Let $(a_k)_{k\in\mathbb{Z}}$ be a sequence of complex numbers, such that $a_0 = 1$, $a_{-k} = a_k^*$ (i.e. Hermitian). We say that $(a_k)_k$ is positive definite if for any $n\in\mathbb{N}$ and $z_0,\ldots,z_n\in\mathbb{C}$, it holds that $$\sum_{j,k=0}^n a_{j-k}z_j z_k^* \geq 0. \tag{1}$$
Fix some $a_1\in\mathbb{C}$, and suppose now that our sequence of coefficients $(a_k)_k$ is recursively defined, for $k\geq 2$, by $$a_k = \frac{1}{k-1}\sum_{l=1}^{k-1} a_l a_{k-l}.$$ The assumption that the sequence is Hermitian fixes the values of $a_{-k}$ for $k\geq 1$.
Question. Is it possible to choose $a_1$ so that the sequence $(a_k)_k$ is positive definite?
I was thinking some sort of induction argument based on the value of $n$ in (1), but I get stuck in the induction step. If it helps, my motivation for this question is about obtaining a probability measure (via Bochner's theorem) from this recursively defined sequence.
You can show by strong induction that $a_k = (a_1)^k$ for every $k\geq 1$. Indeed, the base case is tautological. Suppose there exists some $k_0\geq 1$ such that $a_k = (a_1)^k$ for every $1\leq k \leq k_0$. Then $$a_{k_0+1} = \frac{1}{k_0}\sum_{j=1}^{k_0}a_j a_{k_0+1-j} = \frac{1}{k_0}\sum_{j=1}^{k_0} (a_1)^j (a_1)^{k_0+1-j} = (a_1)^{k_0+1}.$$
Then for any $n\in\mathbb{N}$, $z_0,\ldots,z_n \in\mathbb{C}$, \begin{align}\sum_{j,k=0}^n a_{j-k}z_jz_k^* &= \sum_{j,k=0}^n (a_1)^{j-k}z_jz_k^* = \left|\sum_{j=0}(a_1)^jz_j\right|^2\geq0.\end{align}