What is the minimum size of a partial order containing all partial orders of size 5? This earlier MO question asks to find the minimum size of a partial order that is universal for all partial orders of size $n$, i.e. any partial order of size $n$ embeds into it, preserving the order. In particular, the question asks if the minimum size $f(n)$ has a polynomial upper bound, to which the answer is no.
In this question, I am interested in some concrete values of $f(n)$ for small $n$.
So far, I know that:

*

*$f(0) = 0$


*$f(1) = 1$


*$f(2) = 3$


*$f(3) = 5$


*$f(4) = 8$


*$f(n) \ge 2n - 1$


*$f(n) \in \Omega(n^k)$ for all $k$
Can we compute some additional values in this sequence? In particular, can we compute $f(5)$?
Notes

*

*I was able to verify $f(4) = 8$ using a computer-assisted proof using a SAT solver. I also tried naive enumeration of posets and checking for universality, but this fails at around $f(4)$. Computing $f(5)$ may require smarter enumeration, in particular better symmetry breaking.


*The sequence does not appear to be in OEIS yet (it does not appear to be any of the sequences beginning with 1, 3, 5, 8). I submitted this draft, and it was suggested that the sequence should be posted to MathOverflow to find more terms.
EDIT: New OEIS entry with f(5) = 11 here.
 A: You can solve the problem via integer linear programming as follows.  Let $P$ be the set of $n$-posets to be covered, and for $(i,j)\in [n] \times [n]$ let $a_{p,i,j}$ indicate whether $i \preceq j$ in poset $p$.  We want to find a universal $m$-set if possible.  For $(i,j)\in [m] \times [m]$, let binary decision variable $x_{i,j}$ indicate whether $i \preceq j$ in the universal poset.  For $p \in P$, $i_1\in [n]$, and $i_2\in [m]$, let binary decision variable $y_{p,i_1,i_2}$ indicate whether element $i_1$ in poset $p$ is assigned to element $i_2$ in the universal set.
A universal $m$-poset exists if and only if the following constraints can be satisfied:
\begin{align}
\sum_{i_2 \in [m]} y_{p,i_1,i_2} &= 1 
&&\text{for $p\in P$ and $i_1 \in [n]$} \tag1 \\
\sum_{i_1 \in [n]} y_{p,i_1,i_2} &\le 1 
&&\text{for $p \in P$ and $i_2 \in [m]$} \tag2 \\
y_{p,i_1,i_2} + y_{p,j_1,j_2} - 1 &\le x_{i_2,j_2} 
&&\text{for $p\in P, (i_1,j_1) \in [n] \times [n], (i_2, j_2) \in [m] \times [m]$ with $a_{p,i_1,j_1}=1$} \tag3 \\
y_{p,i_1,i_2} + y_{p,j_1,j_2} - 1 &\le 1 - x_{i_2,j_2}
&&\text{for $p\in P, (i_1,j_1) \in [n] \times [n], (i_2, j_2) \in [m] \times [m]$ with $a_{p,i_1,j_1}=0$} \tag4 \\
x_{i,j} + x_{j,k} - 1 &\le x_{i,k}
&&\text{for $i,j,k \in [m]$} \tag5 \\
\\
\end{align}
Constraint $(1)$ assigns each element in poset $p$ to exactly one element in the universal poset.
Constraint $(2)$ assigns at most one element in poset $p$ to each element in the universal poset.
Constraint $(3)$ enforces $$(y_{p,i_1,i_2} \land y_{p,j_1,j_2} \land a_{p,i_1,j_1}) \implies x_{i_2,j_2}.$$
Constraint $(4)$ enforces $$(y_{p,i_1,i_2} \land y_{p,j_1,j_2} \land \lnot a_{p,i_1,j_1}) \implies \lnot x_{i_2,j_2}.$$
Constraint $(5)$ enforces transitivity in the universal poset.
A: I will try to revive Sagemath's ticket #14110 and provide a Sagemath package for this enumeration (in fact, the C code, corresponding to the paper B. D. McKay and G. Brinkmann, Posets on up to 16 points, Order, 19 (2002) 147-179 -   (mostly) due to Gunnar Brinkmann, which is using Brendan's McKay's nauty, is posted there.
