We introduce a family of rational polynomials, for integer $n \ge 0$, as
$$ r_{n}(x) = \sum_{k=0}^n \frac{ \sum_{j=0}^k x^j \binom{k}{j} (2j + 1)^n }{2^k}. $$
The formula for $r_{n}(x)$ is a variant of a well-known representation of the Euler numbers. Let $\operatorname{E}_n(x)$ denote the Euler polynomials and let us agree to write $ \operatorname{E}_n $ for $ 2^n \operatorname{E}_n(1/2) $. We observe
$$ {r}_{n}(-1) = \operatorname{E}_n \quad (A122045). $$
This leads to the following infinite lower triangular integer matrix $\operatorname{T}$ for $0 \le k \le n$: $$ \operatorname{T}(n, k) = 2^n [x^k] r_{n}(x) $$
$$ 1 $$ $$ 3 \quad 3 $$ $$ 7 \quad 36 \quad 25 $$ $$ 15 \quad 297 \quad 625 \quad 343 $$ $$ 31 \quad 2106 \quad 10000 \quad 14406 \quad 6561 $$ $$ 63 \quad 13851 \quad 131250 \quad 369754 \quad 413343 \quad 161051 $$ $$ \ldots $$
The triangle has the first column $\operatorname{T}(n,0) = 2^{n + 1} - 1$, and the main diagonal $ \operatorname{T}(n, n) = (2n + 1)^n$. But the key statement is
$$ \operatorname{E_n} = \frac{ \sum_{k=0}^n (-1)^k \operatorname{T}(n, k) }{2^n}. $$
Such a representation of the Euler numbers suggests asking for a combinatorial interpretation of the $\operatorname{T}(n, k)$.
Which combinatorial objects does $\operatorname{T}(n,k)$ count?
