Number of compositions of a positive number n, with factors between 1 and a certain number m












2












$begingroup$


I'm trying to find the number of compositions of a positive number $n$, with factors between 1 and a certain number $m$.
That is, all the combinations of limited numbers that add up to $n$



$f(n, m)$



For example:



$f(3, 2) = |(1+1+1), (1+2), (2+1)| = 3$



$f(3, 3) = |(1+1+1), (1+2), (2+1), (3)| = 4$



$f(5, 3) = |(1+1+1+1+1), .... , (3+2), (2+3)| = 13$



Reading online I found this formula for calculating the composition, but it works only in some special cases:
$sum_{k = lceil n/m rceil}^{n} binom{n-1}{k - 1}$



Can someone tell me the general formula?






combinatorics







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I'm trying to find the number of compositions of a positive number $n$, with factors between 1 and a certain number $m$.
    That is, all the combinations of limited numbers that add up to $n$



    $f(n, m)$



    For example:



    $f(3, 2) = |(1+1+1), (1+2), (2+1)| = 3$



    $f(3, 3) = |(1+1+1), (1+2), (2+1), (3)| = 4$



    $f(5, 3) = |(1+1+1+1+1), .... , (3+2), (2+3)| = 13$



    Reading online I found this formula for calculating the composition, but it works only in some special cases:
    $sum_{k = lceil n/m rceil}^{n} binom{n-1}{k - 1}$



    Can someone tell me the general formula?






    combinatorics







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I'm trying to find the number of compositions of a positive number $n$, with factors between 1 and a certain number $m$.
      That is, all the combinations of limited numbers that add up to $n$



      $f(n, m)$



      For example:



      $f(3, 2) = |(1+1+1), (1+2), (2+1)| = 3$



      $f(3, 3) = |(1+1+1), (1+2), (2+1), (3)| = 4$



      $f(5, 3) = |(1+1+1+1+1), .... , (3+2), (2+3)| = 13$



      Reading online I found this formula for calculating the composition, but it works only in some special cases:
      $sum_{k = lceil n/m rceil}^{n} binom{n-1}{k - 1}$



      Can someone tell me the general formula?






      combinatorics







      share|cite|improve this question











      $endgroup$




      I'm trying to find the number of compositions of a positive number $n$, with factors between 1 and a certain number $m$.
      That is, all the combinations of limited numbers that add up to $n$



      $f(n, m)$



      For example:



      $f(3, 2) = |(1+1+1), (1+2), (2+1)| = 3$



      $f(3, 3) = |(1+1+1), (1+2), (2+1), (3)| = 4$



      $f(5, 3) = |(1+1+1+1+1), .... , (3+2), (2+3)| = 13$



      Reading online I found this formula for calculating the composition, but it works only in some special cases:
      $sum_{k = lceil n/m rceil}^{n} binom{n-1}{k - 1}$



      Can someone tell me the general formula?






      combinatorics




      combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 8:29









      Sik Feng Cheong

      1579




      1579










      asked Dec 16 '18 at 8:07









      Gabriele PiccoGabriele Picco

      327




      327






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          There is no general formula. In order to compute $f(n,m)$ quickly, you can write the recurrence relation as a matrix equation. Let $x_n$ be the vector
          $$
          x_n=begin{bmatrix}f(n,m)\f(n-1,m)\f(n-2,m)\vdots\f(n-m+1,m)end{bmatrix}
          $$
          and let $A$ be the $mtimes m$ matrix which has ones just below the diagonal, ones on the top row, and zeroes everywhere else. When $m=4,$
          $$
          A=begin{bmatrix} 1 & 1 & 1 & 1\ 1 & 0 & 0 &0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 end{bmatrix}
          $$



          You can verify that
          $$
          x_n=Ax_{n-1}
          $$
          holds for all $nge 1$. Iterating this, you get
          $$
          x_n=A^nx_0=A^nbegin{bmatrix}1\0\vdots\0end{bmatrix}
          $$
          Therefore, to compute $f(n,m)$ quickly, it suffices to compute $A^n$ quickly, which can be done in $O(m^3log n)$ time using exponentiation by squaring.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic solution! I just think the last line of $ x_ {n} $ should be $ f (n-m, m) $
            $endgroup$
            – Gabriele Picco
            Dec 18 '18 at 9:10



















          2












          $begingroup$

          I was able to callculate $f(m,n)$ quickly by using recurrence:



          $$f(n, m)=1quad text{if} quad nin{0,1}$$
          $$f(n, m)=0quad text{if} quad n<0$$
          $$f(n,m)=sum_{k=1}^m f(n-k,m)$$



          ...or in Mathematica:



          f[n_, m_] := 1 /; n == 0 || n == 1
          

          
f[n_, m_] := 0 /; n < 0
          

          
f[n_, m_] := f[n, m] = Sum[f[n - k, m], {k, 1, m}]


          This gives accurate results for all your examples and calculates the number of combinations for bigger values of $n,m$ fairly quickly. For example:



          $$f(5,3)=13$$



          $$f(20,10)=521472$$



          $$f(100,20)=633800819629853453628932292608$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic solution! I implemented it in Python, unfortunately in some cases it fails to compute it and generates the error: maximum recursion depth exceeded in comparison. For example: $f(99999, 3)$
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:04












          • $begingroup$
            I'm trying to optimize it or find a closed formula that allows me to calculate it even for large numbers
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:12










          • $begingroup$
            @GabrielePicco If you like the solution, please upvote it :) With my approach the recursion depth is equal to $n$ so in your example Python will have to do 99999 recursive calls which will cause the stack to overflow. Even if it had not, the result would have been a number of epic proportions :)
            $endgroup$
            – Oldboy
            Dec 16 '18 at 10:15












          • $begingroup$
            Yes you are right. The number of recursions is $ n$ x $m $ right? in the example $ f (99999,3) $ should be 99999 x 3 recursion.
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:27












          • $begingroup$
            if I know that I will have to apply a module to the result of the function, can I somehow simplify the calculation?
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:28











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          There is no general formula. In order to compute $f(n,m)$ quickly, you can write the recurrence relation as a matrix equation. Let $x_n$ be the vector
          $$
          x_n=begin{bmatrix}f(n,m)\f(n-1,m)\f(n-2,m)\vdots\f(n-m+1,m)end{bmatrix}
          $$
          and let $A$ be the $mtimes m$ matrix which has ones just below the diagonal, ones on the top row, and zeroes everywhere else. When $m=4,$
          $$
          A=begin{bmatrix} 1 & 1 & 1 & 1\ 1 & 0 & 0 &0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 end{bmatrix}
          $$



          You can verify that
          $$
          x_n=Ax_{n-1}
          $$
          holds for all $nge 1$. Iterating this, you get
          $$
          x_n=A^nx_0=A^nbegin{bmatrix}1\0\vdots\0end{bmatrix}
          $$
          Therefore, to compute $f(n,m)$ quickly, it suffices to compute $A^n$ quickly, which can be done in $O(m^3log n)$ time using exponentiation by squaring.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic solution! I just think the last line of $ x_ {n} $ should be $ f (n-m, m) $
            $endgroup$
            – Gabriele Picco
            Dec 18 '18 at 9:10
















          2












          $begingroup$

          There is no general formula. In order to compute $f(n,m)$ quickly, you can write the recurrence relation as a matrix equation. Let $x_n$ be the vector
          $$
          x_n=begin{bmatrix}f(n,m)\f(n-1,m)\f(n-2,m)\vdots\f(n-m+1,m)end{bmatrix}
          $$
          and let $A$ be the $mtimes m$ matrix which has ones just below the diagonal, ones on the top row, and zeroes everywhere else. When $m=4,$
          $$
          A=begin{bmatrix} 1 & 1 & 1 & 1\ 1 & 0 & 0 &0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 end{bmatrix}
          $$



          You can verify that
          $$
          x_n=Ax_{n-1}
          $$
          holds for all $nge 1$. Iterating this, you get
          $$
          x_n=A^nx_0=A^nbegin{bmatrix}1\0\vdots\0end{bmatrix}
          $$
          Therefore, to compute $f(n,m)$ quickly, it suffices to compute $A^n$ quickly, which can be done in $O(m^3log n)$ time using exponentiation by squaring.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic solution! I just think the last line of $ x_ {n} $ should be $ f (n-m, m) $
            $endgroup$
            – Gabriele Picco
            Dec 18 '18 at 9:10














          2












          2








          2





          $begingroup$

          There is no general formula. In order to compute $f(n,m)$ quickly, you can write the recurrence relation as a matrix equation. Let $x_n$ be the vector
          $$
          x_n=begin{bmatrix}f(n,m)\f(n-1,m)\f(n-2,m)\vdots\f(n-m+1,m)end{bmatrix}
          $$
          and let $A$ be the $mtimes m$ matrix which has ones just below the diagonal, ones on the top row, and zeroes everywhere else. When $m=4,$
          $$
          A=begin{bmatrix} 1 & 1 & 1 & 1\ 1 & 0 & 0 &0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 end{bmatrix}
          $$



          You can verify that
          $$
          x_n=Ax_{n-1}
          $$
          holds for all $nge 1$. Iterating this, you get
          $$
          x_n=A^nx_0=A^nbegin{bmatrix}1\0\vdots\0end{bmatrix}
          $$
          Therefore, to compute $f(n,m)$ quickly, it suffices to compute $A^n$ quickly, which can be done in $O(m^3log n)$ time using exponentiation by squaring.






          share|cite|improve this answer









          $endgroup$



          There is no general formula. In order to compute $f(n,m)$ quickly, you can write the recurrence relation as a matrix equation. Let $x_n$ be the vector
          $$
          x_n=begin{bmatrix}f(n,m)\f(n-1,m)\f(n-2,m)\vdots\f(n-m+1,m)end{bmatrix}
          $$
          and let $A$ be the $mtimes m$ matrix which has ones just below the diagonal, ones on the top row, and zeroes everywhere else. When $m=4,$
          $$
          A=begin{bmatrix} 1 & 1 & 1 & 1\ 1 & 0 & 0 &0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 end{bmatrix}
          $$



          You can verify that
          $$
          x_n=Ax_{n-1}
          $$
          holds for all $nge 1$. Iterating this, you get
          $$
          x_n=A^nx_0=A^nbegin{bmatrix}1\0\vdots\0end{bmatrix}
          $$
          Therefore, to compute $f(n,m)$ quickly, it suffices to compute $A^n$ quickly, which can be done in $O(m^3log n)$ time using exponentiation by squaring.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 19:23









          Mike EarnestMike Earnest

          22.6k12051




          22.6k12051












          • $begingroup$
            Fantastic solution! I just think the last line of $ x_ {n} $ should be $ f (n-m, m) $
            $endgroup$
            – Gabriele Picco
            Dec 18 '18 at 9:10


















          • $begingroup$
            Fantastic solution! I just think the last line of $ x_ {n} $ should be $ f (n-m, m) $
            $endgroup$
            – Gabriele Picco
            Dec 18 '18 at 9:10
















          $begingroup$
          Fantastic solution! I just think the last line of $ x_ {n} $ should be $ f (n-m, m) $
          $endgroup$
          – Gabriele Picco
          Dec 18 '18 at 9:10




          $begingroup$
          Fantastic solution! I just think the last line of $ x_ {n} $ should be $ f (n-m, m) $
          $endgroup$
          – Gabriele Picco
          Dec 18 '18 at 9:10











          2












          $begingroup$

          I was able to callculate $f(m,n)$ quickly by using recurrence:



          $$f(n, m)=1quad text{if} quad nin{0,1}$$
          $$f(n, m)=0quad text{if} quad n<0$$
          $$f(n,m)=sum_{k=1}^m f(n-k,m)$$



          ...or in Mathematica:



          f[n_, m_] := 1 /; n == 0 || n == 1
          

          
f[n_, m_] := 0 /; n < 0
          

          
f[n_, m_] := f[n, m] = Sum[f[n - k, m], {k, 1, m}]


          This gives accurate results for all your examples and calculates the number of combinations for bigger values of $n,m$ fairly quickly. For example:



          $$f(5,3)=13$$



          $$f(20,10)=521472$$



          $$f(100,20)=633800819629853453628932292608$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic solution! I implemented it in Python, unfortunately in some cases it fails to compute it and generates the error: maximum recursion depth exceeded in comparison. For example: $f(99999, 3)$
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:04












          • $begingroup$
            I'm trying to optimize it or find a closed formula that allows me to calculate it even for large numbers
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:12










          • $begingroup$
            @GabrielePicco If you like the solution, please upvote it :) With my approach the recursion depth is equal to $n$ so in your example Python will have to do 99999 recursive calls which will cause the stack to overflow. Even if it had not, the result would have been a number of epic proportions :)
            $endgroup$
            – Oldboy
            Dec 16 '18 at 10:15












          • $begingroup$
            Yes you are right. The number of recursions is $ n$ x $m $ right? in the example $ f (99999,3) $ should be 99999 x 3 recursion.
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:27












          • $begingroup$
            if I know that I will have to apply a module to the result of the function, can I somehow simplify the calculation?
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:28
















          2












          $begingroup$

          I was able to callculate $f(m,n)$ quickly by using recurrence:



          $$f(n, m)=1quad text{if} quad nin{0,1}$$
          $$f(n, m)=0quad text{if} quad n<0$$
          $$f(n,m)=sum_{k=1}^m f(n-k,m)$$



          ...or in Mathematica:



          f[n_, m_] := 1 /; n == 0 || n == 1
          

          
f[n_, m_] := 0 /; n < 0
          

          
f[n_, m_] := f[n, m] = Sum[f[n - k, m], {k, 1, m}]


          This gives accurate results for all your examples and calculates the number of combinations for bigger values of $n,m$ fairly quickly. For example:



          $$f(5,3)=13$$



          $$f(20,10)=521472$$



          $$f(100,20)=633800819629853453628932292608$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic solution! I implemented it in Python, unfortunately in some cases it fails to compute it and generates the error: maximum recursion depth exceeded in comparison. For example: $f(99999, 3)$
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:04












          • $begingroup$
            I'm trying to optimize it or find a closed formula that allows me to calculate it even for large numbers
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:12










          • $begingroup$
            @GabrielePicco If you like the solution, please upvote it :) With my approach the recursion depth is equal to $n$ so in your example Python will have to do 99999 recursive calls which will cause the stack to overflow. Even if it had not, the result would have been a number of epic proportions :)
            $endgroup$
            – Oldboy
            Dec 16 '18 at 10:15












          • $begingroup$
            Yes you are right. The number of recursions is $ n$ x $m $ right? in the example $ f (99999,3) $ should be 99999 x 3 recursion.
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:27












          • $begingroup$
            if I know that I will have to apply a module to the result of the function, can I somehow simplify the calculation?
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:28














          2












          2








          2





          $begingroup$

          I was able to callculate $f(m,n)$ quickly by using recurrence:



          $$f(n, m)=1quad text{if} quad nin{0,1}$$
          $$f(n, m)=0quad text{if} quad n<0$$
          $$f(n,m)=sum_{k=1}^m f(n-k,m)$$



          ...or in Mathematica:



          f[n_, m_] := 1 /; n == 0 || n == 1
          

          
f[n_, m_] := 0 /; n < 0
          

          
f[n_, m_] := f[n, m] = Sum[f[n - k, m], {k, 1, m}]


          This gives accurate results for all your examples and calculates the number of combinations for bigger values of $n,m$ fairly quickly. For example:



          $$f(5,3)=13$$



          $$f(20,10)=521472$$



          $$f(100,20)=633800819629853453628932292608$$






          share|cite|improve this answer









          $endgroup$



          I was able to callculate $f(m,n)$ quickly by using recurrence:



          $$f(n, m)=1quad text{if} quad nin{0,1}$$
          $$f(n, m)=0quad text{if} quad n<0$$
          $$f(n,m)=sum_{k=1}^m f(n-k,m)$$



          ...or in Mathematica:



          f[n_, m_] := 1 /; n == 0 || n == 1
          

          
f[n_, m_] := 0 /; n < 0
          

          
f[n_, m_] := f[n, m] = Sum[f[n - k, m], {k, 1, m}]


          This gives accurate results for all your examples and calculates the number of combinations for bigger values of $n,m$ fairly quickly. For example:



          $$f(5,3)=13$$



          $$f(20,10)=521472$$



          $$f(100,20)=633800819629853453628932292608$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 8:55









          OldboyOldboy

          8,1751936




          8,1751936












          • $begingroup$
            Fantastic solution! I implemented it in Python, unfortunately in some cases it fails to compute it and generates the error: maximum recursion depth exceeded in comparison. For example: $f(99999, 3)$
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:04












          • $begingroup$
            I'm trying to optimize it or find a closed formula that allows me to calculate it even for large numbers
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:12










          • $begingroup$
            @GabrielePicco If you like the solution, please upvote it :) With my approach the recursion depth is equal to $n$ so in your example Python will have to do 99999 recursive calls which will cause the stack to overflow. Even if it had not, the result would have been a number of epic proportions :)
            $endgroup$
            – Oldboy
            Dec 16 '18 at 10:15












          • $begingroup$
            Yes you are right. The number of recursions is $ n$ x $m $ right? in the example $ f (99999,3) $ should be 99999 x 3 recursion.
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:27












          • $begingroup$
            if I know that I will have to apply a module to the result of the function, can I somehow simplify the calculation?
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:28


















          • $begingroup$
            Fantastic solution! I implemented it in Python, unfortunately in some cases it fails to compute it and generates the error: maximum recursion depth exceeded in comparison. For example: $f(99999, 3)$
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:04












          • $begingroup$
            I'm trying to optimize it or find a closed formula that allows me to calculate it even for large numbers
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:12










          • $begingroup$
            @GabrielePicco If you like the solution, please upvote it :) With my approach the recursion depth is equal to $n$ so in your example Python will have to do 99999 recursive calls which will cause the stack to overflow. Even if it had not, the result would have been a number of epic proportions :)
            $endgroup$
            – Oldboy
            Dec 16 '18 at 10:15












          • $begingroup$
            Yes you are right. The number of recursions is $ n$ x $m $ right? in the example $ f (99999,3) $ should be 99999 x 3 recursion.
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:27












          • $begingroup$
            if I know that I will have to apply a module to the result of the function, can I somehow simplify the calculation?
            $endgroup$
            – Gabriele Picco
            Dec 16 '18 at 10:28
















          $begingroup$
          Fantastic solution! I implemented it in Python, unfortunately in some cases it fails to compute it and generates the error: maximum recursion depth exceeded in comparison. For example: $f(99999, 3)$
          $endgroup$
          – Gabriele Picco
          Dec 16 '18 at 10:04






          $begingroup$
          Fantastic solution! I implemented it in Python, unfortunately in some cases it fails to compute it and generates the error: maximum recursion depth exceeded in comparison. For example: $f(99999, 3)$
          $endgroup$
          – Gabriele Picco
          Dec 16 '18 at 10:04














          $begingroup$
          I'm trying to optimize it or find a closed formula that allows me to calculate it even for large numbers
          $endgroup$
          – Gabriele Picco
          Dec 16 '18 at 10:12




          $begingroup$
          I'm trying to optimize it or find a closed formula that allows me to calculate it even for large numbers
          $endgroup$
          – Gabriele Picco
          Dec 16 '18 at 10:12












          $begingroup$
          @GabrielePicco If you like the solution, please upvote it :) With my approach the recursion depth is equal to $n$ so in your example Python will have to do 99999 recursive calls which will cause the stack to overflow. Even if it had not, the result would have been a number of epic proportions :)
          $endgroup$
          – Oldboy
          Dec 16 '18 at 10:15






          $begingroup$
          @GabrielePicco If you like the solution, please upvote it :) With my approach the recursion depth is equal to $n$ so in your example Python will have to do 99999 recursive calls which will cause the stack to overflow. Even if it had not, the result would have been a number of epic proportions :)
          $endgroup$
          – Oldboy
          Dec 16 '18 at 10:15














          $begingroup$
          Yes you are right. The number of recursions is $ n$ x $m $ right? in the example $ f (99999,3) $ should be 99999 x 3 recursion.
          $endgroup$
          – Gabriele Picco
          Dec 16 '18 at 10:27






          $begingroup$
          Yes you are right. The number of recursions is $ n$ x $m $ right? in the example $ f (99999,3) $ should be 99999 x 3 recursion.
          $endgroup$
          – Gabriele Picco
          Dec 16 '18 at 10:27














          $begingroup$
          if I know that I will have to apply a module to the result of the function, can I somehow simplify the calculation?
          $endgroup$
          – Gabriele Picco
          Dec 16 '18 at 10:28




          $begingroup$
          if I know that I will have to apply a module to the result of the function, can I somehow simplify the calculation?
          $endgroup$
          – Gabriele Picco
          Dec 16 '18 at 10:28


















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