Given positive integers $n,m$ and the vector of ones $v_n = (1,1,1,\ldots,1) \in \{1\}^n$, is it possible to find an explicit formula for the $i$-th entry of the repeated convolution $C$ of $v_n$ with itself?
$$C = *_{k=1}^m v_n = \underbrace{v_n * v_n * \ldots * v_n}_{m \text{ times}}$$
(Here $*$ denotes the convolution operator.)
For example for $n=2$ we get $C = \left( \binom{n}{0} , \binom{n}{1}, \binom{n}{2},\ldots,\binom{n}{n}\right)$. This is no coincidence as discrete convolution is nothing but polynomial multipication. So $v_n$ could also be represented by the polynomial $\frac{x^{n}-1}{x-1} = x^{n-1}+\ldots+x+1$, and finding the entries of $C$ would mean finding the coefficient of $x^i$ in $\left(\frac{x^n-1}{x-1}\right)^m = (1+x+x^2 +\ldots+x^{n-1})^m$, but I did not succeed finging the $i$-th entry with either approach, except for some specific examples.
There is a nice answer! You want the coefficient of $x^i$ in $$ (1-x^n)^m\cdot (1-x)^{-m}=\sum_i c_ix^i $$ If you write $$ (1-x^n)^m = \sum_j a_jx^j, \qquad (1-x)^{-m} = \sum_j b_jx^j, $$ then you would get $$ c_i = \sum_j a_j b_{i-j} $$ All that remains is to find $a_j$ and $b_j$. These can be found using the (generalized) binomial theorem: $$ (1-x^n)^m = \sum_k \binom{m}{k}(-1)^kx^{nk}\implies a_j=\begin{cases}(-1)^k\binom{m}{k} & j = kn \\ 0 & \text{otherwise}\end{cases} $$ $$ (1-x)^{-m} = \sum_{i=0}^\infty \binom{-m}{k} (-1)^kx^k =\sum_{i=0}^\infty \binom{m+k-1}{k}x^k \implies b_j = \binom{m+j-1}{j} $$ For the $\binom{-m}{k}$ part, note that $$ \binom{-m}k=\frac{(-m)(-m-1)(-m-2)\dots(-m-(k-1))}{k!}=(-1)^k\frac{m(m+1)\dots(m+k-1)}{k!}=(-1)^k\binom{m+k-1}{k} $$