In the frame of a more complicated problem, I need to find the zero of function $$f(x)=\Gamma(k+2-x)-k\,\Gamma(k+1)$$ which is the same as the zero of function $$g(x)=\log \big((\Gamma (k+2-x)\big)-\log \big(k \,\Gamma (k+1)\big)$$ This does not present any difficulty from a numerical point of view.
By pure curiosity, I wondered if it would be possible to find a good formal approximation of the solution.
Newton iterates being $$x_{n+1}=x_{n}-\frac{g(x_n)}{g'(x_n)} \qquad \text{with}\qquad x_0=0$$
this was done rigorously a few times and, only later, were used the asymptotics of the appearing gamma and polygamma functions for large values of $k$,
It very quickly appears that $$\large\color{blue}{x=\sum_{n=1}^\infty (-1)^{n+1} \frac {P_n(t)}{a_n\,(k \,\log (k))^n }}$$ where the $a_n$ are Hirzebruch's numbers (sequence $A091137$ in $OEIS$) and the $P_n$ are polynomials of degree $(n-1)$ in $\color{blue}{t=\log(k)}$, the first ones being $$\left( \begin{array}{cc} n & P_n(t) \\ 1 & 1 \\ 2 & t+3 \\ 3 & 4 t^2+22 t+33 \\ 4 & 6 t^3+49 t^2+135 t+135 \\ 5 & 144 t^4+1604 t^3+6365 t^2+11820 t+8865 \\ 6 & 240 t^5+3456 t^4+18089 t^3+48150 t^2+67725 t+40635 \\ \end{array} \right)$$
where I did not find any pattern beside the fact that the coefficient of the highest power is $\frac {a_n}{n+1}$ and that the constant term is a multiple of $3$.
They cannot factor and they do not seem to have any positive real roots.
Using the given terms, it seems to work decently even for small values of $k$
$$\left( \begin{array}{ccc} k & \text{approximation} & \text{solution} & \text{|difference|}\\ 10 & \color{red}{0.03904}555647 & 0.03904613599 & 5.80\times 10^{-07}\\ 20 & \color{red}{0.01590409}605 & 0.01590409777 & 1.72\times 10^{-09}\\ 30 & \color{red}{0.009504630}69 & 0.00950463076 & 7.01\times 10^{-11}\\ \end{array} \right)$$
Even if it does not mean too much, defining $$x_{(p)}=\sum_{n=1}^p (-1)^{n+1} \frac {P_n(t)}{a_n\,(k \,\log (k))^n }+ O\left( \frac 1 {k^{p+1} \log (k) } \right)$$ $$\left|g(x_{(p)})\right| \sim \frac 1 {(p+1)\,k^{p+1}}$$
What is also "amazing" is that, using @Gary's approximation of the inverse of Stirling approximation $$x \sim k+\frac 32-\frac y {W\left(\frac{y}{e}\right)} \quad \text{where} \quad y=\log \left(\frac{k \, \Gamma (k+1)}{\sqrt{2 \pi }}\right)$$ is quite decent (for the three given examples, the results are $0.04054$, $0.01654$ and $0.00989$).
Now, and again by pure curiosity, are these polynomials $P_n(t)$ already known or encountered in any other problem ? .
Edit
Using as $$x_0=k+\frac 32-\frac y {W\left(\frac{y}{e}\right)} \quad \text{where} \quad y=\log \left(\frac{k \, \Gamma (k+1)}{\sqrt{2 \pi }}\right)$$ for $k=5$ (not covered above), the first iterate of Newton like methods of order $m$ is
$$\left( \begin{array}{cc} m & x_1^{(m)} & \text{method} \\ 2 & \color{red}{0.097746}333900023 & \text{Newton} \\ 3 & \color{red}{0.097746848}623607 & \text{Halley} \\ 4 & \color{red}{0.097746848454}469 & \text{Householder} \\ \cdots & \\ \infty & \color{red}{0.097746848454533} & \text{solution} \\ \end{array} \right)$$
What is impressive is that $$\left|g(x_{2})\right| \sim \frac{1}{5\times 10^3\, k^{10/3}} \qquad \text{and} \qquad \left|g(x_{3})\right| \sim \frac{1}{5\times 10^5\, k^{11/2}}$$
