Following an idea of Dmitrii we make the ansatz:
\( \text{slog}(z)=\frac{1}{\el}\left(\log(z-\el) + r(z-\el)\right) \)
where \( \el \) is the fixed point of \( \log \) in the upper half plane and \( r \) is a powerseries developed at \( \el \).
\( \text{slog} \) must satisfy:
\( \text{slog}(\exp(z))=\text{slog}(z)+1 \)
and hence:
\( \log(e^z-\el)+r(e^z-\el)=\log(z-\el)+r(z-\el)+\el \)
\( \log\left(\frac{e^z-\el}{z-\el}\right)=r(z-\el)-r(e^z-\el)+\el \)
The left side has no singularity at \( \el \) hence we do the transformation \( z\mapsto z+\el \):
\( \log\left(\frac{\el(e^z-1)}{z}\right)=r(z)-r(\el(e^z-1))+\el \)
(*) \( \log\left(\frac{e^z-1}{z}\right)=r(z)-r(\el(e^z-1)) \)
The left side can be developed as:
\( \log\left(\frac{e^z-1}{z}\right)=
\frac{1}{2}z+\frac{1}{24}z^2-\frac{1}{2880}z^4+\frac{1}{181440}z^6 -\frac{1}{9676800}z^8+\frac{1}{479001600}z^{10}+\dots \)
Let \( \lambda_k \) be the coefficients of this series.
And let \( \eta_k \) be the coefficients of the series \( \el(e^z-1)=\el z + \frac{\el}{2} z^2 + \frac{\el}{6} z^3 + \dots \).
Then on the right hand side of (*) we have:
\( (r-r\circ \eta)_n = r_n - \sum_{j=0}^n r_j (\eta^{\cdot j})_n \)
by the composition rule for powerseries (Carleman matrix multiplication) where \( \eta^{\cdot j} \) means the \( j \)-th power (not iteration) of \( \eta \).
This yields a recursive formula for \( r_n \):
\( r_n(1-(\eta^{\cdot n})_n)=\lambda_n + \sum_{j=0}^{n-1} r_j (\eta^{\cdot j})_n \)
Because \( \eta^{\cdot 0}=1 \) we can start the right hand sum at 1 instead of 0 if \( n>0 \), so:
\( r_n = \frac{\lambda_n + \sum_{j=1}^{n-1} r_j (\eta^{\cdot j})_n}{1-(\eta^{\cdot n})_n} \) for \( n>0 \).
Now we know that \( \eta^{\cdot j}(z)=\el^j (e^z-1)^j=\el^j\sum_{k=0}^j \left(j\\k\right)e^{kz}(-1)^{j-k} \)
\( \left(\eta^{\cdot j}\right)_n=\el^j \sum_{k=0}^j \left(j\\k\right)\frac{k^n}{n!}(-1)^{j-k}=\el^j \sum_{k=1}^j \left(j\\k\right)\frac{k^n}{n!}(-1)^{j-k} \)
So for example:
\( r_1=\frac{\lambda_1}{1-\eta_1}=\frac{1/2}{1-\el} \)
\( r_2=\frac{\lambda_2 + \eta_2}{1-(\eta^{\cdot 2})_2}=\frac{\frac{1}{24}+\frac{1}{2-2\el}\frac{\el}{2}}{1-\el^2(-1+2^2/2)}=\frac{\el}{{4 {\el}^{3} } - {4 {\el}^{2} } - {4 \el} + 4} + \frac{1}{24 - {24 {\el}^{2} }} \)
\( r_3=\frac{\el}{{-12 {\el}^{6} } + {12 {\el}^{5} } + {12 {\el}^{4} } - {12 {\el}^{2} } - {12 \el} + 12} + \frac{{\el}^{2} }{{24 {\el}^{5} } - {24 {\el}^{3} } - {24 {\el}^{2} } + 24}+\frac{{\el}^{3} }{{-6 {\el}^{6} } + {6 {\el}^{5} } + {6 {\el}^{4} } - {6 {\el}^{2} } - {6 \el} + 6} \)
\( \text{slog}(z)=\frac{1}{\el}\left(\log(z-\el) + r(z-\el)\right) \)
where \( \el \) is the fixed point of \( \log \) in the upper half plane and \( r \) is a powerseries developed at \( \el \).
\( \text{slog} \) must satisfy:
\( \text{slog}(\exp(z))=\text{slog}(z)+1 \)
and hence:
\( \log(e^z-\el)+r(e^z-\el)=\log(z-\el)+r(z-\el)+\el \)
\( \log\left(\frac{e^z-\el}{z-\el}\right)=r(z-\el)-r(e^z-\el)+\el \)
The left side has no singularity at \( \el \) hence we do the transformation \( z\mapsto z+\el \):
\( \log\left(\frac{\el(e^z-1)}{z}\right)=r(z)-r(\el(e^z-1))+\el \)
(*) \( \log\left(\frac{e^z-1}{z}\right)=r(z)-r(\el(e^z-1)) \)
The left side can be developed as:
\( \log\left(\frac{e^z-1}{z}\right)=
\frac{1}{2}z+\frac{1}{24}z^2-\frac{1}{2880}z^4+\frac{1}{181440}z^6 -\frac{1}{9676800}z^8+\frac{1}{479001600}z^{10}+\dots \)
Let \( \lambda_k \) be the coefficients of this series.
And let \( \eta_k \) be the coefficients of the series \( \el(e^z-1)=\el z + \frac{\el}{2} z^2 + \frac{\el}{6} z^3 + \dots \).
Then on the right hand side of (*) we have:
\( (r-r\circ \eta)_n = r_n - \sum_{j=0}^n r_j (\eta^{\cdot j})_n \)
by the composition rule for powerseries (Carleman matrix multiplication) where \( \eta^{\cdot j} \) means the \( j \)-th power (not iteration) of \( \eta \).
This yields a recursive formula for \( r_n \):
\( r_n(1-(\eta^{\cdot n})_n)=\lambda_n + \sum_{j=0}^{n-1} r_j (\eta^{\cdot j})_n \)
Because \( \eta^{\cdot 0}=1 \) we can start the right hand sum at 1 instead of 0 if \( n>0 \), so:
\( r_n = \frac{\lambda_n + \sum_{j=1}^{n-1} r_j (\eta^{\cdot j})_n}{1-(\eta^{\cdot n})_n} \) for \( n>0 \).
Now we know that \( \eta^{\cdot j}(z)=\el^j (e^z-1)^j=\el^j\sum_{k=0}^j \left(j\\k\right)e^{kz}(-1)^{j-k} \)
\( \left(\eta^{\cdot j}\right)_n=\el^j \sum_{k=0}^j \left(j\\k\right)\frac{k^n}{n!}(-1)^{j-k}=\el^j \sum_{k=1}^j \left(j\\k\right)\frac{k^n}{n!}(-1)^{j-k} \)
So for example:
\( r_1=\frac{\lambda_1}{1-\eta_1}=\frac{1/2}{1-\el} \)
\( r_2=\frac{\lambda_2 + \eta_2}{1-(\eta^{\cdot 2})_2}=\frac{\frac{1}{24}+\frac{1}{2-2\el}\frac{\el}{2}}{1-\el^2(-1+2^2/2)}=\frac{\el}{{4 {\el}^{3} } - {4 {\el}^{2} } - {4 \el} + 4} + \frac{1}{24 - {24 {\el}^{2} }} \)
\( r_3=\frac{\el}{{-12 {\el}^{6} } + {12 {\el}^{5} } + {12 {\el}^{4} } - {12 {\el}^{2} } - {12 \el} + 12} + \frac{{\el}^{2} }{{24 {\el}^{5} } - {24 {\el}^{3} } - {24 {\el}^{2} } + 24}+\frac{{\el}^{3} }{{-6 {\el}^{6} } + {6 {\el}^{5} } + {6 {\el}^{4} } - {6 {\el}^{2} } - {6 \el} + 6} \)