Algorithm for main diagonal of integer coefficients associated with Schroeder numbers

Question

• Let $T_q(n, k)$ be an integer table such that $$T_q(n, k) = \begin{cases} 1 & \textrm{if } n = 0 \vee k = 0 \\ qT_q(n-1, n-1) + T_q(n, n-1) & \textrm{if } n = k > 0 \\ T_q(n, k-1) + T_q(n-1, k) + T_q(n-1, k-1) & \textrm{otherwise} \end{cases}$$

• Start with vector $\nu$ of fixed length $2m$ with elements $\nu_i=1$ (that is, $\nu=\{1,1,\dotsc,1\}$) and vector $t$ of fixed length $m$ with elements $t_i=[i=1]$ and for $i$ from $1$ to $m-1$ and for $j$ from $i+1$ to $2m-i$ apply $\nu_j := \frac{1}{q}\nu_{j+(j-i) \bmod 2} + \nu_{j-1}$ and $t_{i+1} = q^i \nu_{i+1}$ (after ending each cycle for $j$).

Here square bracket denotes Iverson bracket.

I conjecture that after the whole transform we have $$ t_{n} = T_q(n-1, n-1). $$

Here is the PARI/GP program to check it numerically:

upto1(n, m) = my(v1, v2, v3); v1 = vector(n, i, 0); v1[1] = 1; v2 = v1; for(i=1, n-1, v3 = v1; for(j=2, i, v1[j] = v1[j-1] + v3[j] + v3[j-1]); v1[i+1] = m*v3[i] + v1[i]; v2[i+1] = v1[i+1]); v2
upto2(n, m) = my(v1, v2); v1 = vector(2*n, i, 1); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, for(j=i+1, 2*n-i, v1[j] = (1/m)*v1[j+(j-i)%2] + v1[j-1]); v2[i+1] = m^i*v1[i+1]); v2
test(n, m) = upto1(n, m) == upto2(n, m)

Is there a way to prove it? Is there a similar integer coefficients (say, $R_q(n,k)$) for the case where $\nu$ has length $3m$, cycle for $j$ is from $i+1$ to $3m-2i$ and where we change $\nu_{j+(j-i) \bmod 2}$ to $\nu_{j+(j-i) \bmod 3}$?