The Fermat prime 257 and binomial sum $\sum_{n=0}^\infty \frac{(-1)^n}{\binom {8n}{4n}}$?

There is a connection between all of these expressions. Let $$ F_k(z) = \sum_{n=0}^\infty \frac{(2z)^{kn}}{{kn} \choose {kn/2}}$$ In particular $$F_1(z) = \frac{1}{1-z^2} + \frac{z \arcsin(z)}{(1-z^2)^{3/2}} + \frac{\pi z }{2(1-z^2)^{3/2}}$$ Now for $k > 1$, if $\omega_k$ is a primitive $k$'th root of unity we have $\sum_{j=0}^{k-1} \omega_k^n = k$ if $n$ is divisible by $k$, $0$ otherwise. So $$F_k(z) = \frac{1}{k} \sum_{j=0}^{k-1} \sum_{n=0}^\infty \omega_k^{jn} \frac{(2z)^{n}}{n \choose {n/2}}= \frac{1}{k} \sum_{j=0}^{k-1} F_1(\omega_k^j z)$$ Note that $1/(1-z^2) = \sum_{n=0}^\infty z^{2n}$ while $\pi z/(2 (1-z^2)^{3/2}$ is an odd function, so if $k$ is even we get $$F_k(z) = \frac{1}{1-z^{k}} + \frac{1}{k} \sum_{j=0}^{k-1} \frac{\omega_k^j z \arcsin(\omega_k^j z)}{(1 - \omega_k^{2j} z^2)^{3/2}}$$

You're interested in the case $z = \omega_k^{1/2}/2$ so that $(2z)^{kn} = (-1)^n$, thus when $k$ is even $$F_k(\omega_k^{1/2}/2) = \frac{2^k}{2^k + 1} + \frac{1}{k} \sum_{j=0}^{k-1} \frac{\omega_k^{j+1/2} \arcsin(\omega_k^{j+1/2}/2)}{2 (1- \omega_k^{2j+1}/4)^{3/2}}$$

It seems to be the real part of $$ -{\frac {1}{ \left( -1+{\frac {1}{256}}\,({-1})^{1/4}{256}^{3/4} \right) \left( 1+{\frac {1}{256}}\,({-1})^{1/4}{256}^{3/4} \right) }}+{\frac {1}{512}}\,{\frac {({-1})^{1/8}{256}^{{ {7}/{8}}}\sqrt {1-{\frac {1}{256}}\,({-1})^{1/4} {256}^{3/4}}\arcsin \left( {\frac {1 }{256}}\,({-1})^{1/8}{256}^{{ {7}/{8}}} \right) }{ \left( -1+{ \frac {1}{256}}\,({-1})^{1/4}{256}^{3/4} \right) ^{2}}}-{\frac {1}{ 512}}\,{\frac {({-1})^{1/8}{256}^{{ {7}/{8}}}{\rm arcsinh} \left( {\frac {1}{256}}\,({-1})^{1/8}{256}^{{ {7}/ {8}}} \right) }{ \left( 1+{\frac {1}{256}}\,({-1})^{1/4} {256}^{3/4} \right) ^{3/2}} } $$

EDIT: Here's what I did (in Maple 16):

S0:= sum((-1)^n/binomial(8*n,4*n),n=0..infinity);

$$S0 := {\mbox{$_5$F$_4$}(1/4,1/2,3/4,1,1;\,1/8,3/8,5/8,{\frac {7}{8}};\,-{\frac {1}{256}})}$$

S1:= subs(-1/256=z,S0);

S2:= convert(S1,FPS,z);

$$S2 := \sum _{k=0}^{\infty }{\frac { \left( 4\,k \right) !\,{16}^{-k}{z}^{k}} {{\it pochhammer} \left( 1/4,2\,k \right) {\it pochhammer} \left( 3/4, 2\,k \right) }} $$

S3:= value(S2);

$$S3 := -{\frac {1}{ \left( -1+\sqrt [4]{z} \right) \left( 1+\sqrt [4]{z} \right) }}+1/2\,{\frac {\sqrt [8]{z}\sqrt {1-\sqrt [4]{z}}\arcsin \left( \sqrt [8]{z} \right) }{ \left( -1+\sqrt [4]{z} \right) ^{2}}}- 1/2\,{\frac {\sqrt [8]{z}{\it arcsinh} \left( \sqrt [8]{z} \right) }{ \left( 1+\sqrt [4]{z} \right) ^{3/2}}} $$

S4:= eval(S3, z=-1/256);

Somehow this acquired an imaginary part (I suspect because a choice of branch was made in going from S2 to S3).