Is it possible to express $$ A_n = (2n-3)! \sum_{i=0}^{2n} \frac{B_i B_{2n-i}}{i! (2n-i)!} (2^{1-i} -1) (2^{1-2n+i} -1)(2^i -1) $$ in a more compact form, perhaps using the identity $$ E_{2n} = 2^{2n+1} \binom{2n}{2}^{-1} \sum_{i=0}^{2n} B_i B_{2n-i} \binom{2n}{i} (2^{2n-i} -1) (2^{1-i} -1) $$ or a similar one, where B are Bernoulli numbers and E Euler numbers?
$$ \sum_{k=0}^\infty \frac{E_k}{k!} x^k = \frac{1}{\cosh x} $$
We have $$ \sum_{i\geq 0}\frac{B_i}{i!}x^i=\frac{x}{e^x-1},\qquad \sum_{i\geq 0}\frac{B_i}{i}x^i 2^{i-1} = \frac{x}{e^{2x}-1} \tag{1}$$ hence $$\sum_{i\geq 0}\frac{B_i}{i!}(2-2^{i})x^i = \frac{x}{\sinh x},\qquad \sum_{i\geq 0}\frac{B_i}{i!}(2^{1-i}-1)x^i=\frac{x/2}{\sinh(x/2)}\tag{2}$$ and $$ \frac{A(n)}{(2n-3)!}=[x^{2n}]\left(\frac{x^2/2}{\sinh(x)\sinh(x/2)}-\frac{x^2/4}{\sinh^2(x/2)} \right)=[x^{2n}]\left(-\frac{x^2}{8\cosh^2\frac{x}{4}\cosh\frac{x}{2}}\right)\tag{3}$$