Apply [Lagrange reversion](https://en.m.wikipedia.org/wiki/Lagrange_reversion_theorem) to @TheSimpliFire’s equation:

$$\frac{k}{(k-1)^2}+1=e^k\iff k=1+\sqrt{\frac k{e^k-1}}\\\implies k=1+\sum_{n=1}^\infty\frac1{n!}\left.\frac{d^{n-1}}{dx^{n-1}}\left(\frac x{e^x-1}\right)^\frac n2\right|_1$$

Now use the Norlund polynomial, or [generalized Bernoulli number](https://www.fq.math.ca/Papers1/53-4/MollVignat04222015.pdf), [generating function](https://functions.wolfram.com/Polynomials/NorlundB2/02/). One notices the $\frac1{(n-m+1)!}$ truncates the inner sum:

$$\boxed{c=\frac{k^3}{k^2-k+1},k=\sum_{n=0}^\infty\sum_{m=0}^{n+1}\frac{B_n^{(\frac m2)}}{m! (n-m+1)!}}$$

[shown here](https://www.wolframalpha.com/input?i=evaluate+c%3Dk%5E3.0%2F%28k%5E2-k%2B1%29%2Ck%3Dsum%5Bsum%5BNorlundB%5Bm%2Cn%2F2.0%5D%2F%28n%21+%28m-n%2B1%29%21%29%2C%7Bm%2C0%2C20%7D%5D%2C%7Bn%2C0%2C20%7D%5D)