Is there a closed form for the sum of the cubes of the binomial coefficients?












7














We know that
$$
sum_{k=0}^n binom{n}{k} = 2^n;; text{ and };;
sum_{k=0}^n binom{n}{k}^2 = binom{2n}{n}
$$
hold for all $nin mathbb{N}_0$.
Now I tried to find a similar expression for
$$
sum_{k=0}^n binom{n}{k}^3
$$
but didn't get anywhere at all. What I found were only asymptotic estimates (see Sum of cubes of binomial coefficients or Asymptotics of $sum_{k=0}^{n} {binom n k}^a$).



Now is there a closed form for this sum or, what would be even better, for $sum_{k=0}^n binom{n}{k}^alpha$ with any $alpha in mathbb{N}_0$?






combinatorics binomial-coefficients closed-form







share|cite|improve this question




















  • 2




    Its been proven no closed formula can exist
    – Jorge Fernández
    Nov 28 at 18:05






  • 1




    on a positive note, if you add $(-1)^k$ to you sum of cubes, there is closed formula called Dixon's theorem: mathworld.wolfram.com/DixonsTheorem.html
    – Abdelmalek Abdesselam
    Nov 28 at 18:32
















7














We know that
$$
sum_{k=0}^n binom{n}{k} = 2^n;; text{ and };;
sum_{k=0}^n binom{n}{k}^2 = binom{2n}{n}
$$
hold for all $nin mathbb{N}_0$.
Now I tried to find a similar expression for
$$
sum_{k=0}^n binom{n}{k}^3
$$
but didn't get anywhere at all. What I found were only asymptotic estimates (see Sum of cubes of binomial coefficients or Asymptotics of $sum_{k=0}^{n} {binom n k}^a$).



Now is there a closed form for this sum or, what would be even better, for $sum_{k=0}^n binom{n}{k}^alpha$ with any $alpha in mathbb{N}_0$?






combinatorics binomial-coefficients closed-form







share|cite|improve this question




















  • 2




    Its been proven no closed formula can exist
    – Jorge Fernández
    Nov 28 at 18:05






  • 1




    on a positive note, if you add $(-1)^k$ to you sum of cubes, there is closed formula called Dixon's theorem: mathworld.wolfram.com/DixonsTheorem.html
    – Abdelmalek Abdesselam
    Nov 28 at 18:32














7












7








7


3





We know that
$$
sum_{k=0}^n binom{n}{k} = 2^n;; text{ and };;
sum_{k=0}^n binom{n}{k}^2 = binom{2n}{n}
$$
hold for all $nin mathbb{N}_0$.
Now I tried to find a similar expression for
$$
sum_{k=0}^n binom{n}{k}^3
$$
but didn't get anywhere at all. What I found were only asymptotic estimates (see Sum of cubes of binomial coefficients or Asymptotics of $sum_{k=0}^{n} {binom n k}^a$).



Now is there a closed form for this sum or, what would be even better, for $sum_{k=0}^n binom{n}{k}^alpha$ with any $alpha in mathbb{N}_0$?






combinatorics binomial-coefficients closed-form







share|cite|improve this question















We know that
$$
sum_{k=0}^n binom{n}{k} = 2^n;; text{ and };;
sum_{k=0}^n binom{n}{k}^2 = binom{2n}{n}
$$
hold for all $nin mathbb{N}_0$.
Now I tried to find a similar expression for
$$
sum_{k=0}^n binom{n}{k}^3
$$
but didn't get anywhere at all. What I found were only asymptotic estimates (see Sum of cubes of binomial coefficients or Asymptotics of $sum_{k=0}^{n} {binom n k}^a$).



Now is there a closed form for this sum or, what would be even better, for $sum_{k=0}^n binom{n}{k}^alpha$ with any $alpha in mathbb{N}_0$?






combinatorics binomial-coefficients closed-form




combinatorics binomial-coefficients closed-form






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 17:58









amWhy

191k28224439




191k28224439










asked Nov 28 at 17:56









bruderjakob17

1487




1487








  • 2




    Its been proven no closed formula can exist
    – Jorge Fernández
    Nov 28 at 18:05






  • 1




    on a positive note, if you add $(-1)^k$ to you sum of cubes, there is closed formula called Dixon's theorem: mathworld.wolfram.com/DixonsTheorem.html
    – Abdelmalek Abdesselam
    Nov 28 at 18:32














  • 2




    Its been proven no closed formula can exist
    – Jorge Fernández
    Nov 28 at 18:05






  • 1




    on a positive note, if you add $(-1)^k$ to you sum of cubes, there is closed formula called Dixon's theorem: mathworld.wolfram.com/DixonsTheorem.html
    – Abdelmalek Abdesselam
    Nov 28 at 18:32








2




2




Its been proven no closed formula can exist
– Jorge Fernández
Nov 28 at 18:05




Its been proven no closed formula can exist
– Jorge Fernández
Nov 28 at 18:05




1




1




on a positive note, if you add $(-1)^k$ to you sum of cubes, there is closed formula called Dixon's theorem: mathworld.wolfram.com/DixonsTheorem.html
– Abdelmalek Abdesselam
Nov 28 at 18:32




on a positive note, if you add $(-1)^k$ to you sum of cubes, there is closed formula called Dixon's theorem: mathworld.wolfram.com/DixonsTheorem.html
– Abdelmalek Abdesselam
Nov 28 at 18:32










2 Answers
2






active

oldest

votes


















6














These numbers are called the Franel Numbers. It's proven in (Petkovšek, M., Wilf, H. and Zeilberger, D. (1996). A=B. Wellesley, MA: A K Peters. p. 160) that there is no closed form for these numbers, in terms of the sum of a fixed number of hypergeometric terms.



However, as @Robert_Israel points out, the expression could possibly be represented by different types of closed form.






share|cite|improve this answer



















  • 2




    ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form".
    – Robert Israel
    Nov 28 at 18:24










  • @Robert Was just thinking that. Thanks for the suggestion.
    – Jam
    Nov 28 at 18:26



















1














The binomial coefficient for a given pair of $n geq k geq 0$ integers can be expressed in terms of a Pochhammer symbol as the following.




$$
binom n k = frac{(-1)^k(-n)_k} {k!}.
$$




The expression is valid even if $n$ is an arbitrary real number.



Here we note two things.





  1. The Pochhammer symbol $(-n)_k$ is zero, if $n geq 0$ and $k > -n$.

  2. The factorial $k!$ can be written as $(1)_k$.




Using these observations, we can express your sums in terms of a generalized hypergeometric function $_pF_q$ as the following. For the sum of the binomial coefficients, we have



$$
sum_{k=0}^n binom n k = sum_{k=0}^n frac{(-1)^k(-n)_k}{k!} = sum_{k=0}^infty (-n)_k{frac{(-1)^k}{k!}} = {_1F_0}left({{-n}atop{-}}middle|,-1right).
$$
For the sum the square of the binomial coefficients, we have
$$
sum_{k=0}^n {binom n k}^2 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^2 = sum_{k=0}^infty frac{left((-n)_kright)^2}{k!} cdot frac{1}{k!} = {_2F_1}left({{-n, -n}atop{1}}middle|,1right).
$$
And for the sum of the cube of the binomial coefficients $-$ also known as Franel numbers $-$, we have
$$
sum_{k=0}^n {binom n k}^3 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^3 = sum_{k=0}^infty frac{left((-n)_kright)^3}{(k!)^2} cdot frac{(-1)^k}{k!} = {_3F_2}left({{-n, -n, -n}atop{1, 1}}middle|,-1right).
$$
In general, for a positive integer $r$, we have the binomial sum




$$
begin{align*}
sum_{k=0}^n {binom n k}^r &= sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^r = sum_{k=0}^infty frac{left((-n)_kright)^r}{(k!)^{r-1}} cdot frac{(-1)^{rk}}{k!} \ &= {_rF_{r-1}}left({{-n, -n, dots, -n}atop{1, dots, 1}}middle|,(-1)^rright).
end{align*}
$$







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    2 Answers
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    2 Answers
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    6














    These numbers are called the Franel Numbers. It's proven in (Petkovšek, M., Wilf, H. and Zeilberger, D. (1996). A=B. Wellesley, MA: A K Peters. p. 160) that there is no closed form for these numbers, in terms of the sum of a fixed number of hypergeometric terms.



    However, as @Robert_Israel points out, the expression could possibly be represented by different types of closed form.






    share|cite|improve this answer



















    • 2




      ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form".
      – Robert Israel
      Nov 28 at 18:24










    • @Robert Was just thinking that. Thanks for the suggestion.
      – Jam
      Nov 28 at 18:26
















    6














    These numbers are called the Franel Numbers. It's proven in (Petkovšek, M., Wilf, H. and Zeilberger, D. (1996). A=B. Wellesley, MA: A K Peters. p. 160) that there is no closed form for these numbers, in terms of the sum of a fixed number of hypergeometric terms.



    However, as @Robert_Israel points out, the expression could possibly be represented by different types of closed form.






    share|cite|improve this answer



















    • 2




      ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form".
      – Robert Israel
      Nov 28 at 18:24










    • @Robert Was just thinking that. Thanks for the suggestion.
      – Jam
      Nov 28 at 18:26














    6












    6








    6






    These numbers are called the Franel Numbers. It's proven in (Petkovšek, M., Wilf, H. and Zeilberger, D. (1996). A=B. Wellesley, MA: A K Peters. p. 160) that there is no closed form for these numbers, in terms of the sum of a fixed number of hypergeometric terms.



    However, as @Robert_Israel points out, the expression could possibly be represented by different types of closed form.






    share|cite|improve this answer














    These numbers are called the Franel Numbers. It's proven in (Petkovšek, M., Wilf, H. and Zeilberger, D. (1996). A=B. Wellesley, MA: A K Peters. p. 160) that there is no closed form for these numbers, in terms of the sum of a fixed number of hypergeometric terms.



    However, as @Robert_Israel points out, the expression could possibly be represented by different types of closed form.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 29 at 15:03

























    answered Nov 28 at 18:20









    Jam

    4,94711431




    4,94711431








    • 2




      ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form".
      – Robert Israel
      Nov 28 at 18:24










    • @Robert Was just thinking that. Thanks for the suggestion.
      – Jam
      Nov 28 at 18:26














    • 2




      ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form".
      – Robert Israel
      Nov 28 at 18:24










    • @Robert Was just thinking that. Thanks for the suggestion.
      – Jam
      Nov 28 at 18:26








    2




    2




    ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form".
    – Robert Israel
    Nov 28 at 18:24




    ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form".
    – Robert Israel
    Nov 28 at 18:24












    @Robert Was just thinking that. Thanks for the suggestion.
    – Jam
    Nov 28 at 18:26




    @Robert Was just thinking that. Thanks for the suggestion.
    – Jam
    Nov 28 at 18:26











    1














    The binomial coefficient for a given pair of $n geq k geq 0$ integers can be expressed in terms of a Pochhammer symbol as the following.




    $$
    binom n k = frac{(-1)^k(-n)_k} {k!}.
    $$




    The expression is valid even if $n$ is an arbitrary real number.



    Here we note two things.





    1. The Pochhammer symbol $(-n)_k$ is zero, if $n geq 0$ and $k > -n$.

    2. The factorial $k!$ can be written as $(1)_k$.




    Using these observations, we can express your sums in terms of a generalized hypergeometric function $_pF_q$ as the following. For the sum of the binomial coefficients, we have



    $$
    sum_{k=0}^n binom n k = sum_{k=0}^n frac{(-1)^k(-n)_k}{k!} = sum_{k=0}^infty (-n)_k{frac{(-1)^k}{k!}} = {_1F_0}left({{-n}atop{-}}middle|,-1right).
    $$
    For the sum the square of the binomial coefficients, we have
    $$
    sum_{k=0}^n {binom n k}^2 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^2 = sum_{k=0}^infty frac{left((-n)_kright)^2}{k!} cdot frac{1}{k!} = {_2F_1}left({{-n, -n}atop{1}}middle|,1right).
    $$
    And for the sum of the cube of the binomial coefficients $-$ also known as Franel numbers $-$, we have
    $$
    sum_{k=0}^n {binom n k}^3 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^3 = sum_{k=0}^infty frac{left((-n)_kright)^3}{(k!)^2} cdot frac{(-1)^k}{k!} = {_3F_2}left({{-n, -n, -n}atop{1, 1}}middle|,-1right).
    $$
    In general, for a positive integer $r$, we have the binomial sum




    $$
    begin{align*}
    sum_{k=0}^n {binom n k}^r &= sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^r = sum_{k=0}^infty frac{left((-n)_kright)^r}{(k!)^{r-1}} cdot frac{(-1)^{rk}}{k!} \ &= {_rF_{r-1}}left({{-n, -n, dots, -n}atop{1, dots, 1}}middle|,(-1)^rright).
    end{align*}
    $$







    share|cite|improve this answer




























      1














      The binomial coefficient for a given pair of $n geq k geq 0$ integers can be expressed in terms of a Pochhammer symbol as the following.




      $$
      binom n k = frac{(-1)^k(-n)_k} {k!}.
      $$




      The expression is valid even if $n$ is an arbitrary real number.



      Here we note two things.





      1. The Pochhammer symbol $(-n)_k$ is zero, if $n geq 0$ and $k > -n$.

      2. The factorial $k!$ can be written as $(1)_k$.




      Using these observations, we can express your sums in terms of a generalized hypergeometric function $_pF_q$ as the following. For the sum of the binomial coefficients, we have



      $$
      sum_{k=0}^n binom n k = sum_{k=0}^n frac{(-1)^k(-n)_k}{k!} = sum_{k=0}^infty (-n)_k{frac{(-1)^k}{k!}} = {_1F_0}left({{-n}atop{-}}middle|,-1right).
      $$
      For the sum the square of the binomial coefficients, we have
      $$
      sum_{k=0}^n {binom n k}^2 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^2 = sum_{k=0}^infty frac{left((-n)_kright)^2}{k!} cdot frac{1}{k!} = {_2F_1}left({{-n, -n}atop{1}}middle|,1right).
      $$
      And for the sum of the cube of the binomial coefficients $-$ also known as Franel numbers $-$, we have
      $$
      sum_{k=0}^n {binom n k}^3 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^3 = sum_{k=0}^infty frac{left((-n)_kright)^3}{(k!)^2} cdot frac{(-1)^k}{k!} = {_3F_2}left({{-n, -n, -n}atop{1, 1}}middle|,-1right).
      $$
      In general, for a positive integer $r$, we have the binomial sum




      $$
      begin{align*}
      sum_{k=0}^n {binom n k}^r &= sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^r = sum_{k=0}^infty frac{left((-n)_kright)^r}{(k!)^{r-1}} cdot frac{(-1)^{rk}}{k!} \ &= {_rF_{r-1}}left({{-n, -n, dots, -n}atop{1, dots, 1}}middle|,(-1)^rright).
      end{align*}
      $$







      share|cite|improve this answer


























        1












        1








        1






        The binomial coefficient for a given pair of $n geq k geq 0$ integers can be expressed in terms of a Pochhammer symbol as the following.




        $$
        binom n k = frac{(-1)^k(-n)_k} {k!}.
        $$




        The expression is valid even if $n$ is an arbitrary real number.



        Here we note two things.





        1. The Pochhammer symbol $(-n)_k$ is zero, if $n geq 0$ and $k > -n$.

        2. The factorial $k!$ can be written as $(1)_k$.




        Using these observations, we can express your sums in terms of a generalized hypergeometric function $_pF_q$ as the following. For the sum of the binomial coefficients, we have



        $$
        sum_{k=0}^n binom n k = sum_{k=0}^n frac{(-1)^k(-n)_k}{k!} = sum_{k=0}^infty (-n)_k{frac{(-1)^k}{k!}} = {_1F_0}left({{-n}atop{-}}middle|,-1right).
        $$
        For the sum the square of the binomial coefficients, we have
        $$
        sum_{k=0}^n {binom n k}^2 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^2 = sum_{k=0}^infty frac{left((-n)_kright)^2}{k!} cdot frac{1}{k!} = {_2F_1}left({{-n, -n}atop{1}}middle|,1right).
        $$
        And for the sum of the cube of the binomial coefficients $-$ also known as Franel numbers $-$, we have
        $$
        sum_{k=0}^n {binom n k}^3 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^3 = sum_{k=0}^infty frac{left((-n)_kright)^3}{(k!)^2} cdot frac{(-1)^k}{k!} = {_3F_2}left({{-n, -n, -n}atop{1, 1}}middle|,-1right).
        $$
        In general, for a positive integer $r$, we have the binomial sum




        $$
        begin{align*}
        sum_{k=0}^n {binom n k}^r &= sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^r = sum_{k=0}^infty frac{left((-n)_kright)^r}{(k!)^{r-1}} cdot frac{(-1)^{rk}}{k!} \ &= {_rF_{r-1}}left({{-n, -n, dots, -n}atop{1, dots, 1}}middle|,(-1)^rright).
        end{align*}
        $$







        share|cite|improve this answer














        The binomial coefficient for a given pair of $n geq k geq 0$ integers can be expressed in terms of a Pochhammer symbol as the following.




        $$
        binom n k = frac{(-1)^k(-n)_k} {k!}.
        $$




        The expression is valid even if $n$ is an arbitrary real number.



        Here we note two things.





        1. The Pochhammer symbol $(-n)_k$ is zero, if $n geq 0$ and $k > -n$.

        2. The factorial $k!$ can be written as $(1)_k$.




        Using these observations, we can express your sums in terms of a generalized hypergeometric function $_pF_q$ as the following. For the sum of the binomial coefficients, we have



        $$
        sum_{k=0}^n binom n k = sum_{k=0}^n frac{(-1)^k(-n)_k}{k!} = sum_{k=0}^infty (-n)_k{frac{(-1)^k}{k!}} = {_1F_0}left({{-n}atop{-}}middle|,-1right).
        $$
        For the sum the square of the binomial coefficients, we have
        $$
        sum_{k=0}^n {binom n k}^2 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^2 = sum_{k=0}^infty frac{left((-n)_kright)^2}{k!} cdot frac{1}{k!} = {_2F_1}left({{-n, -n}atop{1}}middle|,1right).
        $$
        And for the sum of the cube of the binomial coefficients $-$ also known as Franel numbers $-$, we have
        $$
        sum_{k=0}^n {binom n k}^3 = sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^3 = sum_{k=0}^infty frac{left((-n)_kright)^3}{(k!)^2} cdot frac{(-1)^k}{k!} = {_3F_2}left({{-n, -n, -n}atop{1, 1}}middle|,-1right).
        $$
        In general, for a positive integer $r$, we have the binomial sum




        $$
        begin{align*}
        sum_{k=0}^n {binom n k}^r &= sum_{k=0}^n left(frac{(-1)^k(-n)_k}{k!}right)^r = sum_{k=0}^infty frac{left((-n)_kright)^r}{(k!)^{r-1}} cdot frac{(-1)^{rk}}{k!} \ &= {_rF_{r-1}}left({{-n, -n, dots, -n}atop{1, dots, 1}}middle|,(-1)^rright).
        end{align*}
        $$








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        edited Dec 2 at 17:21

























        answered Dec 2 at 15:34









        user153012

        6,27822277




        6,27822277






























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            Wiesbaden

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            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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