Adventure with infinite series, a curiosity It is easily verifiable that
$$\sum_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$
It is not that difficult to get
$$\sum_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$

Question. Is there something similarly "nice" in computing
  $$\sum_{k\geq0}\binom{8k}{4k}\frac1{2^{10k}}=?$$
  Perhaps the same question about
  $$\sum_{k\geq0}\binom{16k}{8k}\frac1{2^{20k}}=?$$

NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.
 A: Let $$f(x):=\sum_{k\ge 0}\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$$
with $|x|<1/4$. We have for $N,j\in\mathbb{Z}_+$,
\begin{align}
\frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})f\left(xe({r}/{N})\right)&=\frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})\sum_{k\ge 0}\binom{2k}{k}x^ke({kr}/{N})\\
&=\sum_{k\ge 0}\binom{2k}{k}x^k\frac{1}{N}\sum_{r=1}^Ne({(k-j)r}/{N}),
\end{align}
where $e(x):=e^{2\pi{\rm i}x}$. Hence clearly,
$$\sum_{\substack{k\ge 0\\ k\equiv j\pmod N}}\binom{2k}{k}x^k=\frac{1}{N}\sum_{r=1}^N\frac{e(-{jr}/{N})}{\sqrt{1-4xe(r/N)}},$$
holds for all $N,j\in\mathbb{Z}_+$, which is a more general result. 
A: It is moderately nice, I would say. We have $\sum \binom{2k}k x^k=(1-4x)^{-1/2}$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^{-5/2}$, $4x=2^{-1/2}$, we get $$\sum \binom{8k}{4k}2^{-10k}=\frac14\sum_{w^4=1}(1-w/\sqrt{2})^{-1/2}$$
and so on.
A: More generally, it seems that
$$
\sum_{k\ge0}\binom{2^{j+1}k}{2^jk}2^{-a_{j+2} k}=2^{-j}\sum_{w^{2^j}=1}(1-w/\sqrt2)^{-1/2}
$$
where $a_1=2,a_2=3,a_{j+1}=a_j+a_{j-1}+\dots+a_1$ (cf. A257113).
A: Mathematica says, for the first:
$$\,
   _4F_3\left(\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8};\frac{1}{4},\frac{1}{2},\frac{
   3}{4};\frac{1}{4}\right)$$
and
$$\,
   _8F_7\left(\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16},\frac{9}{16},\frac{11}{16}
   ,\frac{13}{16},\frac{15}{16};\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},\frac{5}{8},
   \frac{3}{4},\frac{7}{8};\frac{1}{16}\right)
$$ for the second.
