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Apply Lagrange reversion to @TheSimpliFire’s equationequation:

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

Now use the Norlund polynomial, or generalized Bernoulli number, generating function. 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

Apply Lagrange reversion 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, generating function. 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

Apply Lagrange reversion 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, generating function. 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

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Apply Lagrange reversion to @TheSimpliFire’s equation:

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

Now use the Norlund polynomial, or generalized Bernoulli number, generating function to get. One notices the $\frac1{(n-m+1)!}$ truncates the inner sum:

$$\boxed{c=\frac{k^3}{k^2-k+1},k=\sum_{m,n=0}^\infty\frac{B_m^{(\frac n2)}}{n! (m-n+1)!}}$$$$\boxed{c=\frac{k^3}{k^2-k+1},k=\sum_{n=0}^\infty\sum_{m=0}^{n+1}\frac{B_n^{(\frac m2)}}{m! (n-m+1)!}}$$

shown here

Apply Lagrange reversion to @TheSimpliFire’s equation:

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

Now use the Norlund polynomial, or generalized Bernoulli number, generating function to get:

$$\boxed{c=\frac{k^3}{k^2-k+1},k=\sum_{m,n=0}^\infty\frac{B_m^{(\frac n2)}}{n! (m-n+1)!}}$$

shown here

Apply Lagrange reversion 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, generating function. 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

deleted 2 characters in body
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Apply Lagrange reversion to @TheSimpli’s@TheSimpliFire’s equation:

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

ThereforeNow use the Norlund polynomial, or Norlund polynomials/generalized Bernoulli numbersnumber appear and we, generating function to get:

$$\boxed{c=\frac{k^3}{k^2-k+1},k=\sum_{m,n=0}^\infty\frac{B_m^{(\frac n2)}}{n! (m-n+1)!}}$$

shown here

Apply Lagrange reversion to @TheSimpli’s equation:

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

Therefore Norlund polynomials/generalized Bernoulli numbers appear and we get:

$$\boxed{c=\frac{k^3}{k^2-k+1},k=\sum_{m,n=0}^\infty\frac{B_m^{(\frac n2)}}{n! (m-n+1)!}}$$

shown here

Apply Lagrange reversion to @TheSimpliFire’s equation:

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

Now use the Norlund polynomial, or generalized Bernoulli number, generating function to get:

$$\boxed{c=\frac{k^3}{k^2-k+1},k=\sum_{m,n=0}^\infty\frac{B_m^{(\frac n2)}}{n! (m-n+1)!}}$$

shown here

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