F=(a,b=a)=>a?F(--a,b)*~b+F(a,b-=2)*(a-b):+!b
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The code consists of only basic mathematics, but the mathematics behind the code requires calculus. The recursion formula is derived from the expansion of the derivatives of \$\mathrm{sech}(x)\$ of different orders.
Explanation
Here I'll use some convenient notation. Let \$T^n:=\mathrm{tanh}^n(t)\$ and \$S^n:=\mathrm{sech}^n(t)\$. Then we have
$$\frac{d^nS}{dt^n}=\sum_{i=0}^nF(n,i)T^{n-i}S^{i+1}$$
Since \$\frac{dT}{dt}=S^2\$ and \$\frac{dS}{dt}=-TS\$, we can deduce that
$$
\begin{equation}
\begin{aligned}
\frac{d}{dt}(T^aS^b)&=aT^{a-1}(S^2)(S^b)+bS^{b-1}(-TS)(T^a) \\
&=aT^{a-1}S^{b+2}-bT^{a+1}S^b
\end{aligned}
\end{equation}
$$
Let \$b=i+1\$ and \$a=n-i\$, we can rewrite the relation above as
$$
\begin{equation}
\begin{aligned}
\frac{d}{dt}(T^{n-i}S^{i+1})&=(n-i)T^{n-i-1}S^{i+3}-(i+1)T^{n-i+1}S^{i+1}\\
&=(n-i)T^{(n+1)-(i+2)}S^{(i+2)+1}-(i+1)T^{(n+1)-i}S^{i+1}
\end{aligned}
\end{equation}
$$
That is, \$F(n,i)\$ contributes to both \$F(n+1,i+2)\$ and \$F(n+1,i)\$. As a result, we can write \$F(n,i)\$ in terms of \$F(n-1,i-2)\$ and \$F(n-1,i)\$:
$$F(n,i)=(n-i+1)F(n-1,i-2)-(i+1)F(n-1,i)$$
with initial condition \$F(0,0)=1\$ and \$F(0,i)=0\$ where \$i\neq 0\$.
The related part of the code a?F(--a,b)*~b+F(a,b-=2)*(a-b):+!b is exactly calculating using the above recurrence formula. Here's the breakdown:
F(--a,b) // F(n-1, i) [ a = n-1, b = i ]
*~b // *-(i+1) [ a = n-1, b = i ]
+F(a,b-=2) // +F(n-1, i-2) [ a = n-1, b = i-2 ]
*(a-b) // *((n-1)-(i-2)) [ a = n-1, b = i-2 ]
// which is equivalent to *(n-i+1)
Since \$T(0)=0\$ and \$S(0)=1\$, \$E_n\$ equals the coefficient of \$S^{n+1}\$ in the expansion of \$\frac{d^nS}{dt^n}\$, which is \$F(n,n)\$.
For branches that \$F(0,0)\$ can never be reached, the recurrences always end at 0, so \$F(n,i)=0\$ where \$i<0\$ or \$i\$ is odd. The latter one, particularly, implies that \$E_n=0\$ for all odd \$n\$s. For even \$i\$s strictly larger than \$n\$, the recurrence may eventually allow \$0\leq i\leq n\$ to happen at some point, but before that step it must reach a point where \$i=n+1\$, and the recurrence formula shows that the value must be 0 at that point (since the first term is multiplied by \$n-i+1=n-(n+1)+1=0\$, and the second term is farther from the "triangle" of \$0\leq i\leq n\$). As a result, \$F(n,i)=0\$ where \$i > n\$. This completes the proof of the validity of the algorithm.
Extensions
The code can be modified to calculate three more related sequences:
Tangent Numbers (46 bytes)
F=(a,b=a)=>a?F(--a,b)*++b+F(a,b-=3)*(a-b):+!~b
Secant Numbers (45 bytes)
F=(a,b=a)=>a?F(--a,b)*++b+F(a,b-=3)*(a-b):+!b
Euler Zigzag Numbers (48 bytes)
F=(a,b=a)=>a?F(--a,b)*++b+F(a,b-=3)*(a-b):!b+!~b
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