<s>[INST] find a function satisfying the recurrence find a function satisfying the recurrence
$$F (n) = 2F (\sqrt{n}) + 1$$
replace $n$ by $2^m$
Thus getting the answer as $$F(n)=\frac{1}{2}c \log(n) + \log(n) - 1$$
Is this  correct
 [/INST] Looks good:
\begin{align}
2 F(\sqrt{n}) + 1 
&=
2\left(
\left(\frac{c}{2} + 1 \right) \log(\sqrt{n}) - 1
\right) + 1 \\
&=
2\left(
\left(\frac{c}{2} + 1 \right) \frac{1}{2} \log(n) - 1
\right) + 1 \\
&=
\left(\frac{c}{2} + 1 \right) \log(n) - 1 \\
&= F(n)
\end{align}
You could replace the constant factor $c/2 + 1$ by another constant $d$.
In case you asked about the derivation as well (the wording of your question is somewhat ambigious to me):
Setting $n = 2^m$ gives the equation
$$
F(2^m) = 2 F(2^{m/2}) + 1
$$
setting $F(2^m) = G(m)$ gives
$$
G(m) = 2 G(m/2) + 1 \iff 
G(2k) = 2 G(k) + 1
$$
That looks pretty linear. Trying $G(k) = a k + b$ gives
$$
2 a k + b = 2 a k + 2 b + 1 \iff
b = - 1 
$$
So $G(m) = a m - 1$ and $F(2^m) = a m - 1 \iff F(n) = a \log_2(n) - 1 = c \ln(n) - 1$ for some constant $c$.
</s>