(Edited several times from earlier partial answer, which gave $f(5) \ge 11$.)
We have exact results $f(5) = 11$ and $f(6)=16$, and bounds $16 \le f(7) \le 25$.
1. Proving $f(5)=11$
A short proof shows that $f(5) \ge 10$. To be 5-universal (i.e. universal forcontain isomorphic copies of all partial orders of 5 elements), our poset must contain a 5-chain. Also it must contain two incomparable 2-chains, only one of which can be in the 5-chain. Also it must contain 5 incomparable elements (only two of which could be in the two chains). So at least 5+2+1+1+1 = 10 elements. I believe this is essentially the kind of lower-bound argument that was mentioned in the earlier question. This "multiple chains" argument says nothing about branching structures in the 5-posets, so perhaps one could consider them and work out an improved lower bound.
(Beware of bad layout, vertex 9 is not covered by 8, but by 10.)
To go further to $f(6)$, this code is probably not practical, mainly because Poset() generates the posets pretty slowly. In SageMath 8.8, it seems >90% of our computation time is spent there, and not in testing for universality. The best road forward is probably to use the C program by Brinkmann & McKay to generate the candidate posets. It should be lightning fast compared to Posets(), so then the bottleneck probably moves to the universality check. The B&M program can be found as an attachment to an old enhancement request for faster poset generation in SageMath. (Also McKay says in a comment here that he can send the code.)
# Find ana u-poset that contains all n-posets as induced posetssubposets.
def find_universal_poset(n,u):
PP = list(Posets(n))
for U in Posets(u):
ok = True
for P in PP:
if not U.has_isomorphic_subposet(P):
ok = False
break
if ok:
return U
return None
2. Proving $f(6)=16$
Update. For $f(6)$ the SageMath code is too slow. We also havecan do faster brute-force in two phases: $15 \le f(6) \le 17$(1) list the candidate posets using "posets.c" by Brinkmann & McKay, available in an old SageMath enhancement request, and (2) check them for 6-universality by C code corresponding to the SageMath code listed above.
The multiple-chains argument gives easily $f(6) \ge 14$, because a 6-universal poset must contain a 6-chain, two mutually incomparable 3-chains, three such 2-chains, and six incomparable elements; these can overlap but at least 6+3+2+1+1+1=14 elements are required.
But $f(6)=14$ was
I have ruled out $f(6)=14$ by exhaustive search over all $1.34 \times 10^{12}$ 14-posets, using the C code "posets.c" by Brinkmann & McKay to list the posets, and some more C code to test for universality (about 16 cpu-days of computation;computation), and ruled out $f(6)=15$ similarly (about 1200 cpu-days). The result rests on heavy computation, so it would be nice to have been impossible in Sage)a more succint lower bound proof, perhaps from a more elaborate version of the multiple-chains argument. Doing exhaustive
Exhaustive search over 15all 16- orposets would take about 500 cpu-years, but some solutions were found after just 190 cpu-hours, that is, having done about 1/20000 of the search space. (There must be quite a lot of 6-universal 16-element posets does not seem promising — I hopeposets out there are better ways, to explain this luck.) One of proving lower boundsthe solutions has cover relation
[[2, 0], [2, 1], [3, 0], [3, 1], [4, 0], [4, 1], [5, 0], [5, 1], [6, 0], [7, 0], [8, 2], [8, 3], [8, 4], [8, 6], [8, 7], [9, 6], [9, 7], [10, 6], [11, 9], [11, 10], [12, 2], [12, 3], [12, 10], [13, 9], [13, 12], [14, 5], [14, 7], [14, 12], [15, 11], [15, 13], [15, 14]].
So we have $f(6) = 16$.
Another computational approach, which gives upper bounds, for upper bounds is to start from a known 6-universal poset, such as the Boolean lattice $B_6$ (= power set with inclusion relation), and remove elements one by one, if possible without breaking the universality. The idea of removing some unneeded elements is already implicit in the old question. This is potentially much faster than brute-force for finding positive examples -- if they exist! Not knowing any better, I removed elements in random order until impossible, and restarted 100 times. Already here I got one 17-poset and seventeen 18-posets. This 6-universal 17-poset has cover relation
[[0, 11], [0, 13], [0, 15], [1, 2], [1, 3], [1, 5], [2, 8], [2, 11], [3, 11], [3, 12], [4, 5], [4, 10], [5, 6], [5, 7], [6, 9], [6, 11], [6, 14], [7, 8], [7, 12], [8, 9], [8, 13], [9, 16], [10, 11], [10, 12], [10, 15], [11, 16], [12, 13], [12, 14], [13, 16], [14, 16], [15, 16]].
(Again bad layout, sorry; check the listed cover relation for reference.) This was done with very simple SageMath code (below), certainly one could do much more searching with faster C code.
3. Bounds for $f(7)$
Update (12.1.2021). You can knock me over with a feather. I had the brute Brute-force search for 6-universal 16is pretty much out of question (note that nobody has even counted all 17-posets running at a low priority "just in case we are lucky"yet). For some loose bounds:
The full search would take about 500 cpumultiple-yearschains argument gives $f(7) \ge 16$, but some solutions were found after just 190 cpubecause you need one 7-hourschain, that istwo 3-chains, having done about 1/20000 of the search spacethree 2-chains and seven incomparable elements, 7+3+2+1+1+1+1=16. There must be quite a lot
Removing random elements from $B_7$, we find easily (in less than ten random restarts) an example of 6a 7-universal 1625-posets out thereposet, to explain this luck. One of the solutions haswith cover relation
[[2[[0, 0]7], [0, 8], [0, 14], [1, 2], [1, 5], [2, 1]
6], [2, 11], [3, 0]4], [3, 1]5], [3, 8], [3, 14], [4, 0]7], [4, 1]18], [5, 0]6],
[5, 7], [5, 1]12], [6, 0]9], [7[6, 0]13], [8[6, 2]19], [8[7, 3]22], [8[7, 4]23], [8, 6]9], [8
[9, 7]15], [9, 6]23], [9[10, 7]11], [10, 6]12], [10, 14], [11, 9]13], [11, 10]15],
[11, 20], [12, 2]13], [12, 3]15], [12, 10]16], [13[12, 9]20], [13, 12]21], [14, 5]15], [14
[14, 7]16], [14[15, 12]22], [15[16, 11]24], [15[17, 13]18], [15[18, 14]]19], [19, 20], [19, 23],
[20, 21], [20, 22], [21, 24], [22, 24], [23, 24]].
(The question of a 15-element 6-universal poset is still open
So we have $f(7) \le 25$. Unless someone hasThis might be improved by trying more random restarts, perhaps with faster methodsC code. I'm not planning to do that now, but it willshould be settled in about week from now by brute forcestraightforward.)
