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












1












$begingroup$


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











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – ajotatxe
    Dec 9 '18 at 18:10












  • $begingroup$
    @ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
    $endgroup$
    – user
    Dec 9 '18 at 19:30






  • 1




    $begingroup$
    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).
    $endgroup$
    – darij grinberg
    Dec 11 '18 at 3:16






  • 1




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






  • 1




    $begingroup$
    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.
    $endgroup$
    – darij grinberg
    Dec 11 '18 at 3:45


















1












$begingroup$


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











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – ajotatxe
    Dec 9 '18 at 18:10












  • $begingroup$
    @ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
    $endgroup$
    – user
    Dec 9 '18 at 19:30






  • 1




    $begingroup$
    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).
    $endgroup$
    – darij grinberg
    Dec 11 '18 at 3:16






  • 1




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






  • 1




    $begingroup$
    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.
    $endgroup$
    – darij grinberg
    Dec 11 '18 at 3:45
















1












1








1


0



$begingroup$


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











$endgroup$




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




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 '18 at 6:06







user

















asked Dec 9 '18 at 18:08









useruser

3,9051627




3,9051627












  • $begingroup$
    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.
    $endgroup$
    – ajotatxe
    Dec 9 '18 at 18:10












  • $begingroup$
    @ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
    $endgroup$
    – user
    Dec 9 '18 at 19:30






  • 1




    $begingroup$
    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).
    $endgroup$
    – darij grinberg
    Dec 11 '18 at 3:16






  • 1




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






  • 1




    $begingroup$
    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.
    $endgroup$
    – darij grinberg
    Dec 11 '18 at 3:45




















  • $begingroup$
    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.
    $endgroup$
    – ajotatxe
    Dec 9 '18 at 18:10












  • $begingroup$
    @ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
    $endgroup$
    – user
    Dec 9 '18 at 19:30






  • 1




    $begingroup$
    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).
    $endgroup$
    – darij grinberg
    Dec 11 '18 at 3:16






  • 1




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






  • 1




    $begingroup$
    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.
    $endgroup$
    – darij grinberg
    Dec 11 '18 at 3:45


















$begingroup$
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.
$endgroup$
– ajotatxe
Dec 9 '18 at 18:10






$begingroup$
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.
$endgroup$
– ajotatxe
Dec 9 '18 at 18:10














$begingroup$
@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
$endgroup$
– user
Dec 9 '18 at 19:30




$begingroup$
@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
$endgroup$
– user
Dec 9 '18 at 19:30




1




1




$begingroup$
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).
$endgroup$
– darij grinberg
Dec 11 '18 at 3:16




$begingroup$
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).
$endgroup$
– darij grinberg
Dec 11 '18 at 3:16




1




1




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




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




1




1




$begingroup$
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.
$endgroup$
– darij grinberg
Dec 11 '18 at 3:45






$begingroup$
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.
$endgroup$
– darij grinberg
Dec 11 '18 at 3:45












3 Answers
3






active

oldest

votes


















4












$begingroup$

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











$endgroup$













  • $begingroup$
    I think I have found such a closed polynomial formula. See my answer below.
    $endgroup$
    – user
    Dec 27 '18 at 12:06










  • $begingroup$
    @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 $n$.
    $endgroup$
    – Markus Scheuer
    Dec 27 '18 at 14:05












  • $begingroup$
    What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
    $endgroup$
    – user
    Dec 27 '18 at 16:14










  • $begingroup$
    @user: I agree, you're right. (+1) for your nice approach.
    $endgroup$
    – Markus Scheuer
    Dec 27 '18 at 16:27



















1












$begingroup$

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









$endgroup$





















    1












    $begingroup$

    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:
    $$
    F (n,k)=sum_{l=1}^n T (k,l)n^underline{l},tag {1}
    $$
    where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size, and $n^underline{l}$ is falling factorial. $T(k,l)$ can be recognized as the OEIS sequence A156289 with known close-form and recurrence expressions.




    Note added: 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












      $begingroup$

      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











      $endgroup$













      • $begingroup$
        I think I have found such a closed polynomial formula. See my answer below.
        $endgroup$
        – user
        Dec 27 '18 at 12:06










      • $begingroup$
        @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 $n$.
        $endgroup$
        – Markus Scheuer
        Dec 27 '18 at 14:05












      • $begingroup$
        What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
        $endgroup$
        – user
        Dec 27 '18 at 16:14










      • $begingroup$
        @user: I agree, you're right. (+1) for your nice approach.
        $endgroup$
        – Markus Scheuer
        Dec 27 '18 at 16:27
















      4












      $begingroup$

      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











      $endgroup$













      • $begingroup$
        I think I have found such a closed polynomial formula. See my answer below.
        $endgroup$
        – user
        Dec 27 '18 at 12:06










      • $begingroup$
        @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 $n$.
        $endgroup$
        – Markus Scheuer
        Dec 27 '18 at 14:05












      • $begingroup$
        What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
        $endgroup$
        – user
        Dec 27 '18 at 16:14










      • $begingroup$
        @user: I agree, you're right. (+1) for your nice approach.
        $endgroup$
        – Markus Scheuer
        Dec 27 '18 at 16:27














      4












      4








      4





      $begingroup$

      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











      $endgroup$



      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 '18 at 19:53

























      answered Dec 9 '18 at 19:40









      Markus ScheuerMarkus Scheuer

      60.8k456145




      60.8k456145












      • $begingroup$
        I think I have found such a closed polynomial formula. See my answer below.
        $endgroup$
        – user
        Dec 27 '18 at 12:06










      • $begingroup$
        @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 $n$.
        $endgroup$
        – Markus Scheuer
        Dec 27 '18 at 14:05












      • $begingroup$
        What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
        $endgroup$
        – user
        Dec 27 '18 at 16:14










      • $begingroup$
        @user: I agree, you're right. (+1) for your nice approach.
        $endgroup$
        – Markus Scheuer
        Dec 27 '18 at 16:27


















      • $begingroup$
        I think I have found such a closed polynomial formula. See my answer below.
        $endgroup$
        – user
        Dec 27 '18 at 12:06










      • $begingroup$
        @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 $n$.
        $endgroup$
        – Markus Scheuer
        Dec 27 '18 at 14:05












      • $begingroup$
        What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
        $endgroup$
        – user
        Dec 27 '18 at 16:14










      • $begingroup$
        @user: I agree, you're right. (+1) for your nice approach.
        $endgroup$
        – Markus Scheuer
        Dec 27 '18 at 16:27
















      $begingroup$
      I think I have found such a closed polynomial formula. See my answer below.
      $endgroup$
      – user
      Dec 27 '18 at 12:06




      $begingroup$
      I think I have found such a closed polynomial formula. See my answer below.
      $endgroup$
      – user
      Dec 27 '18 at 12:06












      $begingroup$
      @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 $n$.
      $endgroup$
      – Markus Scheuer
      Dec 27 '18 at 14:05






      $begingroup$
      @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 $n$.
      $endgroup$
      – Markus Scheuer
      Dec 27 '18 at 14:05














      $begingroup$
      What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
      $endgroup$
      – user
      Dec 27 '18 at 16:14




      $begingroup$
      What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
      $endgroup$
      – user
      Dec 27 '18 at 16:14












      $begingroup$
      @user: I agree, you're right. (+1) for your nice approach.
      $endgroup$
      – Markus Scheuer
      Dec 27 '18 at 16:27




      $begingroup$
      @user: I agree, you're right. (+1) for your nice approach.
      $endgroup$
      – Markus Scheuer
      Dec 27 '18 at 16:27











      1












      $begingroup$

      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









      $endgroup$


















        1












        $begingroup$

        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









        $endgroup$
















          1












          1








          1





          $begingroup$

          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









          $endgroup$



          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 '18 at 21:48









          marty cohenmarty cohen

          73.2k549128




          73.2k549128























              1












              $begingroup$

              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:
              $$
              F (n,k)=sum_{l=1}^n T (k,l)n^underline{l},tag {1}
              $$
              where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size, and $n^underline{l}$ is falling factorial. $T(k,l)$ can be recognized as the OEIS sequence A156289 with known close-form and recurrence expressions.




              Note added: 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











              $endgroup$


















                1












                $begingroup$

                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:
                $$
                F (n,k)=sum_{l=1}^n T (k,l)n^underline{l},tag {1}
                $$
                where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size, and $n^underline{l}$ is falling factorial. $T(k,l)$ can be recognized as the OEIS sequence A156289 with known close-form and recurrence expressions.




                Note added: 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











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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:
                  $$
                  F (n,k)=sum_{l=1}^n T (k,l)n^underline{l},tag {1}
                  $$
                  where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size, and $n^underline{l}$ is falling factorial. $T(k,l)$ can be recognized as the OEIS sequence A156289 with known close-form and recurrence expressions.




                  Note added: 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











                  $endgroup$



                  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:
                  $$
                  F (n,k)=sum_{l=1}^n T (k,l)n^underline{l},tag {1}
                  $$
                  where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size, and $n^underline{l}$ is falling factorial. $T(k,l)$ can be recognized as the OEIS sequence A156289 with known close-form and recurrence expressions.




                  Note added: 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.







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                  edited Jan 1 at 10:58

























                  answered Dec 26 '18 at 19:53









                  useruser

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