11/23/2007, 08:22 AM
This is indeed an interesting connection.
Now my quick 2 cents about it.
First, we want the explicit function \( f \) of which the sums of the powers of the inverted fixed points are the coefficients. We compare Jay's beginnin with index 1 in the first row with Sloane's beginning at index 1 in the second row:
\( \begin{pmatrix}
-1 & -1 & 2 & 9 & -6 & -155 & -232 & 3969 & 20870
& -118779 & -1655028\\
1 & 1 & -1 & -2 & 9 & 6 & -155 & 232 & 3969 & -20870 & -118779 & 1655028\end{pmatrix} \)
Obviously we have to move the lower row to the left which is the same as dividing Sloane's function \( S(z)=\ln\left(1+ze^z\right) \) by \( z \). We get then
\( \begin{pmatrix}
-1 & -1 & 2 & 9 & -6 & -155 & -232 & 3969 & 20870
& -118779 & -1655028\\
1 & -1 & -2 & 9 & 6 & -155 & 232 & 3969 & -20870 & -118779 & 1655028\end{pmatrix} \)
and see that the sign is swapped for each uneven power, which can be achieved by using \( -z \) instead of \( z \).
So we get
\( f(z)=-\frac{1}{z} \ln\left(1-\frac{z}{e^z}\right) \) with \( f_n = \sum_{k=0} \left(\frac{1}{\overline{c_k}^n}+\frac{1}{c_k^n}\right) \).
\( f(z)=\sum_{n=1}^\infty z^n\sum_{k=0}^\infty \left(\frac{1}{\overline{c_k}^n}+\frac{1}{c_k^n}\right)
=\sum_{k=0}^\infty \sum_{n=1}^\infty \left(\frac{z}{\overline{c_k}}\right)^n+\left(\frac{z}{c_k}\right)^n
=\sum_{k=0}^\infty \frac{\frac{z}{\overline{c_k}}}{1-\frac{z}{\overline{c_k}}}+
\frac{ \frac{z}{c_k} }{ 1-\frac{z}{c_k}}=\sum_{k=0}^\infty \frac{z}{\overline{c_k}-z}+\frac{z}{c_k-z} \)
for \( |z|<|c_k| \). If we transform this further via \( e^{-zf(z)}=1-z/e^z \) we get
\( \prod_{k=0}^\infty e^{\frac{z^2}{z-\overline{c_k}}} e^{ \frac{z^2}{z-c_k}}=1-z/e^z \) to prove.
Looks strange, perhaps I made an error somewhere.
Now my quick 2 cents about it.
First, we want the explicit function \( f \) of which the sums of the powers of the inverted fixed points are the coefficients. We compare Jay's beginnin with index 1 in the first row with Sloane's beginning at index 1 in the second row:
\( \begin{pmatrix}
-1 & -1 & 2 & 9 & -6 & -155 & -232 & 3969 & 20870
& -118779 & -1655028\\
1 & 1 & -1 & -2 & 9 & 6 & -155 & 232 & 3969 & -20870 & -118779 & 1655028\end{pmatrix} \)
Obviously we have to move the lower row to the left which is the same as dividing Sloane's function \( S(z)=\ln\left(1+ze^z\right) \) by \( z \). We get then
\( \begin{pmatrix}
-1 & -1 & 2 & 9 & -6 & -155 & -232 & 3969 & 20870
& -118779 & -1655028\\
1 & -1 & -2 & 9 & 6 & -155 & 232 & 3969 & -20870 & -118779 & 1655028\end{pmatrix} \)
and see that the sign is swapped for each uneven power, which can be achieved by using \( -z \) instead of \( z \).
So we get
\( f(z)=-\frac{1}{z} \ln\left(1-\frac{z}{e^z}\right) \) with \( f_n = \sum_{k=0} \left(\frac{1}{\overline{c_k}^n}+\frac{1}{c_k^n}\right) \).
\( f(z)=\sum_{n=1}^\infty z^n\sum_{k=0}^\infty \left(\frac{1}{\overline{c_k}^n}+\frac{1}{c_k^n}\right)
=\sum_{k=0}^\infty \sum_{n=1}^\infty \left(\frac{z}{\overline{c_k}}\right)^n+\left(\frac{z}{c_k}\right)^n
=\sum_{k=0}^\infty \frac{\frac{z}{\overline{c_k}}}{1-\frac{z}{\overline{c_k}}}+
\frac{ \frac{z}{c_k} }{ 1-\frac{z}{c_k}}=\sum_{k=0}^\infty \frac{z}{\overline{c_k}-z}+\frac{z}{c_k-z} \)
for \( |z|<|c_k| \). If we transform this further via \( e^{-zf(z)}=1-z/e^z \) we get
\( \prod_{k=0}^\infty e^{\frac{z^2}{z-\overline{c_k}}} e^{ \frac{z^2}{z-c_k}}=1-z/e^z \) to prove.
Looks strange, perhaps I made an error somewhere.
