of solution has been obtained in [@Maeda:2006hj].

### The case of $m=4$, $p=2$, $D\geq 6$

In this case, Eq. (\[lineq\]) give $$\label{c0c2r0Dm4} c_0=\frac{1}{2}(D^2-5D-2)r_{0}^{-2},\quad
c_2=\frac{1}{2(D-4)(D-5)}r_{0}^{2}\, .$$ We then have $$\widehat{\alpha}_0=\frac{2D-11}{3(D-4)}r_{0}^{-2}\, ,\quad
\widehat{\alpha}_1=\frac{2}{D-4}\, ,\quad
\widehat{\alpha}_2=\frac{1}{(D-4)(D-5)}r_{0}^{2}\, .$$ The equation for $F$ becomes $$\widehat{\alpha}_0+\widehat{\alpha}_1
F+\widehat{\alpha}_2F^2=\frac{M}{r^3}+\frac{Q}{r^4}\, ,$$ which has the solution $$F(r)=-\frac{\widehat{\alpha}_1}{2\widehat{\alpha}_2}
\left(~1\mp\sqrt{1-\frac{4\widehat{\alpha}_0
\widehat{\alpha}_2}{\widehat{\alpha}_1^2}+\frac{4
\widehat{\alpha}_2M}{\widehat{\alpha}_1^2~r^3}+\frac{4
\widehat{\alpha}_2Q}{\widehat{\alpha}_1^2~r^4}}~\right)\, ,$$ $$\label{m4solution} f(r)=\delta+\frac{\widehat{\alpha}_1
r^2}{2\widehat{\alpha}_2}\left(~1\mp\sqrt{1-\frac{4\widehat{\alpha}_0
\widehat{\alpha}_2}{\widehat{\alpha}_1^2}+\frac{4
\widehat{\alpha}_2M}{\widehat{\alpha}_1^2~r^3}+\frac{4
\widehat{\alpha}_2Q}{\widehat{\alpha}_1^2~r^4}}~\right)\, .$$ This solution with two branches is just the one recently obtained by Maeda and Dadhich in [@Maeda:2006iw]. It is easy to see that for $D\ge 6$, we have $$\widehat{\alpha}_1^2-4\widehat{\alpha}_0\widehat{\alpha}_2=\frac{4}{3(D-4)(D-5)}>0\,
,$$ So the vacuum AdS solution ($M=Q=0$) always exists.

### The case of $m=5$, $p=3$, $D\geq 7$

Equations (\[lineq\]) in this case lead to $$\begin{aligned}
c_0&=&\frac{D^4-10D^3+11D^2+22D+360}{3(D^2-7D+4)}r_{0}^{-2}\, ,\nonumber \\
c_2&=&\frac{1}{(D^2-7D+4)}r_{0}^{2}\, ,\nonumber\\
c_3&=&\frac{1}{3(D^2-7D+4)(D-5)(D-6)}r_{0}^{4}\, .\end{aligned}$$ Then the corresponding $\widehat{\alpha}$ are $$\begin{aligned}
\widehat{\alpha}_0&=&\frac{4(5D^2-67D+225)}{5(D^2-7D+4)(D-5)}r_{0}^{-2}\, ,\nonumber \\
\widehat{\alpha}_1&=&\frac{8(2D-13)}{(D^2-7D+4)(D-5)}\, ,\nonumber \\
\widehat{\alpha}_2&=&\frac{12}{(D^2-7D+4)(D-5)}r_{0}^{2}\, .\end{aligned}$$ The equation for $F$ is still in second order and has the solution $$F(r)=-\frac{\widehat{\alpha}_1}{2\widehat{\alpha}_2}
\left(~1\mp\sqrt{1-\frac{4\widehat{\alpha}_0
\widehat{\alpha}_2}{\widehat{\alpha}_1^2}+\frac{4
\widehat{\alpha}_2M}{\widehat{\alpha}_1^2~r^4}+\frac{4
\widehat{\alpha}_2Q}{\widehat{\alpha}_1^2~r^5}}~\right)\, ,$$ $$f(r)=\delta+\frac{\widehat{\alpha}_1
r^2}{2\widehat{\alpha}_2}\left(~1\mp\sqrt{1-\frac{4\widehat{\alpha}_0
\widehat{\alpha}_2}{\widehat{\alpha}_1^2}+\frac{4
\widehat{\alpha}_2M}{\widehat{\alpha}_1^2~r^4}+\frac{4
\widehat{\alpha}_2Q}{\widehat{\alpha}_1^2~r^5}}~\right)\, .$$ This solution is an example with the third order Lovelock term. Note that when $m=5$, the solution also has two branches. In addition, it is also easy to see that the vacuum AdS solution exists, because $$\widehat{\alpha}_1-4\widehat{\alpha}_0\widehat{\alpha}_1
=\frac{64 (5D-34)}{5(D^2-7D+4)(D-5)}>0$$ for $D\ge 