Strange Radical

$\Large\mathfrak T(\color{#3D99F6}{a})=\color{#3D99F6}{a}^{\overbrace{\frac{F_1}{2}-\frac{F_2}{2^2}+\frac{F_3}{2^3}+\cdots}^{\color{#D61F06}{\mathfrak{\Large P}}}}$

$\begin{aligned}\color{#D61F06}{\mathfrak{\large P}}=&\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{2}{2^3}-\dfrac{3}{2^4}+\cdots\\-\dfrac{\color{#D61F06}{\mathfrak{\large P}}}{2}=&~~~-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{2}{2^4}+\cdots\end{aligned}$

Subtracting and multiplying by $2$ we get:

$3\color{#D61F06}{\mathfrak{\large P}}=1+\underbrace{\color{#D61F06}{\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{2}{2^4}+\cdots}}_{\dfrac{\color{#D61F06}{\mathfrak{\large P}}}{2}}$ $\Large\implies\color{#D61F06}{\mathfrak{\large P}}=\dfrac 25$

$\Large\implies \mathfrak T(\color{#3D99F6}{a})=\color{#3D99F6}{a}^{\frac 25}$ $\Large \mathfrak T(\color{#3D99F6}{4})=4^{\frac 25}=2^{\frac 45}$

$\huge \therefore 2+4+5=\boxed{11}$

The generating function of $F_n$ is $G(x)=\sum_{n=1}^{\infty}F_nx^n=\frac{x}{1-x-x^2}$ , and $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{F_n}{2^n}$ $=-\sum_{n=1}^{\infty}F_n(-\frac{1}{2})^n=-G\left(-\frac{1}{2}\right)=\frac{2}{5}$ . Thus $\mathfrak{T}(4)=4^{2/5}=2^{4/5}$ . The answer is $2+4+5=\boxed{11}$ .