# Finding certain posets in SAGE

Distributive lattices can be characterised as lattices having no sublattice isomorphic to the diamond or the pentagon (see for example https://en.wikipedia.org/wiki/Distrib... ). Now I try to obtain in SAGE all connected posets with a global maximum and a global minimum that do not contain the diamond or the pentagon as a subposet. How to do that in SAGE in the quickest possible way? And do such (maybe not necessarily connected) posets have a name or are enumerated in the OEIS?

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The oeis is here: A006982

It is a highly combinatorial job to compute the numbers in the sequence.

Their strategy is to compute first the numbers of the vertically indecomposable, distributive lattices. Which are lattices that cannot be decomposed as a vertical sum of at least two distributive lattices. The vertical sum of $$(D_A;0_A,1_A;\le_A)$$ (placed in our minds vertically below) and $$(D_B;0_A,1_A;\le_A)$$ (placed in our minds vertically above the previous structure) is the lattice on the set $D=D_A\sqcup D_B/1_A\sim 0_B$ with $0=0_A$ and $1=1_B$ and the induced order that makes $D_A\le D_B$, elementwise.

Each distributive lattice appears uniquely as a vertical sum of vertically independent distributive lattices. The generating series for the vertically independent ones is: $V(t) = t^2+t^4+t^6+3t^8+t^9+6t^{10}+2t^{11}+16t^{12}+8t^{13}+42t^{10}+28t^{15}+112t^{16}+\dots$ and form here on the numbers start to grow "also for the human eye in exponential taste". The coefficient in power $t^{49}$ is $V_{49}= 5 045 200 597$. Having this list, we can generate the list for the distribute lattices, we will get $$D = \sum_{k\ge 0}D_kt^k$$ with $D_{49}=908 414 736 485$.

At the level of enumeration, if we know the coefficients in $V$, then we can find the ones in $D$ simply. Let $W=V/t$. Then $$D=t(1+V+V^2+\dots)=\frac t{1-V}$$ with computations in $\mathbb Z[[ t ]]$ . Sage code for this:

R.<t> = PolynomialRing( QQ )
V = ( t^2 + t^4 + t^6 + 3*t^8 + t^9
+ 6*t^10 + 2*t^11 + 16*t^12 + 8*t^13 + 42*t^14 + 28*t^15 + 112*t^16 + 93*t^17 + 311*t^18 + 295*t^19
+ 869*t^20 + 939*t^21 + 2454*t^22 + 2931*t^23 + 7032*t^24 + 9036*t^25 + 20301*t^26 + 27701*t^27 + 58929*t^28 + 84413*t^29
+ O(t^30) )
W = V/t
D = t/(1-W)
for c in D.coefficients():
print c


This gives:

1
1
1
2
3
5
8
15
26
47
82
151
269
494
891
1639
2978
5483
10006
18428
33749
62162
114083
210189
386292
711811
1309475
2413144
4442221


In accordance with A006982.

So it is enough to write code that generates the vertically independent, distributive lattices.

I will try to do this till degree $64$ in the following days... This is then computing $D$ also till degree $64$.

more

Thank you for your answer. Sorry im confused. Do you enumerate the distributive lattices here? My question was about connected posets with top and buttom that do not contain the pentagon and the diamond, which should be more general than distributive lattice (the lattice property is not included).

( 2017-09-23 12:24:16 -0500 )edit

The equivalence holds for lattices

https://en.wikipedia.org/wiki/Distributive_lattice

where they claim

A lattice is distributive if and only if none of its sublattices is isomorphic to $M_3$ or $N_5$.

(My relation to this field of research is rather tangential. The only intersection is related to standard tableaux and the corresponding order.)

Let $X$ with existing $0=\min X$ and $1=\max X$ be a poset without pentagons as subposets. Assume $a,b\in X$ have two different least upper bounds $s,t$. Then the subposet with elements $0;a,s;b;1$ is a pentagon. Contradiction. So the supremum of $(a,b)$ exists, since $1$ is an upper bound. Same / dually for infimum.

Is this ok?!

( 2017-09-23 13:19:50 -0500 )edit

I have no experience with posets or lattice, I am more or less just interesting in homological things related to incidence algebras. So I might do stupid things here. I thought I had a poset with top and bottum without pentagon or diamond poset that is not a lattice, but of course I might have a thinking error. I do not see a pentagon in 0;a,s;b;1, shouldnt it look like number 6 in http://math.chapman.edu/~jipsen/tikzs... ?

( 2017-09-23 13:52:56 -0500 )edit
1

Yes, my argument above is false, an error, $s$ is connected to both $a$ and $b$. But if we allow configurations like

  1
/ \
s   t
|\ /|
| \ |
|/ \|
a   b
\ /
0


then the world becomes very rich! Even if we assume a level structure, the combinatorial data between levels may become very complicated. At this point i have no idea how to sort the structure...

( 2017-09-24 12:40:59 -0500 )edit