Polynomial expression for $frac 1{2^n} sum_{i=0}^n binom{n}{i}(2i-n)^{2k}$












1














Let
$$F (n,k)=frac {1}{2^n}sum_{i=0}^n binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.



By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$

and so on.



Is there a simple general expression for the polynomial?










share|cite|improve this question
























  • That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
    – ajotatxe
    Dec 9 at 18:10












  • @ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
    – user
    Dec 9 at 19:30






  • 1




    This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
    – darij grinberg
    Dec 11 at 3:16






  • 1




    If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
    – darij grinberg
    Dec 11 at 3:18






  • 1




    Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
    – darij grinberg
    Dec 11 at 3:45


















1














Let
$$F (n,k)=frac {1}{2^n}sum_{i=0}^n binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.



By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$

and so on.



Is there a simple general expression for the polynomial?










share|cite|improve this question
























  • That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
    – ajotatxe
    Dec 9 at 18:10












  • @ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
    – user
    Dec 9 at 19:30






  • 1




    This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
    – darij grinberg
    Dec 11 at 3:16






  • 1




    If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
    – darij grinberg
    Dec 11 at 3:18






  • 1




    Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
    – darij grinberg
    Dec 11 at 3:45
















1












1








1


0





Let
$$F (n,k)=frac {1}{2^n}sum_{i=0}^n binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.



By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$

and so on.



Is there a simple general expression for the polynomial?










share|cite|improve this question















Let
$$F (n,k)=frac {1}{2^n}sum_{i=0}^n binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.



By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$

and so on.



Is there a simple general expression for the polynomial?







sequences-and-series combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 at 6:06

























asked Dec 9 at 18:08









user

3,6361627




3,6361627












  • That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
    – ajotatxe
    Dec 9 at 18:10












  • @ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
    – user
    Dec 9 at 19:30






  • 1




    This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
    – darij grinberg
    Dec 11 at 3:16






  • 1




    If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
    – darij grinberg
    Dec 11 at 3:18






  • 1




    Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
    – darij grinberg
    Dec 11 at 3:45




















  • That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
    – ajotatxe
    Dec 9 at 18:10












  • @ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
    – user
    Dec 9 at 19:30






  • 1




    This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
    – darij grinberg
    Dec 11 at 3:16






  • 1




    If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
    – darij grinberg
    Dec 11 at 3:18






  • 1




    Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
    – darij grinberg
    Dec 11 at 3:45


















That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
– ajotatxe
Dec 9 at 18:10






That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
– ajotatxe
Dec 9 at 18:10














@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
– user
Dec 9 at 19:30




@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
– user
Dec 9 at 19:30




1




1




This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
– darij grinberg
Dec 11 at 3:16




This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
– darij grinberg
Dec 11 at 3:16




1




1




If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
– darij grinberg
Dec 11 at 3:18




If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
– darij grinberg
Dec 11 at 3:18




1




1




Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
– darij grinberg
Dec 11 at 3:45






Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
– darij grinberg
Dec 11 at 3:45












3 Answers
3






active

oldest

votes


















4














It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance



begin{align*}
n![z^n]e^{jz}=j^ntag{1}
end{align*}




We obtain
begin{align*}
color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
&=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
&=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
&,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
end{align*}



We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.




Comment:




  • In (2) we apply the coefficient of operator according to (1).


  • In (3) use the linearity of the coefficient of operator.


  • In (4) we apply the binomial theorem.


  • In (5) we write the expression somewhat more conveniently.







share|cite|improve this answer























  • I think I have found such a closed polynomial formula. See my answer below.
    – user
    Dec 27 at 12:06










  • @user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
    – Markus Scheuer
    Dec 27 at 14:05












  • What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
    – user
    Dec 27 at 16:14










  • @user: I agree, you're right. (+1) for your nice approach.
    – Markus Scheuer
    Dec 27 at 16:27



















1














I'll play with the cosh and
see if I get
anything other than
the original problem.



$begin{array}\
cosh^n(x)
&=frac1{2^n}(e^x+e^{-x})^n\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
&=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
&=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
end{array}
$



And this is the OP.



Oh well.






share|cite|improve this answer





























    1














    Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
    positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
    $$
    F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
    =frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
    =sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
    =sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
    $$



    To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
    $$
    F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
    $$

    where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.



    By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
    $$
    F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
    $$

    with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.






    share|cite|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance



      begin{align*}
      n![z^n]e^{jz}=j^ntag{1}
      end{align*}




      We obtain
      begin{align*}
      color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
      &=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
      &=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
      &=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
      &=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
      &,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
      end{align*}



      We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.




      Comment:




      • In (2) we apply the coefficient of operator according to (1).


      • In (3) use the linearity of the coefficient of operator.


      • In (4) we apply the binomial theorem.


      • In (5) we write the expression somewhat more conveniently.







      share|cite|improve this answer























      • I think I have found such a closed polynomial formula. See my answer below.
        – user
        Dec 27 at 12:06










      • @user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
        – Markus Scheuer
        Dec 27 at 14:05












      • What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
        – user
        Dec 27 at 16:14










      • @user: I agree, you're right. (+1) for your nice approach.
        – Markus Scheuer
        Dec 27 at 16:27
















      4














      It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance



      begin{align*}
      n![z^n]e^{jz}=j^ntag{1}
      end{align*}




      We obtain
      begin{align*}
      color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
      &=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
      &=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
      &=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
      &=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
      &,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
      end{align*}



      We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.




      Comment:




      • In (2) we apply the coefficient of operator according to (1).


      • In (3) use the linearity of the coefficient of operator.


      • In (4) we apply the binomial theorem.


      • In (5) we write the expression somewhat more conveniently.







      share|cite|improve this answer























      • I think I have found such a closed polynomial formula. See my answer below.
        – user
        Dec 27 at 12:06










      • @user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
        – Markus Scheuer
        Dec 27 at 14:05












      • What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
        – user
        Dec 27 at 16:14










      • @user: I agree, you're right. (+1) for your nice approach.
        – Markus Scheuer
        Dec 27 at 16:27














      4












      4








      4






      It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance



      begin{align*}
      n![z^n]e^{jz}=j^ntag{1}
      end{align*}




      We obtain
      begin{align*}
      color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
      &=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
      &=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
      &=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
      &=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
      &,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
      end{align*}



      We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.




      Comment:




      • In (2) we apply the coefficient of operator according to (1).


      • In (3) use the linearity of the coefficient of operator.


      • In (4) we apply the binomial theorem.


      • In (5) we write the expression somewhat more conveniently.







      share|cite|improve this answer














      It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance



      begin{align*}
      n![z^n]e^{jz}=j^ntag{1}
      end{align*}




      We obtain
      begin{align*}
      color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
      &=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
      &=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
      &=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
      &=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
      &,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
      end{align*}



      We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.




      Comment:




      • In (2) we apply the coefficient of operator according to (1).


      • In (3) use the linearity of the coefficient of operator.


      • In (4) we apply the binomial theorem.


      • In (5) we write the expression somewhat more conveniently.








      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 9 at 19:53

























      answered Dec 9 at 19:40









      Markus Scheuer

      60k455143




      60k455143












      • I think I have found such a closed polynomial formula. See my answer below.
        – user
        Dec 27 at 12:06










      • @user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
        – Markus Scheuer
        Dec 27 at 14:05












      • What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
        – user
        Dec 27 at 16:14










      • @user: I agree, you're right. (+1) for your nice approach.
        – Markus Scheuer
        Dec 27 at 16:27


















      • I think I have found such a closed polynomial formula. See my answer below.
        – user
        Dec 27 at 12:06










      • @user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
        – Markus Scheuer
        Dec 27 at 14:05












      • What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
        – user
        Dec 27 at 16:14










      • @user: I agree, you're right. (+1) for your nice approach.
        – Markus Scheuer
        Dec 27 at 16:27
















      I think I have found such a closed polynomial formula. See my answer below.
      – user
      Dec 27 at 12:06




      I think I have found such a closed polynomial formula. See my answer below.
      – user
      Dec 27 at 12:06












      @user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
      – Markus Scheuer
      Dec 27 at 14:05






      @user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
      – Markus Scheuer
      Dec 27 at 14:05














      What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
      – user
      Dec 27 at 16:14




      What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
      – user
      Dec 27 at 16:14












      @user: I agree, you're right. (+1) for your nice approach.
      – Markus Scheuer
      Dec 27 at 16:27




      @user: I agree, you're right. (+1) for your nice approach.
      – Markus Scheuer
      Dec 27 at 16:27











      1














      I'll play with the cosh and
      see if I get
      anything other than
      the original problem.



      $begin{array}\
      cosh^n(x)
      &=frac1{2^n}(e^x+e^{-x})^n\
      &=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
      &=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
      &=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
      &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
      &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
      &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
      &=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
      &=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
      &=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
      &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
      &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
      &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
      &=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
      end{array}
      $



      And this is the OP.



      Oh well.






      share|cite|improve this answer


























        1














        I'll play with the cosh and
        see if I get
        anything other than
        the original problem.



        $begin{array}\
        cosh^n(x)
        &=frac1{2^n}(e^x+e^{-x})^n\
        &=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
        &=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
        &=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
        &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
        &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
        &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
        &=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
        &=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
        &=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
        &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
        &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
        &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
        &=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
        end{array}
        $



        And this is the OP.



        Oh well.






        share|cite|improve this answer
























          1












          1








          1






          I'll play with the cosh and
          see if I get
          anything other than
          the original problem.



          $begin{array}\
          cosh^n(x)
          &=frac1{2^n}(e^x+e^{-x})^n\
          &=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
          &=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
          &=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
          &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
          &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
          &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
          &=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
          &=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
          end{array}
          $



          And this is the OP.



          Oh well.






          share|cite|improve this answer












          I'll play with the cosh and
          see if I get
          anything other than
          the original problem.



          $begin{array}\
          cosh^n(x)
          &=frac1{2^n}(e^x+e^{-x})^n\
          &=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
          &=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
          &=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
          &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
          &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
          &=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
          &=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
          &=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
          &=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
          end{array}
          $



          And this is the OP.



          Oh well.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 at 21:48









          marty cohen

          72.5k549127




          72.5k549127























              1














              Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
              positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
              $$
              F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
              =frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
              =sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
              =sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
              $$



              To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
              $$
              F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
              $$

              where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.



              By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
              $$
              F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
              $$

              with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.






              share|cite|improve this answer




























                1














                Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
                positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
                $$
                F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
                =frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
                =sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
                =sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
                $$



                To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
                $$
                F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
                $$

                where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.



                By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
                $$
                F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
                $$

                with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.






                share|cite|improve this answer


























                  1












                  1








                  1






                  Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
                  positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
                  $$
                  F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
                  =frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
                  =sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
                  =sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
                  $$



                  To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
                  $$
                  F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
                  $$

                  where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.



                  By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
                  $$
                  F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
                  $$

                  with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.






                  share|cite|improve this answer














                  Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
                  positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
                  $$
                  F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
                  =frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
                  =sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
                  =sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
                  $$



                  To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
                  $$
                  F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
                  $$

                  where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.



                  By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
                  $$
                  F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
                  $$

                  with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 27 at 11:41

























                  answered Dec 26 at 19:53









                  user

                  3,6361627




                  3,6361627






























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