Sum of terms with recurrence relation











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I have the following sequence, where $s$ is some positive multiple of 4:



begin{equation}
L_n = begin{cases}
frac{(s-2)!}{2^{s/4-1}(s/2)!(s/4-1)!}, & text{for $n=1$} \ \
L_{n-1}cdotfrac{2(s/4-n+1)}{s-2n+1}, & text{for $n=2,3,...frac{s}{4}$}
end{cases}
end{equation}



I would like to find the value of $sum_{k=1}^{s/4}L_k$.



So far I have tried factoring, but this can't be done repeatedly in any effective way. I'm mainly stumped by how messy these terms are; simplification just doesn't seem possible. I might be willing to settle for upper and lower bounds, provided that they're better than the trivial ones ($s/4$ times the smallest and biggest terms). Asymptotic results might be ok as well. Or even just a suggestion of a strategy which might work.



This is not a homework problem, so there's no reason to believe that a solution will be simple, unfortunately.



Edit: I'm currently trying to obtain an upper and lower bound by making estimates on the coefficient on $L_{n-1}$. I know that for $n=2$ it's very close to $1/2$, and then decreases to 0 as $n$ gets larger.






combinatorics discrete-mathematics summation recurrence-relations upper-lower-bounds







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  • 1




    Is $s$ a multiple of $4$? If not, are you summing to $lfloor{frac s4}rfloor$ and how are you defining $(frac s4)!$?
    – Rhys Hughes
    Dec 4 at 19:36












  • Sorry, $s$ is a multiple of 4. I'll edit the question to note that.
    – Alex
    Dec 4 at 19:37










  • @RhysHughes I've also made a small edit of the definition of the terms, since I had mislabeled the index variable initially.
    – Alex
    Dec 5 at 18:45















up vote
2
down vote

favorite
1












I have the following sequence, where $s$ is some positive multiple of 4:



begin{equation}
L_n = begin{cases}
frac{(s-2)!}{2^{s/4-1}(s/2)!(s/4-1)!}, & text{for $n=1$} \ \
L_{n-1}cdotfrac{2(s/4-n+1)}{s-2n+1}, & text{for $n=2,3,...frac{s}{4}$}
end{cases}
end{equation}



I would like to find the value of $sum_{k=1}^{s/4}L_k$.



So far I have tried factoring, but this can't be done repeatedly in any effective way. I'm mainly stumped by how messy these terms are; simplification just doesn't seem possible. I might be willing to settle for upper and lower bounds, provided that they're better than the trivial ones ($s/4$ times the smallest and biggest terms). Asymptotic results might be ok as well. Or even just a suggestion of a strategy which might work.



This is not a homework problem, so there's no reason to believe that a solution will be simple, unfortunately.



Edit: I'm currently trying to obtain an upper and lower bound by making estimates on the coefficient on $L_{n-1}$. I know that for $n=2$ it's very close to $1/2$, and then decreases to 0 as $n$ gets larger.






combinatorics discrete-mathematics summation recurrence-relations upper-lower-bounds







share|cite|improve this question




















  • 1




    Is $s$ a multiple of $4$? If not, are you summing to $lfloor{frac s4}rfloor$ and how are you defining $(frac s4)!$?
    – Rhys Hughes
    Dec 4 at 19:36












  • Sorry, $s$ is a multiple of 4. I'll edit the question to note that.
    – Alex
    Dec 4 at 19:37










  • @RhysHughes I've also made a small edit of the definition of the terms, since I had mislabeled the index variable initially.
    – Alex
    Dec 5 at 18:45













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I have the following sequence, where $s$ is some positive multiple of 4:



begin{equation}
L_n = begin{cases}
frac{(s-2)!}{2^{s/4-1}(s/2)!(s/4-1)!}, & text{for $n=1$} \ \
L_{n-1}cdotfrac{2(s/4-n+1)}{s-2n+1}, & text{for $n=2,3,...frac{s}{4}$}
end{cases}
end{equation}



I would like to find the value of $sum_{k=1}^{s/4}L_k$.



So far I have tried factoring, but this can't be done repeatedly in any effective way. I'm mainly stumped by how messy these terms are; simplification just doesn't seem possible. I might be willing to settle for upper and lower bounds, provided that they're better than the trivial ones ($s/4$ times the smallest and biggest terms). Asymptotic results might be ok as well. Or even just a suggestion of a strategy which might work.



This is not a homework problem, so there's no reason to believe that a solution will be simple, unfortunately.



Edit: I'm currently trying to obtain an upper and lower bound by making estimates on the coefficient on $L_{n-1}$. I know that for $n=2$ it's very close to $1/2$, and then decreases to 0 as $n$ gets larger.






combinatorics discrete-mathematics summation recurrence-relations upper-lower-bounds







share|cite|improve this question















I have the following sequence, where $s$ is some positive multiple of 4:



begin{equation}
L_n = begin{cases}
frac{(s-2)!}{2^{s/4-1}(s/2)!(s/4-1)!}, & text{for $n=1$} \ \
L_{n-1}cdotfrac{2(s/4-n+1)}{s-2n+1}, & text{for $n=2,3,...frac{s}{4}$}
end{cases}
end{equation}



I would like to find the value of $sum_{k=1}^{s/4}L_k$.



So far I have tried factoring, but this can't be done repeatedly in any effective way. I'm mainly stumped by how messy these terms are; simplification just doesn't seem possible. I might be willing to settle for upper and lower bounds, provided that they're better than the trivial ones ($s/4$ times the smallest and biggest terms). Asymptotic results might be ok as well. Or even just a suggestion of a strategy which might work.



This is not a homework problem, so there's no reason to believe that a solution will be simple, unfortunately.



Edit: I'm currently trying to obtain an upper and lower bound by making estimates on the coefficient on $L_{n-1}$. I know that for $n=2$ it's very close to $1/2$, and then decreases to 0 as $n$ gets larger.






combinatorics discrete-mathematics summation recurrence-relations upper-lower-bounds




combinatorics discrete-mathematics summation recurrence-relations upper-lower-bounds






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share|cite|improve this question













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edited Dec 5 at 19:02

























asked Dec 4 at 19:25









Alex

673318




673318








  • 1




    Is $s$ a multiple of $4$? If not, are you summing to $lfloor{frac s4}rfloor$ and how are you defining $(frac s4)!$?
    – Rhys Hughes
    Dec 4 at 19:36












  • Sorry, $s$ is a multiple of 4. I'll edit the question to note that.
    – Alex
    Dec 4 at 19:37










  • @RhysHughes I've also made a small edit of the definition of the terms, since I had mislabeled the index variable initially.
    – Alex
    Dec 5 at 18:45














  • 1




    Is $s$ a multiple of $4$? If not, are you summing to $lfloor{frac s4}rfloor$ and how are you defining $(frac s4)!$?
    – Rhys Hughes
    Dec 4 at 19:36












  • Sorry, $s$ is a multiple of 4. I'll edit the question to note that.
    – Alex
    Dec 4 at 19:37










  • @RhysHughes I've also made a small edit of the definition of the terms, since I had mislabeled the index variable initially.
    – Alex
    Dec 5 at 18:45








1




1




Is $s$ a multiple of $4$? If not, are you summing to $lfloor{frac s4}rfloor$ and how are you defining $(frac s4)!$?
– Rhys Hughes
Dec 4 at 19:36






Is $s$ a multiple of $4$? If not, are you summing to $lfloor{frac s4}rfloor$ and how are you defining $(frac s4)!$?
– Rhys Hughes
Dec 4 at 19:36














Sorry, $s$ is a multiple of 4. I'll edit the question to note that.
– Alex
Dec 4 at 19:37




Sorry, $s$ is a multiple of 4. I'll edit the question to note that.
– Alex
Dec 4 at 19:37












@RhysHughes I've also made a small edit of the definition of the terms, since I had mislabeled the index variable initially.
– Alex
Dec 5 at 18:45




@RhysHughes I've also made a small edit of the definition of the terms, since I had mislabeled the index variable initially.
– Alex
Dec 5 at 18:45










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For simplicity put $r=s/4$ and $S(r)=sum_{k=1}^r L_k$. For $n=2dots, r$ we have $L_n<frac 12L_{n-1}$, so $L_n$ decreases to zero very quickly (and $L_1<S(r)<2L_1$). Thus a sum of a few first $L_k$ is a good approximation for $S(r)$. A computational evidence suggests that



$$S(r)=L_1left(2-frac 3{2r}+frac 3{4r^2}-frac{T(r)}{r^3}right),$$



with $0le T(r)=O(1)$.






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    For simplicity put $r=s/4$ and $S(r)=sum_{k=1}^r L_k$. For $n=2dots, r$ we have $L_n<frac 12L_{n-1}$, so $L_n$ decreases to zero very quickly (and $L_1<S(r)<2L_1$). Thus a sum of a few first $L_k$ is a good approximation for $S(r)$. A computational evidence suggests that



    $$S(r)=L_1left(2-frac 3{2r}+frac 3{4r^2}-frac{T(r)}{r^3}right),$$



    with $0le T(r)=O(1)$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      For simplicity put $r=s/4$ and $S(r)=sum_{k=1}^r L_k$. For $n=2dots, r$ we have $L_n<frac 12L_{n-1}$, so $L_n$ decreases to zero very quickly (and $L_1<S(r)<2L_1$). Thus a sum of a few first $L_k$ is a good approximation for $S(r)$. A computational evidence suggests that



      $$S(r)=L_1left(2-frac 3{2r}+frac 3{4r^2}-frac{T(r)}{r^3}right),$$



      with $0le T(r)=O(1)$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        For simplicity put $r=s/4$ and $S(r)=sum_{k=1}^r L_k$. For $n=2dots, r$ we have $L_n<frac 12L_{n-1}$, so $L_n$ decreases to zero very quickly (and $L_1<S(r)<2L_1$). Thus a sum of a few first $L_k$ is a good approximation for $S(r)$. A computational evidence suggests that



        $$S(r)=L_1left(2-frac 3{2r}+frac 3{4r^2}-frac{T(r)}{r^3}right),$$



        with $0le T(r)=O(1)$.






        share|cite|improve this answer












        For simplicity put $r=s/4$ and $S(r)=sum_{k=1}^r L_k$. For $n=2dots, r$ we have $L_n<frac 12L_{n-1}$, so $L_n$ decreases to zero very quickly (and $L_1<S(r)<2L_1$). Thus a sum of a few first $L_k$ is a good approximation for $S(r)$. A computational evidence suggests that



        $$S(r)=L_1left(2-frac 3{2r}+frac 3{4r^2}-frac{T(r)}{r^3}right),$$



        with $0le T(r)=O(1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 at 8:39









        Alex Ravsky

        37.8k32079




        37.8k32079






























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