from mpmath import mp, gammainc, findroot, exp, factorial
mp.dps = 30; k = 3
T = lambda x: gammainc(k - 1, 0, x, regularized=True) # P(Po(x) >= k-1)
x = findroot(lambda x: T(x) - exp(-x)*x**(k-1)/factorial(k-2), k)
print(x / T(x)) # c_kEach entry is the value at the minimiser $\lambda_k$ of $f(\lambda)=\lambda/\mathbb{P}(\mathrm{Po}(\lambda)\geq k-1)$, computed in ball arithmetic with 64 guard bits beyond the 100 digits written.
That $\lambda_k$ is the unique minimiser and the minimum is global follows from the sign of $f'$, which is the sign of $g$ in (2): $g$ is negative on $(0,\lambda_k)$ and positive on $(\lambda_k,\infty)$, so $f$ decreases and then increases. The computation encloses $\lambda_k$ by bisection on the sign of $g$ between $k-2$ and $4k+10$, each sign decided in ball arithmetic, down to a bracket of half-width $10^{-106}$ whose ends are then checked to give $g$ opposite signs, and evaluates $f$ on that bracket as a ball; the Poisson tail is the finite sum $1-e^{-\lambda}\sum_{j<k-1}\lambda^j/j!$. Every ball has radius about $6\cdot10^{-106}$. Before any entry was written, all ten values were compared with a computation in mpmath at 120 digits that writes the tail as the regularised incomplete gamma function $\gamma(k-1,\lambda)/\Gamma(k-1)$ and finds the minimiser as a root of the numerical derivative of $f$, which agreed to the last written digit in every row; $c_3\approx3.35$ and the size $0.27\,n$ of the newborn $3$-core were compared with the values Pittel, Spencer and Wormald state.