Double sum over $leftlfloor{ac+bdover k}rightrfloor$












2












$begingroup$


We have
$$leftlfloor{ac+bdover k}rightrfloor-leftlfloor{ac+bd-1over k}rightrfloor=1-leftlceil{ (ac+bd)mod{k}over k}rightrceil$$
for $a,b,c,d,k$ - integers, $ageqslant0$, $bgeqslant0$, $c>0$, $d>0$, $k>0$. Next we want to calculate
$$sumlimits_{a=0}^{n}sumlimits_{b=0}^{m}leftlfloor{ac+bdover k}rightrfloor$$
or
$$sumlimits_{a=0}^{n}sumlimits_{b=0}^{m}leftlceil{(ac+bd)mod{k}over k}rightrceil$$
Is there a nice closed form for it?






summation closed-form floor-function ceiling-function







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    $begingroup$


    We have
    $$leftlfloor{ac+bdover k}rightrfloor-leftlfloor{ac+bd-1over k}rightrfloor=1-leftlceil{ (ac+bd)mod{k}over k}rightrceil$$
    for $a,b,c,d,k$ - integers, $ageqslant0$, $bgeqslant0$, $c>0$, $d>0$, $k>0$. Next we want to calculate
    $$sumlimits_{a=0}^{n}sumlimits_{b=0}^{m}leftlfloor{ac+bdover k}rightrfloor$$
    or
    $$sumlimits_{a=0}^{n}sumlimits_{b=0}^{m}leftlceil{(ac+bd)mod{k}over k}rightrceil$$
    Is there a nice closed form for it?






    summation closed-form floor-function ceiling-function







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    $endgroup$















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      $begingroup$


      We have
      $$leftlfloor{ac+bdover k}rightrfloor-leftlfloor{ac+bd-1over k}rightrfloor=1-leftlceil{ (ac+bd)mod{k}over k}rightrceil$$
      for $a,b,c,d,k$ - integers, $ageqslant0$, $bgeqslant0$, $c>0$, $d>0$, $k>0$. Next we want to calculate
      $$sumlimits_{a=0}^{n}sumlimits_{b=0}^{m}leftlfloor{ac+bdover k}rightrfloor$$
      or
      $$sumlimits_{a=0}^{n}sumlimits_{b=0}^{m}leftlceil{(ac+bd)mod{k}over k}rightrceil$$
      Is there a nice closed form for it?






      summation closed-form floor-function ceiling-function







      share|cite|improve this question











      $endgroup$




      We have
      $$leftlfloor{ac+bdover k}rightrfloor-leftlfloor{ac+bd-1over k}rightrfloor=1-leftlceil{ (ac+bd)mod{k}over k}rightrceil$$
      for $a,b,c,d,k$ - integers, $ageqslant0$, $bgeqslant0$, $c>0$, $d>0$, $k>0$. Next we want to calculate
      $$sumlimits_{a=0}^{n}sumlimits_{b=0}^{m}leftlfloor{ac+bdover k}rightrfloor$$
      or
      $$sumlimits_{a=0}^{n}sumlimits_{b=0}^{m}leftlceil{(ac+bd)mod{k}over k}rightrceil$$
      Is there a nice closed form for it?






      summation closed-form floor-function ceiling-function




      summation closed-form floor-function ceiling-function






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 3 at 17:00









      Key Flex

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      8,31261233










      asked Jan 3 at 16:05









      user514787user514787

      744310




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          1 Answer
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          $begingroup$

          Here we give a closed form expression for a special case of the double sum indicating that there is no nice formula in the general case.




          Let $N, M, c$ be positive integer with
          begin{align*}
          N=&cleftlfloorfrac{N}{c}rightrfloor+nqquad 0leq n<c\
          M=&cleftlfloorfrac{M}{c}rightrfloor+mqquad 0leq m<c
          end{align*}
          then the following is valid
          begin{align*}
          sum_{k=0}^N&sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloortag{1}\
          &=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor\
          &quad+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor\
          &quad+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2\
          &quad+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)\
          &quad+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloortag{2}
          end{align*}




          We observe the special case $leftlfloorfrac{ak+bl}{c}rightrfloor$ in (1) with $a=b=1$ results in a rather complex formula (2) indicating the general case is also rather complex.



          In order to prove (2) we need some preliminary results, namely three types of single sums with floor functions given below.




          Let $n,a,c$ be integer $ngeq 0, c>0, 0leq a <c$.



          The following is valid



          begin{align*}
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor
          &=left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2tag{3}\
          sum_{k=0}^n kleftlfloor frac{k+a}{c}rightrfloor
          &=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3tag{4}\
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor^2
          &=frac{c}{6}leftlfloorfrac{n+a}{c}rightrfloor+left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{2c}{3}leftlfloorfrac{n+a}{c}rightrfloor^3tag{5}
          end{align*}



          Ad (3):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{leftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]tag{6}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}1
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n 1\
          &=csum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &=frac{c}{2}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &,,color{blue}{=left(n+a+1-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2}
          end{align*}



          and (3) follows.




          Comment:




          • In (6) we use Iverson brackets to get rid of the floor function.



          Ad (4):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{kleftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}k
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n k\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jleft(frac{1}{2}left(c(j+1)-a-1right)(c(j+1)-a)-frac{1}{2}left(cj-a-1right)(cj-a)right)\
          &qquad+leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{1}{2}sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j(-2ac+2c^2j+c^2-c)
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=c^2sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j^2+left(frac{c^2-c}{2}-acright)sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{c^2}{6}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloorleft(2leftlfloorfrac{n+a}{c}rightrfloor-1right)\
          &qquad+frac{c}{4}(c-1-2a)left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{1}{2}left(n(n+1)-left(cleftlfloorfrac{n+a}{c}rightrfloor-aright)left(cleftlfloorfrac{n+a}{c}rightrfloor-a-1right)right)leftlfloorfrac{n+a}{c}rightrfloor\
          &,,color{blue}{=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor}\
          &qquadcolor{blue}{+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3}
          end{align*}



          and (4) follows.




          Ad (5):



          The proof of this case is similar to (3) and is omitted.




          The main formula (2):



          The proof of formula (2) is really cumbersome and somewhat lengthy. Here I will show the first steps and everything else follows by consequently applying (3) to (5).



          We obtain
          begin{align*}
          color{blue}{sum_{k=0}^N}&color{blue}{sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloor}\
          &=sum_{k=0}^{cleftlfloorfrac{N}{c}rightrfloor+n} sum_{l=0}^{cleftlfloorfrac{M}{c}rightrfloor+m}leftlfloorfrac{k+l}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{ck_1+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          leftlfloorfrac{ck_1+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+l_1right)\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+l_1right)\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+leftlfloorfrac{M}{c}rightrfloorright)\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleft(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +c^2leftlfloorfrac{M}{c}rightrfloorsum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1
          +c^2leftlfloorfrac{N}{c}rightrfloorsum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor
          +(n+1)csum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          end{align*}
          begin{align*}
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)
          +c^2leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor+(n+1)cleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)leftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)tag{7}\
          \
          &=cdotstag{8}\
          \
          &,,color{blue}{=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor}\
          &quadcolor{blue}{+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2}\
          &quadcolor{blue}{+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)}\
          &quadcolor{blue}{+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloor}
          end{align*}



          Note: I've checked the formula programmatically for small values of $M,N$ and $c$.




          There are in fact some steps which are needed to fill in (8), since we have double sums in (7) which have to be substituted iteratively by (3) to (5). So, the steps up to (7) are rather the beginning of the main calculation.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very for such detail answer! Formulas are really amazing!
            $endgroup$
            – user514787
            Feb 16 at 8:02










          • $begingroup$
            @user514787: You're welcome. Probably the results of the three single sums will be sometimes useful. As you may have already noticed, I've added this question as addendum to my answer. Regards,
            $endgroup$
            – Markus Scheuer
            Feb 16 at 10:12











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          $begingroup$

          Here we give a closed form expression for a special case of the double sum indicating that there is no nice formula in the general case.




          Let $N, M, c$ be positive integer with
          begin{align*}
          N=&cleftlfloorfrac{N}{c}rightrfloor+nqquad 0leq n<c\
          M=&cleftlfloorfrac{M}{c}rightrfloor+mqquad 0leq m<c
          end{align*}
          then the following is valid
          begin{align*}
          sum_{k=0}^N&sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloortag{1}\
          &=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor\
          &quad+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor\
          &quad+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2\
          &quad+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)\
          &quad+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloortag{2}
          end{align*}




          We observe the special case $leftlfloorfrac{ak+bl}{c}rightrfloor$ in (1) with $a=b=1$ results in a rather complex formula (2) indicating the general case is also rather complex.



          In order to prove (2) we need some preliminary results, namely three types of single sums with floor functions given below.




          Let $n,a,c$ be integer $ngeq 0, c>0, 0leq a <c$.



          The following is valid



          begin{align*}
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor
          &=left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2tag{3}\
          sum_{k=0}^n kleftlfloor frac{k+a}{c}rightrfloor
          &=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3tag{4}\
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor^2
          &=frac{c}{6}leftlfloorfrac{n+a}{c}rightrfloor+left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{2c}{3}leftlfloorfrac{n+a}{c}rightrfloor^3tag{5}
          end{align*}



          Ad (3):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{leftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]tag{6}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}1
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n 1\
          &=csum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &=frac{c}{2}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &,,color{blue}{=left(n+a+1-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2}
          end{align*}



          and (3) follows.




          Comment:




          • In (6) we use Iverson brackets to get rid of the floor function.



          Ad (4):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{kleftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}k
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n k\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jleft(frac{1}{2}left(c(j+1)-a-1right)(c(j+1)-a)-frac{1}{2}left(cj-a-1right)(cj-a)right)\
          &qquad+leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{1}{2}sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j(-2ac+2c^2j+c^2-c)
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=c^2sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j^2+left(frac{c^2-c}{2}-acright)sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{c^2}{6}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloorleft(2leftlfloorfrac{n+a}{c}rightrfloor-1right)\
          &qquad+frac{c}{4}(c-1-2a)left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{1}{2}left(n(n+1)-left(cleftlfloorfrac{n+a}{c}rightrfloor-aright)left(cleftlfloorfrac{n+a}{c}rightrfloor-a-1right)right)leftlfloorfrac{n+a}{c}rightrfloor\
          &,,color{blue}{=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor}\
          &qquadcolor{blue}{+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3}
          end{align*}



          and (4) follows.




          Ad (5):



          The proof of this case is similar to (3) and is omitted.




          The main formula (2):



          The proof of formula (2) is really cumbersome and somewhat lengthy. Here I will show the first steps and everything else follows by consequently applying (3) to (5).



          We obtain
          begin{align*}
          color{blue}{sum_{k=0}^N}&color{blue}{sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloor}\
          &=sum_{k=0}^{cleftlfloorfrac{N}{c}rightrfloor+n} sum_{l=0}^{cleftlfloorfrac{M}{c}rightrfloor+m}leftlfloorfrac{k+l}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{ck_1+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          leftlfloorfrac{ck_1+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+l_1right)\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+l_1right)\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+leftlfloorfrac{M}{c}rightrfloorright)\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleft(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +c^2leftlfloorfrac{M}{c}rightrfloorsum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1
          +c^2leftlfloorfrac{N}{c}rightrfloorsum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor
          +(n+1)csum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          end{align*}
          begin{align*}
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)
          +c^2leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor+(n+1)cleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)leftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)tag{7}\
          \
          &=cdotstag{8}\
          \
          &,,color{blue}{=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor}\
          &quadcolor{blue}{+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2}\
          &quadcolor{blue}{+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)}\
          &quadcolor{blue}{+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloor}
          end{align*}



          Note: I've checked the formula programmatically for small values of $M,N$ and $c$.




          There are in fact some steps which are needed to fill in (8), since we have double sums in (7) which have to be substituted iteratively by (3) to (5). So, the steps up to (7) are rather the beginning of the main calculation.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very for such detail answer! Formulas are really amazing!
            $endgroup$
            – user514787
            Feb 16 at 8:02










          • $begingroup$
            @user514787: You're welcome. Probably the results of the three single sums will be sometimes useful. As you may have already noticed, I've added this question as addendum to my answer. Regards,
            $endgroup$
            – Markus Scheuer
            Feb 16 at 10:12
















          2












          $begingroup$

          Here we give a closed form expression for a special case of the double sum indicating that there is no nice formula in the general case.




          Let $N, M, c$ be positive integer with
          begin{align*}
          N=&cleftlfloorfrac{N}{c}rightrfloor+nqquad 0leq n<c\
          M=&cleftlfloorfrac{M}{c}rightrfloor+mqquad 0leq m<c
          end{align*}
          then the following is valid
          begin{align*}
          sum_{k=0}^N&sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloortag{1}\
          &=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor\
          &quad+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor\
          &quad+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2\
          &quad+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)\
          &quad+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloortag{2}
          end{align*}




          We observe the special case $leftlfloorfrac{ak+bl}{c}rightrfloor$ in (1) with $a=b=1$ results in a rather complex formula (2) indicating the general case is also rather complex.



          In order to prove (2) we need some preliminary results, namely three types of single sums with floor functions given below.




          Let $n,a,c$ be integer $ngeq 0, c>0, 0leq a <c$.



          The following is valid



          begin{align*}
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor
          &=left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2tag{3}\
          sum_{k=0}^n kleftlfloor frac{k+a}{c}rightrfloor
          &=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3tag{4}\
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor^2
          &=frac{c}{6}leftlfloorfrac{n+a}{c}rightrfloor+left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{2c}{3}leftlfloorfrac{n+a}{c}rightrfloor^3tag{5}
          end{align*}



          Ad (3):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{leftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]tag{6}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}1
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n 1\
          &=csum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &=frac{c}{2}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &,,color{blue}{=left(n+a+1-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2}
          end{align*}



          and (3) follows.




          Comment:




          • In (6) we use Iverson brackets to get rid of the floor function.



          Ad (4):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{kleftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}k
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n k\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jleft(frac{1}{2}left(c(j+1)-a-1right)(c(j+1)-a)-frac{1}{2}left(cj-a-1right)(cj-a)right)\
          &qquad+leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{1}{2}sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j(-2ac+2c^2j+c^2-c)
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=c^2sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j^2+left(frac{c^2-c}{2}-acright)sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{c^2}{6}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloorleft(2leftlfloorfrac{n+a}{c}rightrfloor-1right)\
          &qquad+frac{c}{4}(c-1-2a)left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{1}{2}left(n(n+1)-left(cleftlfloorfrac{n+a}{c}rightrfloor-aright)left(cleftlfloorfrac{n+a}{c}rightrfloor-a-1right)right)leftlfloorfrac{n+a}{c}rightrfloor\
          &,,color{blue}{=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor}\
          &qquadcolor{blue}{+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3}
          end{align*}



          and (4) follows.




          Ad (5):



          The proof of this case is similar to (3) and is omitted.




          The main formula (2):



          The proof of formula (2) is really cumbersome and somewhat lengthy. Here I will show the first steps and everything else follows by consequently applying (3) to (5).



          We obtain
          begin{align*}
          color{blue}{sum_{k=0}^N}&color{blue}{sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloor}\
          &=sum_{k=0}^{cleftlfloorfrac{N}{c}rightrfloor+n} sum_{l=0}^{cleftlfloorfrac{M}{c}rightrfloor+m}leftlfloorfrac{k+l}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{ck_1+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          leftlfloorfrac{ck_1+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+l_1right)\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+l_1right)\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+leftlfloorfrac{M}{c}rightrfloorright)\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleft(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +c^2leftlfloorfrac{M}{c}rightrfloorsum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1
          +c^2leftlfloorfrac{N}{c}rightrfloorsum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor
          +(n+1)csum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          end{align*}
          begin{align*}
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)
          +c^2leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor+(n+1)cleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)leftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)tag{7}\
          \
          &=cdotstag{8}\
          \
          &,,color{blue}{=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor}\
          &quadcolor{blue}{+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2}\
          &quadcolor{blue}{+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)}\
          &quadcolor{blue}{+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloor}
          end{align*}



          Note: I've checked the formula programmatically for small values of $M,N$ and $c$.




          There are in fact some steps which are needed to fill in (8), since we have double sums in (7) which have to be substituted iteratively by (3) to (5). So, the steps up to (7) are rather the beginning of the main calculation.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very for such detail answer! Formulas are really amazing!
            $endgroup$
            – user514787
            Feb 16 at 8:02










          • $begingroup$
            @user514787: You're welcome. Probably the results of the three single sums will be sometimes useful. As you may have already noticed, I've added this question as addendum to my answer. Regards,
            $endgroup$
            – Markus Scheuer
            Feb 16 at 10:12














          2












          2








          2





          $begingroup$

          Here we give a closed form expression for a special case of the double sum indicating that there is no nice formula in the general case.




          Let $N, M, c$ be positive integer with
          begin{align*}
          N=&cleftlfloorfrac{N}{c}rightrfloor+nqquad 0leq n<c\
          M=&cleftlfloorfrac{M}{c}rightrfloor+mqquad 0leq m<c
          end{align*}
          then the following is valid
          begin{align*}
          sum_{k=0}^N&sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloortag{1}\
          &=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor\
          &quad+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor\
          &quad+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2\
          &quad+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)\
          &quad+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloortag{2}
          end{align*}




          We observe the special case $leftlfloorfrac{ak+bl}{c}rightrfloor$ in (1) with $a=b=1$ results in a rather complex formula (2) indicating the general case is also rather complex.



          In order to prove (2) we need some preliminary results, namely three types of single sums with floor functions given below.




          Let $n,a,c$ be integer $ngeq 0, c>0, 0leq a <c$.



          The following is valid



          begin{align*}
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor
          &=left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2tag{3}\
          sum_{k=0}^n kleftlfloor frac{k+a}{c}rightrfloor
          &=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3tag{4}\
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor^2
          &=frac{c}{6}leftlfloorfrac{n+a}{c}rightrfloor+left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{2c}{3}leftlfloorfrac{n+a}{c}rightrfloor^3tag{5}
          end{align*}



          Ad (3):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{leftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]tag{6}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}1
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n 1\
          &=csum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &=frac{c}{2}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &,,color{blue}{=left(n+a+1-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2}
          end{align*}



          and (3) follows.




          Comment:




          • In (6) we use Iverson brackets to get rid of the floor function.



          Ad (4):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{kleftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}k
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n k\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jleft(frac{1}{2}left(c(j+1)-a-1right)(c(j+1)-a)-frac{1}{2}left(cj-a-1right)(cj-a)right)\
          &qquad+leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{1}{2}sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j(-2ac+2c^2j+c^2-c)
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=c^2sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j^2+left(frac{c^2-c}{2}-acright)sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{c^2}{6}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloorleft(2leftlfloorfrac{n+a}{c}rightrfloor-1right)\
          &qquad+frac{c}{4}(c-1-2a)left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{1}{2}left(n(n+1)-left(cleftlfloorfrac{n+a}{c}rightrfloor-aright)left(cleftlfloorfrac{n+a}{c}rightrfloor-a-1right)right)leftlfloorfrac{n+a}{c}rightrfloor\
          &,,color{blue}{=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor}\
          &qquadcolor{blue}{+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3}
          end{align*}



          and (4) follows.




          Ad (5):



          The proof of this case is similar to (3) and is omitted.




          The main formula (2):



          The proof of formula (2) is really cumbersome and somewhat lengthy. Here I will show the first steps and everything else follows by consequently applying (3) to (5).



          We obtain
          begin{align*}
          color{blue}{sum_{k=0}^N}&color{blue}{sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloor}\
          &=sum_{k=0}^{cleftlfloorfrac{N}{c}rightrfloor+n} sum_{l=0}^{cleftlfloorfrac{M}{c}rightrfloor+m}leftlfloorfrac{k+l}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{ck_1+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          leftlfloorfrac{ck_1+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+l_1right)\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+l_1right)\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+leftlfloorfrac{M}{c}rightrfloorright)\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleft(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +c^2leftlfloorfrac{M}{c}rightrfloorsum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1
          +c^2leftlfloorfrac{N}{c}rightrfloorsum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor
          +(n+1)csum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          end{align*}
          begin{align*}
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)
          +c^2leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor+(n+1)cleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)leftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)tag{7}\
          \
          &=cdotstag{8}\
          \
          &,,color{blue}{=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor}\
          &quadcolor{blue}{+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2}\
          &quadcolor{blue}{+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)}\
          &quadcolor{blue}{+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloor}
          end{align*}



          Note: I've checked the formula programmatically for small values of $M,N$ and $c$.




          There are in fact some steps which are needed to fill in (8), since we have double sums in (7) which have to be substituted iteratively by (3) to (5). So, the steps up to (7) are rather the beginning of the main calculation.






          share|cite|improve this answer











          $endgroup$



          Here we give a closed form expression for a special case of the double sum indicating that there is no nice formula in the general case.




          Let $N, M, c$ be positive integer with
          begin{align*}
          N=&cleftlfloorfrac{N}{c}rightrfloor+nqquad 0leq n<c\
          M=&cleftlfloorfrac{M}{c}rightrfloor+mqquad 0leq m<c
          end{align*}
          then the following is valid
          begin{align*}
          sum_{k=0}^N&sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloortag{1}\
          &=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor\
          &quad+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor\
          &quad+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2\
          &quad+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor\
          &quad+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)\
          &quad+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloortag{2}
          end{align*}




          We observe the special case $leftlfloorfrac{ak+bl}{c}rightrfloor$ in (1) with $a=b=1$ results in a rather complex formula (2) indicating the general case is also rather complex.



          In order to prove (2) we need some preliminary results, namely three types of single sums with floor functions given below.




          Let $n,a,c$ be integer $ngeq 0, c>0, 0leq a <c$.



          The following is valid



          begin{align*}
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor
          &=left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2tag{3}\
          sum_{k=0}^n kleftlfloor frac{k+a}{c}rightrfloor
          &=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3tag{4}\
          sum_{k=0}^nleftlfloorfrac{k+a}{c}rightrfloor^2
          &=frac{c}{6}leftlfloorfrac{n+a}{c}rightrfloor+left(n+1+a-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{2c}{3}leftlfloorfrac{n+a}{c}rightrfloor^3tag{5}
          end{align*}



          Ad (3):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{leftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]tag{6}\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nsum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}1
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n 1\
          &=csum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &=frac{c}{2}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor+leftlfloorfrac{n+a}{c}rightrfloorleft(n-cleftlfloorfrac{n+a}{c}rightrfloor+a+1right)\
          &,,color{blue}{=left(n+a+1-frac{c}{2}right)leftlfloorfrac{n+a}{c}rightrfloor-frac{c}{2}leftlfloorfrac{n+a}{c}rightrfloor^2}
          end{align*}



          and (3) follows.




          Comment:




          • In (6) we use Iverson brackets to get rid of the floor function.



          Ad (4):



          Let $n,a,c$ integer $ngeq 0, c>0, 0leq a <c$. We obtain



          begin{align*}
          color{blue}{sum_{k=0}^n}&color{blue}{kleftlfloorfrac{k+a}{c}rightrfloor}\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[j=leftlfloorfrac{k+a}{c}rightrfloorright]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ jleq frac{k+a}{c} < j+1right]\
          &=sum_{k=0}^nksum_{j=1}^{leftlfloorfrac{k+a}{c}rightrfloor} jleft[ cj-aleq k < c(j+1)-aright]\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jsum_{k=cj-a}^{c(j+1)-a-1}k
          +leftlfloorfrac{n+a}{c}rightrfloorsum_{k=cleftlfloorfrac{n+a}{c}rightrfloor-a}^n k\
          &=sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}jleft(frac{1}{2}left(c(j+1)-a-1right)(c(j+1)-a)-frac{1}{2}left(cj-a-1right)(cj-a)right)\
          &qquad+leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{1}{2}sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j(-2ac+2c^2j+c^2-c)
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=c^2sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j^2+left(frac{c^2-c}{2}-acright)sum_{j=1}^{leftlfloorfrac{n+a}{c}rightrfloor-1}j
          +leftlfloorfrac{n+a}{c}rightrfloorleft(binom{n+1}{2}-binom{cleftlfloorfrac{n+a}{c}rightrfloor-a}{2}right)\
          &=frac{c^2}{6}left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloorleft(2leftlfloorfrac{n+a}{c}rightrfloor-1right)\
          &qquad+frac{c}{4}(c-1-2a)left(leftlfloorfrac{n+a}{c}rightrfloor-1right)leftlfloorfrac{n+a}{c}rightrfloor\
          &qquad+frac{1}{2}left(n(n+1)-left(cleftlfloorfrac{n+a}{c}rightrfloor-aright)left(cleftlfloorfrac{n+a}{c}rightrfloor-a-1right)right)leftlfloorfrac{n+a}{c}rightrfloor\
          &,,color{blue}{=frac{1}{12}left(6n(n+1)-6a^2+6a(c-1)-c^2+3cright)leftlfloorfrac{n+a}{c}rightrfloor}\
          &qquadcolor{blue}{+frac{c}{4}(2a-c+1)leftlfloorfrac{n+a}{c}rightrfloor^2-frac{c^2}{6}leftlfloorfrac{n+a}{c}rightrfloor^3}
          end{align*}



          and (4) follows.




          Ad (5):



          The proof of this case is similar to (3) and is omitted.




          The main formula (2):



          The proof of formula (2) is really cumbersome and somewhat lengthy. Here I will show the first steps and everything else follows by consequently applying (3) to (5).



          We obtain
          begin{align*}
          color{blue}{sum_{k=0}^N}&color{blue}{sum_{l=0}^Mleftlfloorfrac{k+l}{c}rightrfloor}\
          &=sum_{k=0}^{cleftlfloorfrac{N}{c}rightrfloor+n} sum_{l=0}^{cleftlfloorfrac{M}{c}rightrfloor+m}leftlfloorfrac{k+l}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{ck_1+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          leftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cl_1+l_2}{c}rightrfloor\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          leftlfloorfrac{ck_1+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleftlfloorfrac{cleftlfloorfrac{N}{c}rightrfloor+k_2+cleftlfloorfrac{M}{c}rightrfloor+l_2}{c}rightrfloor\
          &=sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+l_1right)\
          &qquad+sum_{k_2=0}^n sum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1} sum_{l_2=0}^{c-1}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+l_1right)\
          &qquad+sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1} sum_{k_2=0}^{c-1} sum_{l_2=0}^{m}
          left(leftlfloorfrac{k_2+l_2}{c}rightrfloor+k_1+leftlfloorfrac{M}{c}rightrfloorright)\
          &qquad+sum_{k_2=0}^nsum_{l2=0}^nleft(leftlfloorfrac{k_2+l_2}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +c^2leftlfloorfrac{M}{c}rightrfloorsum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1
          +c^2leftlfloorfrac{N}{c}rightrfloorsum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor
          +(n+1)csum_{l_1=0}^{leftlfloorfrac{M}{c}rightrfloor-1}l_1\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor
          +leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)sum_{k_1=0}^{leftlfloorfrac{N}{c}rightrfloor-1}k_1\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)\
          end{align*}
          begin{align*}
          &=leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)
          +c^2leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{M}{c}rightrfloorsum_{k_2=0}^nsum_{l_2=0}^{c-1}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+(n+1)leftlfloorfrac{M}{c}rightrfloor cleftlfloorfrac{N}{c}rightrfloor+(n+1)cleftlfloorfrac{M}{c}rightrfloorleft(leftlfloorfrac{M}{c}rightrfloor-1right)\
          &qquad+leftlfloorfrac{N}{c}rightrfloorsum_{k_2=0}^{c-1}sum_{l_2=0}^{m}leftlfloorfrac{k_2+l_2}{c}rightrfloor\
          &qquadqquad+leftlfloorfrac{N}{c}rightrfloor c (m+1)leftlfloorfrac{M}{c}rightrfloor
          +c(m+1)leftlfloorfrac{N}{c}rightrfloorleft(leftlfloorfrac{N}{c}rightrfloor-1right)\
          &qquad+sum_{k_2=0}^{n}sum_{l_2=0}^mleftlfloorfrac{k_2+l_2}{c}rightrfloor+(n+1)(m+1)left(leftlfloorfrac{N}{c}rightrfloor+leftlfloorfrac{M}{c}rightrfloorright)tag{7}\
          \
          &=cdotstag{8}\
          \
          &,,color{blue}{=-frac{1}{2}c(1-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{2c-2}{2}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}m(m+1)leftlfloorfrac{N}{c}rightrfloorleftlfloorfrac{c+m-1}{c}rightrfloor
          +frac{1}{2}n(n+1)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{c+n-1}{c}rightrfloor}\
          &quadcolor{blue}{+left(m+1-frac{c}{2}right)(n+1)leftlfloorfrac{M}{c}rightrfloor+left(n+1-frac{c}{2}right)(m+1)leftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}(n+1)cleftlfloorfrac{M}{c}rightrfloor^2+frac{1}{2}(m+1)cleftlfloorfrac{N}{c}rightrfloor^2}\
          &quadcolor{blue}{+c(m+n+2-c)leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{N}{c}rightrfloor}\
          &quadcolor{blue}{+frac{1}{2}c^2leftlfloorfrac{M}{c}rightrfloorleftlfloorfrac{M}{c}rightrfloor
          left(leftlfloorfrac{M}{c}rightrfloor+leftlfloorfrac{N}{c}rightrfloorright)}\
          &quadcolor{blue}{+left(frac{1}{2}((m+n)^2+3(m+n)+2)-left(m+n+frac{3}{2}right)c+frac{1}{2}c^2right)leftlfloorfrac{m+n}{c}rightrfloor}
          end{align*}



          Note: I've checked the formula programmatically for small values of $M,N$ and $c$.




          There are in fact some steps which are needed to fill in (8), since we have double sums in (7) which have to be substituted iteratively by (3) to (5). So, the steps up to (7) are rather the beginning of the main calculation.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 16 at 10:14

























          answered Feb 15 at 16:01









          Markus ScheuerMarkus Scheuer

          62.2k459149




          62.2k459149












          • $begingroup$
            Thank you very for such detail answer! Formulas are really amazing!
            $endgroup$
            – user514787
            Feb 16 at 8:02










          • $begingroup$
            @user514787: You're welcome. Probably the results of the three single sums will be sometimes useful. As you may have already noticed, I've added this question as addendum to my answer. Regards,
            $endgroup$
            – Markus Scheuer
            Feb 16 at 10:12


















          • $begingroup$
            Thank you very for such detail answer! Formulas are really amazing!
            $endgroup$
            – user514787
            Feb 16 at 8:02










          • $begingroup$
            @user514787: You're welcome. Probably the results of the three single sums will be sometimes useful. As you may have already noticed, I've added this question as addendum to my answer. Regards,
            $endgroup$
            – Markus Scheuer
            Feb 16 at 10:12
















          $begingroup$
          Thank you very for such detail answer! Formulas are really amazing!
          $endgroup$
          – user514787
          Feb 16 at 8:02




          $begingroup$
          Thank you very for such detail answer! Formulas are really amazing!
          $endgroup$
          – user514787
          Feb 16 at 8:02












          $begingroup$
          @user514787: You're welcome. Probably the results of the three single sums will be sometimes useful. As you may have already noticed, I've added this question as addendum to my answer. Regards,
          $endgroup$
          – Markus Scheuer
          Feb 16 at 10:12




          $begingroup$
          @user514787: You're welcome. Probably the results of the three single sums will be sometimes useful. As you may have already noticed, I've added this question as addendum to my answer. Regards,
          $endgroup$
          – Markus Scheuer
          Feb 16 at 10:12


















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