Treating j as an $n$-dimensional vector $j[t]$ is its $t$-th entry.

Thanks. I should have fixed this earlier but can't edit the comment. The multiplicativity of rank only holds as an inequality $rank(A \dagger B) \ge rank(A) rank(B)$.

Thanks, just for the rank argument tensor product is definitely enough, I was just wondering about the matrix itself.

I just need the fact that $rank(A \dagger B) = rank(A)rank(B)$ which is easy to show directly. However, it would be helpful to know if this operation and its properties are already known so that I can just cite an appropriate source. I like your "super-slam" idea though :)

Good point, but a closed approximation up to low order terms is fine, i.e. $P[X_2 = 1] = \frac{\ln n }{ n} + \frac{c}{n} + o\left(\frac{1}{n}\right)$.

Thanks, but this still doesn't seem like a closed form. If I take $\frac{d^t g_{nj}(z) }{ d z^t}$ at $z = 0$ then it still looks like an expansion over all possible paths.