If we know $\,a_n\,$ for certain $\,n \gg 1,$ then $a_{n-1}=y(a_n, n),\;$
wherein
$$y(x)=\dfrac{\ln(e+x)}{\ln(n)} = \dfrac{1+\ln(1+x/e)}{\ln(n)}
\approx \dfrac{{1+\dfrac xe-\dfrac{x^2}{2e^2}}}{{{\ln(n)}}}, \tag1 $$
Since denominators and numerators of $(1)$ changes slowly, then asymptotically $a_{n-1}\rightarrow a_n.$
Assuming $y= x$, easily to get the table of the approximated and exact estimations
$$a_{n}\approx
a_{n-1}\approx x\approx y\approx e\sqrt{e^2\ln^2 n - 2e \ln n+3} - e\ln n -1)\tag2$$
Exact solution has the closed form via analytic continuation of the Lambert W-function and has not essential differences from $(2).$

Iterations from the starting value $a_{20}$ (Table1) to $a_2$ (Table2) are shown at the right columns (at the inverse order).
Wolfram Alpha suffices all described calculations.