Immediate problem
I have a sequence of numbers $(a_{i})_{0 \leq i \leq k}$ satisfying the recurrence relation
$$
a_0 = 1, \qquad a_{i+1} = -\sum_{j=1}^{i} a_{j}\frac{1}{(i+2 – j)!}.
$$
Thus, the recurrence relation satisfies, up to a sign, a convolution with the sequence and the reciprocal of the factorial.
The goal is to find an explicit expression of the $a_i$, independent of the $a_j$. I don't really know where to start. I tried looking for some solutions using ordinary or exponential generating functions (but have little experience in using these).
Background of the problem
Suppose we have a Taylor series of some function $v:\mathbb{R}\rightarrow \mathbb{R}^n$ and an exponential generating function, both having real-valued functions as coefficients (domain also $\mathbb{R}$), which we want to be equal up to some truncation error, say truncated to order 2 in parameter $h \in \mathbb{R}$ (we assume that the first function coefficients $v(t) = d_0(t)$ are equal):
$$
v'(t) + \frac{h}{2!}v''(t) + \frac{h^2}{3!}v'''(t) = d_1(t) + \frac{h}{2!}d_2(t) + \frac{h^2}{3!}d_3(t) + \mathcal{O}(h^3).
$$
(where the $\mathcal{O}$ is defined in some straightforward extension which is applicable to function-valued coefficients).
The goal is to turn this equation into an expression in $v'(t)$ by removing the derivatives of $v$ by substitution.
The equations used for this substitution are found by differentiating the original equation $1 \leq k \leq 2$ times, finding the expression (surpressing $t \in \mathbb{R}$)
\begin{align}
v^{(k+1)} &= – \frac{h}{2} v^{(k+2)} – \frac{h^2}{6}v^{(k+3)} + d^{(k)}_1 +\frac{h}{2} d^{(k)}_2 + \frac{h^2}{6}d^{(k)}_3 + \mathcal{O}(h^3).\\
\end{align}
Substituting this equation into the original equation equation in order of increasing $k$, one finds after some manipulations the equation
$$
v' + \mathcal{O}(h^3) = \sum_{i=1}^3 \sum_{j=i}^3 h^{j-1} \frac{a_i}{(j-(i-1))!} d^{(i-1)}_{j-(i-1)} = \sum_{i=1}^3 h^{i-1} a_i\sum_{j=1}^{4-i} \frac{h^{j-1}}{j!} d^{({j+i-2})}_{j},
$$
where the $a_k$ are defined as in the immediate problem.
Best Answer
After revisiting the problem after a few months. I noticed that the coefficients are similar to the Bernoulli numbers. I should have added the first few numbers $a_i$ in the question, which are, for example $$ \left\{1,-\frac{1}{2},\frac{1}{12},0,-\frac{1}{720},0,\frac{1}{30240},0,-\frac{1}{1209600},0,\frac{1}{47900160},0,-\frac{691}{1307674368000},0,\frac{1}{747 24249600},0,-\frac{3617}{10670622842880000},0,\frac{43867}{5109094217170944000},0\right\}. $$ In fact these are just the scaled Bernoulli numbers: $$ a_k = \frac{1}{k!}B_k. $$