A376025 Number of elements of the free multiplicatively idempotent rig on n generators.

4, 13, 284, 510605

Offset: 0

Author

Morgan Rogers, Sep 06 2024

Keywords

• hard
• more
• nonn

Comments

Let T_n denote the free idempotent monoid on n generators and alpha: T_n -> P([n]) the function sending an element to the generators appearing in it. Then this sequence is computed as the number of triples (S,D,p), where S is a "replete" subsemigroup of T_n (i.e. a subsemigroup for which (xuy) and (xvy) in S implies (xuvy) is in S), D is a "sparse" subset of T_n (i.e., for s and t in D, alpha(s) a subset of alpha(t) implies s = t), S "dominates" D (i.e. for s in S and t in D, alpha(s) is not a subset of alpha(t) and S is closed under multiplication by elements of D) and p is a "parity" function from the image of S under alpha to {0,1}. See Rogers 2024.

Examples

			For n = 0, the free idempotent rig on zero generators is the quotient of the natural numbers by the congruence generated by x ~ x^2. Considering different values of x, this yields the trivial relations 0 ~ 0 and 1 ~ 1, then 2 ~ 4, whence x+2 ~ x+4 for every x. By a parity argument and induction, this entirely determines the congruence: 2 is related to every larger even number and 3 is related to every larger odd number. Thus the resulting rig thus has 4 elements: 0, 1 and the equivalence classes of 2 and 3.
		

Links

• Morgan Rogers, From free idempotent monoids to free multiplicatively idempotent rigs, arXiv:2408.17440 [math.RA], 2024, pages 28-30.

Crossrefs

Cf. A005345.