NOTATION
I'm a physicist studying higher-spin theory. In my research, we work with fully symmetric tensors using a notation which is implicit both in the dimension of space and the order of the tensor, i.e. an order-$s$ fully symmetric tensor is simply written as
$$ \phi_{\mu_1 \cdots \mu_s} \equiv \phi. $$
The $n$-th gradient of $\phi$ is written as $\partial^n \phi$, the $n$-th divergence, as $\partial^n \cdot \phi$ and the $n$-th trace as $\phi^{[n]}$. Lower traces are simply written with a prime, e.g. $\phi''$ for the second trace.
All indices are implicitly symmetrized, without weight factors, using the minimal number of terms.
For example, if $s=2$,
$$ \partial^2 \phi \equiv \partial_\mu \partial_\nu \phi_{\sigma \rho} + \partial_\mu \partial_\sigma \phi_{\nu \rho} + \partial_\mu \partial_\rho \phi_{\sigma \nu} + \partial_\nu \partial_\sigma \phi_{\mu \rho} + \partial_\nu \partial_\rho \phi_{\sigma \mu} + \partial_\sigma \partial_\rho \phi_{\mu \nu} $$
$$ \partial (\partial \cdot \phi) \equiv \partial_\nu (\partial^\lambda \phi_{\mu \lambda}) + \partial_\mu (\partial^\lambda \phi_{\lambda \nu})$$
$$ \eta \partial^2 \cdot \phi \equiv \eta_{\mu\nu} \partial^\rho \partial^\sigma \phi_{\rho \sigma} $$
The formalism implies the following set of rules
$$ ( \partial^p \phi )' = \Box \partial^{p-2} \phi + 2 \partial^{p-1} \left( \partial \cdot \phi \right) + \partial^p \phi' $$
$$ \partial \cdot (\partial^p \phi) = \Box \partial^{p-1} \phi + \partial^p \left( \partial \cdot \phi \right) $$
$$ \partial^p \partial^q = {{p+q}\choose{q}} \partial^{p+q} $$
which you can simply take as axioms of the game. By the way, $\Box = \partial \cdot \partial$, which we are allowed to invert to $\frac{1}{\Box}$, but don't worry about what this means, you can just use the three rules written above.
PROBLEM
So, the thing I'm interested in is the recursion relation $$\mathcal{F}_{n+1} = \mathcal{F}_{n} - \frac{1}{n+1} \frac{\partial}{\Box} \left( \partial \cdot \mathcal{F}_{n} \right) + \frac{1}{(n+1)(2n+1)} \frac{\partial^2}{\Box} \mathcal{F}_{n}{}'$$ with the initial condition $$ \mathcal{F}_0 = \Box \phi.$$
I would like solve the recursion fully, i.e. to express $\mathcal{F}_{n+1}$ only in terms of $\phi$'s and not $\mathcal{F}_{k}'s$ with $k<n+1$.
It's easy to rewrite $\mathcal{F}_{n+1}$ as $$\mathcal{F}_{n+1} = \mathcal{F}_{n} - \frac{1}{n(2n+1)} \frac{\partial^2}{\Box} \mathcal{F}_{n}' + \frac{1+2n}{n} \frac{\partial^{2n+2}}{\Box^n} \phi^{[n+1]}$$ but I don't know how to proceed from there and I've been trying for days. I could really use some help.
Please do tell me if you want me to clarify anything.
EDIT 1
By lower traces I simply mean a number of traces low enough to actually write out the primes explicitly, so, for example, $\phi''''$ could just be written as $\phi^{[4]}$.
EDIT 2
I provide here the explicit form of first two $\mathcal{F}_n$ after $n=0$.
$$ \mathcal{F}_1 = \Box \phi - \partial (\partial \cdot \phi) + \partial^2 \phi'$$
$$ \mathcal{F}_2 = \Box \phi - \partial (\partial \cdot \phi) + \frac{2}{3 \Box} \partial^2 (\partial^2 \cdot \phi) + \frac{1}{3} \partial^2 \phi' - \frac{1}{\Box} \partial^3 (\partial \cdot \phi') + \frac{1}{\Box} \partial^4 \phi''$$
