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I have been motivated by answer on this question: Special sum of multinomial coefficients!

$$ S=\sum_{\substack{a_1+a_2+\dots+a_k=2n\\ a_i\text{ even}}}\frac{(2n)!}{a_1!a_2!\dots a_k!}=\frac{1}{2^k}\sum_{j=0}^{k}\binom{k}{j}(k-2j)^{2n}. $$

I am wondering if it is possible to find an explicit formula for

$$ S=\sum_{\substack{a_1+a_2+\dots+a_k=2n\\ \text{only $m\leq k$-components of } a_i\text{odd}}}\frac{(2n)!}{a_1!a_2!\dots a_k!} $$

Forbs
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  • What does "only $m\le 2n$-components of $a_i$ odd" mean? – anon Aug 03 '20 at 10:02
  • @runway44: Sorry, it was a typo. It should be that $m \leq k$ numbers of $a_i$ are odd. – Forbs Aug 03 '20 at 14:41
  • This may help: https://arxiv.org/abs/1710.02956 You want to choose one part and separate by even blocks and the other part by odd blocks. – Phicar Aug 03 '20 at 18:42
  • Ah, I think you're saying "exactly $m$ of the numbers $a_1,\cdots,a_k$ are odd." That was not clear to me from the phrasing "components of $a_i$" (the $a_i$s do not have "components"...) – anon Aug 04 '20 at 00:03

1 Answers1

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Let's use $n$ instead of $2n$ for greater generality.

Next, let's restrict to the first $m$ of the $a_i$s being odd:

$$ \sum_{m~\mathrm{of}~a_i~\mathrm{odd}}\binom{n}{a_1,\cdots,a_k} = \binom{n}{m}\sum_{\substack{a_1,\cdots,a_m~\mathrm{odd} \\ a_{m+1},\cdots,a_k~\mathrm{even}}}\binom{n}{a_1,\cdots,a_k} $$

If $f(x)=\sum f_k x^k$ is a polynomial then we have (anti-)symmetrization operators

$$ \frac{f(x)+f(-x)}{2}=\sum_{k~\mathrm{even}} f_kx^k, \quad \frac{f(x)-f(-x)}{2}=\sum_{k~\mathrm{odd}} f_kx^k. $$

So, for instance, we could write

$$ \sum_{k~\mathrm{odd}}\binom{n}{k} = \left.\frac{(1+x)^n-(1-x)^n}{2}\right|_{x=1}=2^{n-1}. $$

For us, though, we need $k$ variables $x_1,\cdots,x_k$. Define operators $R_i$ by

$$ R_i f(x_1,\cdots,x_i,\cdots,x_k) = f(x_1,\cdots,-x_i,\cdots,x_n) $$

with (anti-)symmetrizations $\frac{1}{2}(\mathrm{Id}\pm R_i)$. Then

$$ \sum_{\substack{a_1,\cdots,a_m~\mathrm{odd} \\ a_{m+1},\cdots,a_k~\mathrm{even}}}\binom{n}{a_1,\cdots,a_k} =\left.\frac{1}{2^k} \prod_{i=1}^m (\mathrm{Id}-R_i)\prod_{j=m+1}^k (\mathrm{Id}+R_j) (x_1+\cdots+x_k)^n\right|_{x_1,\cdots,x_k=1} $$

Abbreviate any product of $t$ distinct $R_i$s as $R^t$ for simplicity, in which case

$$ \left. R^t(x_1+\cdots+x_k)^n\right|_{x_1,\cdots,x_k=1} = (k-2t)^n. $$

The product of operators becomes

$$ \prod_{i=1}^m (\mathrm{Id}-R_i)\prod_{j=m+1}^k (\mathrm{Id}+R_j) = \sum_{t=0}^m \binom{m}{t} (-R)^t \sum_{s=0}^{k-m} \binom{k-m}{s}R^s. $$

In conclusion,

$$ \sum_{m~\mathrm{of}~a_i~\mathrm{odd}}\binom{n}{a_1,\cdots,a_k} = \frac{1}{2^k}\binom{n}{m}\sum_{t=0}^m \sum_{s=0}^{k-m}\binom{m}{t}\binom{k-m}{s} (-1)^t (k-2t-2s)^n. $$

anon
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  • May I clarify one thing here. Would the same procedure work if instead of operator $R_i$ we would take operator $R_i^l$-operator which would "switch" the sign of $l$ elements instead of only one element in the $i$-th position. Thank you. – Forbs Aug 14 '20 at 05:03
  • @Forbs I don't know what procedure you envision. – anon Aug 14 '20 at 05:31
  • Sorry, I will clarify: you are defining operator $R_i$ by $R_if(x_1,\ldots, x_i, \ldots, x_k)=(x_1, \ldots, -x_i, \ldots, x_k)$. For my specific case I need to define the operator $R=R_{i_i, \ldots, i_l}$ by $Rf(x_1, \ldots, x_k)=(x_1, \ldots, -x_{i_1},\ldots, -x_{i_l}, \ldots x_k)$. Would the final formula in your post depend on which operator $R$ I am chosing. Thank you – Forbs Aug 15 '20 at 01:52
  • Well, with the notation $[m]={1,\cdots,m}$, we have $$ \prod_{i=1}^m (I-R_i)=\sum_{T\subseteq[m]} (-1)^{|T|} R_T, \quad \prod_{j=m+1}^k (I+R_j)=\sum_{S\subseteq[k]\backslash[m]} R_S $$ where $R_J=\prod_{j\in J}R_j$. This doesn't give a different proof, it just fills in details of this proof (since "$R^t$" is a bit handwavy in this proof). I find your question strangely worded though: all true statements are true regardless of how you prove them, so yes the final formula doesn't "depend" on how you got the formula... – anon Aug 15 '20 at 04:11