Closed form for the sums $S(n)={(2n-1)!\over \sqrt2}\cdot{\left(4/ \pi\right)^{2n}}\cdot\sum\limits_{k=0}^{\infty}(-1)^{k(k+1)/2}(2k+1)^{-2n}$ Consider the sums

$$S(n)={(2n-1)!\over \sqrt2}\cdot{\left(4\over \pi\right)^{2n}}\cdot\sum_{k=0}^{\infty}{(-1)^{k(k+1)\over 2}\over (2k+1)^{2n}}$$

We have $S(1)=1$, $S(2)=11$, $S(3)=361$, $S(4)=24611$.
I cannot spot any pattern within this sequence.
How can we work out a closed form for $S(n)$?

We may notice that
$$ T(n)=\sum_{k\geq 0}\frac{(-1)^{k(k+1)/2}}{(2k+1)^{2n}} = \sum_{m\geq 1}\frac{\chi(m)}{m^{2n}} = L(\chi,2n)\tag{1}$$
is a Dirichlet $L$-function associated with the multiplicative function $\chi(m)$, that equals $0$ if $m$ is even, $1$ if $m\equiv \pm 1\pmod{8}$ and $-1$ if $m\equiv \pm 3\pmod{8}$. In particular
$$ T(n) = \prod_{p>2}\left(1-\frac{\left(\frac{2}{p}\right)}{p^{2n}}\right)^{-1}\tag{2} $$
where $\left(\frac{2}{p}\right)$ is Legendre's symbol. From Hazem Orabi's integral representation
$$ S(n) = \frac{1}{\sqrt{2}(2\pi)^{2n}}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\left(e^{x/8}-e^{3x/8}-e^{5x/8}+e^{7x/8}\right)\,dx \tag{3} $$
we also have:
$$ S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\int_{1}^{+\infty}\log(x)^{2n-1}\frac{x^2-1}{1+x^4}\,dx \tag{4}$$
or
$$ S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\left.\frac{d^{2n-1}}{d\alpha^{2n-1}}\int_{1}^{+\infty}\frac{x^{2+\alpha}-x^{\alpha}}{1+x^4}\,dx\, \right|_{\alpha=0^+}\tag{5}$$
so $S(n)$ depends on $\psi^{(2n-1)}(z)$ (the $(2n-1)$-th derivative of the digamma function, i.e. the $2n$-th derivative of $\log\Gamma$) evaluated at $z=\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}$. By the reflection formulas for the $\psi^{(2n-1)}(z)$ function, everything boils down to the derivatives of the function $\cot(\pi z)$ at $z=\frac{1}{8}$ and $z=\frac{3}{8}$. This proves G.H.Hardy's claim: 

$$ S(n) = (2n-1)!\cdot [z^{2n-1}]\frac{\sin(z)}{\cos(2z)}.\tag{6} $$

In particular, $S(n)$ can be expressed in terms of Euler's polynomials $E_n(z)$ evaluated at $z=\frac{1}{4}$ and $z=\frac{3}{4}$.
