By multinomial expansion formula, we know that $$ \sum_{p_1 + \cdots + p_k = r} \binom{r}{p_1,\ldots,p_k} = k^r, $$ where the multinomial coefficient is defined by $ \binom{r}{p_1, \ldots, p_k} := \frac{r!}{p_1!\cdots p_k!}$. Here is my question:
How can we find the sum $$ \sum_{p_1 + \cdots + p_k=r} \binom{r}{p_1,\ldots,p_k} $$ with the restriction that all $ p_j $'s are even? This sum shows up in some multiple commutators of Hilbert space operators. Any hint is greatly appreciated.
If you take the averaged sum over all choices of signs $$\frac{1}{2^k} \sum_{\varepsilon_i = \pm 1} (\varepsilon_1x_1 + \cdots + \varepsilon_kx_k)^r$$ we see that only the terms with even exponents survive. If we place all $x_i=1$ we get the quantity that you are interested in. This is more explicitly equal to $$ \frac{1}{2^k} \left( \sum_{m=0}^k {k \choose m} (k-2m)^r \right).$$
– Gjergji Zaimi, Aug 24 at 0:45
The sum is the coefficient of $x^r/r!$ in $\cosh^kx$.
– Ira Gessel, Aug 24 at 3:53
– Max Alekseyev Aug 24 at 9:38