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add definitions here
\begin{align*} T(a,b;c,d) &= \max_{\pi \in \Pi_{(a,b)\to(c,d)}} \sum_{(i,j)\in \pi} X_{i,j} \quad \textrm{(last passage percolation time from }(a,b)\textrm{ to }(c,d) \textrm{)} \\ \mathbf{V}^n &= (V^n_1,\ldots,V^n_{n-1}) \quad \qquad \qquad \qquad \qquad \qquad \qquad \textrm{(block finishing times)} \\ \mathbf{U}^n &= (T(1,1; n,1), T(1,1; n-1,2), \ldots, T(1,1; 1,n)) \qquad \textrm{(point to line last passage percolation)} \\ \mathbf{W}^n &= (T(1,1; n,1), T(2,1; n,2), \ldots, T(n,1; n,n)) \qquad \textrm{(line to line last passage percolation)} \end{align*}
Theorem. $ \mathbf{U}^n \stackrel{D}{=} \mathbf{W}^n $
Proof. add this
Conjecture. $ \mathbf{V}^n \stackrel{D}{=} \mathbf{W}^n $
Proposition. Let $f_n(x_1,\ldots,x_n)$ denote the joint density function of the random vector $\mathbf{U}^n$. Then we have the recurrence relation $$ f_{n+1}(x_1,\ldots,x_{n+1}) = \exp\left( - \sum_{k=1}^{n+1} x_k \right) \int_0^{x_1 \wedge x_2} \int_0^{x_2 \wedge x_3} \ldots \int_0^{x_n \wedge x_{n+1}} f_n(y_1,\ldots,y_n) \exp\left( \sum_{k=1}^{n+1} y_{k-1} \vee y_k \right) dy_n \, dy_{n-1} \ldots dy_2 \, dy_1 $$
add this
add an explanation of the algorithm for computing the joint density of the finishing times
Weak version of the main conjecture. For $n\ge2$ and $1\le k< n$, we have the equality in distribution $$ (V^n(k), V^n(k+1)) \stackrel{D}{=} (U^n_k, U^n_{k+1}). $$
Consider as before an array $(X_{i,j)})_{i,j\ge 1}$ of iid $\operatorname{Exp}(1)$ random variables. Denote $$ X_{i,j}' = \begin{cases} 0 & \textrm{if }(i,j)=(1,1) \\ X_{i,j} & \textrm{otherwise.} \end{cases} $$ Let $T(a,b;c,d)$ denote as before the last passage percolation times associated with the array $(X_{i,j})_{i,j}$. We have \begin{align*} U^n_k &= T(1,1; k, n+1-k), \\ U^n_{k+1} &= T(1,1; k+1, n-k). \end{align*} Let $T'(a,b;c,d)$ denote the last passage percolation times associated with the modified array $(X_{i,j}')_{i,j}$. Now consider the rectangular $(k+1)\times (n-k)$ subarray $$ \mathbf{\tau}_{n,k} = (T'(1,1; i,j))_{1\le i\le k+1, 1\le j\le n-k} $$