Assume $b_0=1$ to simplify things. You want a closed formula for the recursively defined sequence $$d_0=1$$ $$d_n=-\sum_{k=1}^nb_kd_{n-k}. $$ Let $\alpha=(\alpha_1,\dots,\alpha_r)\in \mathbb{N}_ +^\omega$ be a multi-index with length $l(\alpha):=r$ and weight $|\alpha|:=\sum_{j=1}^r\alpha_j$. Let's denote $b_\alpha:=b_{\alpha_1}\dots b_{\alpha_r}$.

We have (induction) $$d_n:=\sum_{|\alpha|=n}(-1)^{l(\alpha)}b_\alpha. $$

There are of course sereral equal terms in the sum, due to the commutativity; summing equal terms a corresponding smaller set of indices would be the increasing multi-indices (the number of terms in the sum would then be the number of partitions $p(n)$).

Assume $b_0=1$ to simplify things. You want a closed formula for the recursively defined sequence $$d_0=1$$ $$d_n=-\sum_{k=0}^{n-1}d_kb_{n-k}. $$ Let $\alpha=(\alpha_1,\dots,\alpha_r)\in \mathbb{N}_ +^\omega$ be a multi-index with length $l(\alpha):=r$ and weight $|\alpha|:=\sum_{j=1}^r\alpha_j$. Let's denote $b_\alpha:=b_{\alpha_1}\dots b_{\alpha_r}$.

We have (induction) $$d_n:=\sum_{|\alpha|=n}(-1)^{l(\alpha)}b_\alpha. $$

There are of course several equal terms in the sum, due to the commutativity; summing equal terms, a corresponding smaller set of indices would be the increasing multi-indices (the number of terms in the sum would then be the number of partitions $p(n)$).