The closed-form of $1-5\left(\frac{1}{2}\right)^k+9\left(\frac{1\cdot3}{2\cdot4}\right)^k-13\left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right)^k+\dots$?

Question

(A related MSE question by P. Singh.) First define,

$$F_k = 1-5\left(\frac{1}{2}\right)^k+9\left(\frac{1\cdot 3}{2\cdot 4}\right)^k-13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^k+17\left(\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\right)^k-\cdots$$

Or more concisely,

$$F_k = \sum_{n=0}^\infty\, (-1)^n\,(4n+1) \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{\Gamma\big(n+1\big)\;\Gamma\big(\tfrac{1}{2}\big)}\right)^k$$

The gamma quotient is also,

$$ \frac{\Gamma\big(n+\tfrac{1}{2}\big)}{\Gamma\big(n+1\big)\;\Gamma\big(\tfrac{1}{2}\big)} = \frac1{2^{2n}}\frac{(2n)!}{\,n!^2}$$

Then we have the nice closed-forms,

\begin{align} F_1 &= 0\\[4pt] F_2 &= \dfrac{2\sqrt2\,\Gamma\big(\tfrac{1}{2}\big)}{\Gamma^2\big(\tfrac{1}{4}\big)}\\[4pt] F_3 &= \dfrac{2}{\Gamma^2\big(\tfrac{1}{2}\big)}=\dfrac{2}{\pi} \\[4pt] F_4 &= \; \color{red}{??}\\[4pt] F_5 &= \dfrac{2}{\Gamma^4\big(\tfrac{3}{4}\big)} \end{align}

I found $F_2$ empirically. The closed-form of $F_3=\large\frac{2}{\pi}$ was included by Ramanujan in his letter to Hardy, and $F_5$ is also by him. (I presume a version of $F_2$ may be in his Notebooks.)

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Alternatively, while the $F_k$ are clearly a sum of two generalized hypergeometric functions $_pF_q$, we learn from this post that they can also be expressed as just one $_pF_q$,

\begin{align} F_1 &= \,_2F_1\big(\tfrac12,\tfrac54;\tfrac14;-1\big)\\[4pt] F_2 &= \,_3F_2\big(\tfrac12,\tfrac12,\tfrac54;\tfrac14,1;-1\big)\\[4pt] F_3 &= \,_4F_3\big(\tfrac12,\tfrac12,\tfrac12,\tfrac54;\tfrac14,1,1;-1\big)\\[4pt] F_4 &= \,_5F_4\big(\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac54;\tfrac14,1,1,1;-1\big)= \; \color{red}{??}\\[4pt] F_5 &= \,_6F_5\big(\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac54;\tfrac14,1,1,1,1;-1\big) \end{align}

and so on.

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Question: What is the closed-form of $F_4,\,F_6$ and others, if any?