Let $S_n = \{ 1, 3, ... \}$ be the set of odd numbers not greater than $n$ and $(x)_n$ be the falling factorial $x(x-1)\cdots(x-n+1)$. I am interested in the following partial sum: $$ \sum_{i \in S_n} \binom{x}{i} i! {n \brace i} = \sum_{i \in S_n} {n \brace i} (x)_i, $$ where ${n \brace i}$ is a Stirling numbers of the second kind.
I know that $\sum_{i = 1}^{n} \binom{x}{i} i! {n \brace i} = \sum_{i = 1}^{n} {n \brace i} (x)_i = x^n$ but it is not clear for me how to simplify this partial sum with respect to $S_n$ only. It would be appreciated if someone could give me any hint to simplify or bound this partial sum.
Not an answer, merely a remark. Let $$O_n(x):=\sum_{i \ odd} {n \brace i} (x)_i.$$ Then by using the basic recurrences for binomial coefficients and Stirling numbers, we see that $O_n$ satisfies the recursion: $$O_n(x)=x\left(O_{n-1}(x)-2O_{n-1}(x-1)+(x-1)^{n-1}\right)$$ with initial condition $O_1(x)=1$. Not sure whether it is useful for the question though.
Edit
This recursion can be used to derive the first terms of an expansion for $O_n(x)$: $$O_n(x)=-\frac{1-(-1)^n}{2}(-x)^n-\frac{n(n-1)}{2}(-x)^{n-1}-\frac{n(n-1)^2(n-2)}{4}(-x)^{n-2}- \cdot\cdot\cdot$$ The calculations become increasingly laborious by this method.
Now, by using the expression of the falling factorial in terms of Stirling numbers of the first kind ${n \brack k}$, one obtains $$ O_n(x)= -\sum_{h\ge0} P_h(n)(-x)^{n-h} $$ where $$ P_h(n)= \sum_{i\ge 0} {n \brace 2i+1}{2i+1 \brack n-h} $$ which is indeed a polynomial function of $n$ of degree $2h$. This can be seen when written as $$ P_h(n)= \sum_{i= 0}^h {n \brace n-i}{n-i \brack n-i-(h-i)} \frac{1-(1-)^{n-i}}{2}.$$ Each summand in the former is the product of a polynomial in $n$ of degree $2i$ and a polynomial in $n$ of degree $2h-2i$.