The Bernoulli-Barnes polynomials $B_n^{(2)}(x;a_1,a_2)$ with generating function
$$ \sum_{n\geqslant0}\frac{B_n^{(2)} (x;a_1,a_2)}{n!} z^n = \left(\frac{a_1 z}{e^{a_1 z}-1}\right)\left(\frac{a_2 z}{e^{a_2 z}-1}\right) e^{zx} $$ satisfy, for $0\leqslant a_1,a_2\leqslant1$ and $a_1+a_2=1,$ the identity $$ B_n^{(2)}(x;a_1,a_2) = a_1 B_n^{(2)} (x;1,a_2) + a_2 B_n^{(2)} (x;1,a_1)+a_1 a_2 nB_{n-1}(x) $$ where $B_n(x)$ is the usual Bernoulli polynomial with generating function $$ \sum_{n\geqslant0}\frac{B_n(x)}{n!}z^n=\frac{z}{e^{z}-1}e^{zx}. $$ Since we always assume $a_1+a_2=1,$ we may simplify the notations and write $B_n^{(2)}(x;a_1,a_2) = B_n(x;a_1).$ Moreover, since $$ B_n^{(2)}(x;1,a)=(1+a)^n B_n^{(2)}\left(\frac{x}{1+a};\frac{1}{1+a},\frac{a}{1+a}\right) = (1+a)^{n}B_n \left(\frac{x}{1+a};\frac{1}{1+a}\right) $$ with $\frac{1}{1+a}+\frac{a}{1+a}=1,$ we have \begin{align} B_n(x;a_1) = {} & a_1(1+a_2)^n B_n\left(\frac{x}{1+a_2};\frac{a_2}{1+a_{2}}\right) \\[8pt] & {} + a_2 (1+a_1)^n B_n \left(\frac{x}{1+a_1};\frac{a_1}{1+a_1}\right) +a_1(1-a_1) nB_{n-1} (x) \tag1\label{1} \end{align} Now this can be iterated: each $\displaystyle B_n\left(\frac{x}{1+a_2};\frac{a_2}{1+a_2}\right)$ and $\displaystyle B_n\left(\frac{x}{1+a_{1}};\frac{a_1}{1+a_1}\right)$ satisfies the recurrence \eqref{1} and can in turn be expressed as the sum of $3$ terms.
My main question: Can we write an explicit expression for the infinitely iterated version of \eqref{1}?
Notice that each iteration transforms $a_{1}$ into $\frac{1}{1+a_{1}}$ and then into $\cfrac{1}{1+\cfrac{1}{1+a_1}}\dots$ which is related to the sequence of convergents $\left[1,1,1,1,\dots\right]$ of the continued fraction associated to $\varphi=\frac{1+\sqrt{5}}{2}.$