x^10)*(1-x^20)*(1-x^50)*(1-x^100)*(1-x^500)),{x,0,2000}] (same command should work on Mathematica), the number you want is 86950230.
$endgroup$
– achille hui
Dec 18 '18 at 23:52














$begingroup$
@MathNewbie No, sorry. This is related to "partitions" and that is a very difficult combinatorial problem. en.wikipedia.org/wiki/Partition_(number_theory)
$endgroup$
– Fly by Night
Dec 19 '18 at 0:46






$begingroup$
@MathNewbie No, sorry. This is related to "partitions" and that is a very difficult combinatorial problem. en.wikipedia.org/wiki/Partition_(number_theory)
$endgroup$
– Fly by Night
Dec 19 '18 at 0:46












1 Answer
1






active

oldest

votes


















2












$begingroup$

So you have to find the number of non-negative integer solutions to:



$$a+5b+10c+20d+50e+100f+500g = 2000$$



A short C++ program with no more than 20 lines of code and a simple recurrence will serve for the purpose. 



#include <iostream>



using namespace std;



int numSolutions(int *list, int size, int sum) {

    int num = list[0];

    if(size == 1) {

        return (sum % num == 0)? 1: 0;

    }

    int count = 0, cases = sum / num;

    for(int i = 0; i <= cases; i++) {

        count += numSolutions(list + 1, size - 1, sum - i * num);

    }

    return count;

}



int main() {

    int list = {500, 100, 50, 20, 10, 5, 1};

    int size = 7, sum = 2000;

    cout << numSolutions(list, size, sum);

}


...and the answer is: 86950230. Execution time is less than a second.






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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    So you have to find the number of non-negative integer solutions to:



    $$a+5b+10c+20d+50e+100f+500g = 2000$$



    A short C++ program with no more than 20 lines of code and a simple recurrence will serve for the purpose. 



    #include <iostream>
    

    
using namespace std;
    

    
int numSolutions(int *list, int size, int sum) {
    
    int num = list[0];
    
    if(size == 1) {
    
        return (sum % num == 0)? 1: 0;
    
    }
    
    int count = 0, cases = sum / num;
    
    for(int i = 0; i <= cases; i++) {
    
        count += numSolutions(list + 1, size - 1, sum - i * num);
    
    }
    
    return count;
    
}
    

    
int main() {
    
    int list = {500, 100, 50, 20, 10, 5, 1};
    
    int size = 7, sum = 2000;
    
    cout << numSolutions(list, size, sum);
    
}


    ...and the answer is: 86950230. Execution time is less than a second.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      So you have to find the number of non-negative integer solutions to:



      $$a+5b+10c+20d+50e+100f+500g = 2000$$



      A short C++ program with no more than 20 lines of code and a simple recurrence will serve for the purpose. 



      #include <iostream>
      

      
using namespace std;
      

      
int numSolutions(int *list, int size, int sum) {
      
    int num = list[0];
      
    if(size == 1) {
      
        return (sum % num == 0)? 1: 0;
      
    }
      
    int count = 0, cases = sum / num;
      
    for(int i = 0; i <= cases; i++) {
      
        count += numSolutions(list + 1, size - 1, sum - i * num);
      
    }
      
    return count;
      
}
      

      
int main() {
      
    int list = {500, 100, 50, 20, 10, 5, 1};
      
    int size = 7, sum = 2000;
      
    cout << numSolutions(list, size, sum);
      
}


      ...and the answer is: 86950230. Execution time is less than a second.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        So you have to find the number of non-negative integer solutions to:



        $$a+5b+10c+20d+50e+100f+500g = 2000$$



        A short C++ program with no more than 20 lines of code and a simple recurrence will serve for the purpose. 



        #include <iostream>
        

        
using namespace std;
        

        
int numSolutions(int *list, int size, int sum) {
        
    int num = list[0];
        
    if(size == 1) {
        
        return (sum % num == 0)? 1: 0;
        
    }
        
    int count = 0, cases = sum / num;
        
    for(int i = 0; i <= cases; i++) {
        
        count += numSolutions(list + 1, size - 1, sum - i * num);
        
    }
        
    return count;
        
}
        

        
int main() {
        
    int list = {500, 100, 50, 20, 10, 5, 1};
        
    int size = 7, sum = 2000;
        
    cout << numSolutions(list, size, sum);
        
}


        ...and the answer is: 86950230. Execution time is less than a second.






        share|cite|improve this answer









        $endgroup$



        So you have to find the number of non-negative integer solutions to:



        $$a+5b+10c+20d+50e+100f+500g = 2000$$



        A short C++ program with no more than 20 lines of code and a simple recurrence will serve for the purpose. 



        #include <iostream>
        

        
using namespace std;
        

        
int numSolutions(int *list, int size, int sum) {
        
    int num = list[0];
        
    if(size == 1) {
        
        return (sum % num == 0)? 1: 0;
        
    }
        
    int count = 0, cases = sum / num;
        
    for(int i = 0; i <= cases; i++) {
        
        count += numSolutions(list + 1, size - 1, sum - i * num);
        
    }
        
    return count;
        
}
        

        
int main() {
        
    int list = {500, 100, 50, 20, 10, 5, 1};
        
    int size = 7, sum = 2000;
        
    cout << numSolutions(list, size, sum);
        
}


        ...and the answer is: 86950230. Execution time is less than a second.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 11:56









        OldboyOldboy

        8,1651936




        8,1651936






























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