I'm interested in the sequence $y_n$, defined for $n=0,1,2,\dots$ according to $$ y_n=\lim_{x\to\infty}\left((-x)^n\left[\frac1e\left(1+\frac1x\right)^x-\sum_{k=0}^{n-1}y_k (-x)^{-k}\right]\right). $$ The first few values are $y_0=1$, $y_1=\frac12$, $y_2=\frac{11}{24}$, $y_3=\frac{7}{16}$, which are the Taylor coefficients of the function $f(x)=\frac1e(1-x)^{-\frac 1x}$ at $x=0$.
Does this sequence converge as $n\to\infty$, and if so, to what? Is there a simple expression for $y_n$ that does not involve a derivative?
Some answers are given in the article referenced in this OEIS entry (given in comments to the "duplicated" question). An expression for $y_n$ given there is a double sum involving Stirling numbers (of the first kind). For computation, I would probably stay with the recurrence $(n+1)y_{n+1}=\sum\limits_{k=0}^{n}\frac{k+1}{k+2}y_{n-k}$ coming from [multiplying power series in] $f'(x)=f(x)g'(x)$ where $g(x)=-\ln(1-x)/x-1$.
If we know that $y_n$ converges, then $\lim\limits_{n\to\infty}y_n=\lim\limits_{x\to1^-}(1-x)f(x)=1/e$ is easy.