Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Let $a$ be the number of 2's in the sequences. So, the number of -1's in the sequence is $100-a$.

Now, let $H_A$ be the $hocus-pocus$ sum of the sequence $A$. So,

\[H_A=4\binom{a}{2}+\binom{100-a}{2}-2a(100-a)\]

\[=2a^2-2a+\frac{(100-a)(100-a-1)}{2}-200a+2a^2\]

\[=4.5a^2-301.5a+4950\]

The value of a quadratic equation $ax^2+bx+c$ is the lowest (or highest if $a$ is negative) when the variable is at the vertex, i.e. $a=\frac{-b}{2a}$

The vertex of this equation is $\frac{-(-301.5)}{2\cdot 4.5}=33.5$

But, $a$ must be an integer. So, taking the integers closest to the vertex, taking $a=33$, we see $H_A=-99$

Taking $a=34$, again, $H_A=-99$

So, the least value of $H_A$ is $-99$.