We consider algebras with a single

One of the most demanding tasks of combinatorics consists in counting finite algebraic structures with certain properties or even in classifying them up to isomorphism. Various types of structures can be enumerated by the number of their elements using the methods of Pólya [

This is the standard notion of isomorphism for algebras, (cf. [

Hence we have asked how many associative mono-2-ary algebras with

In fact, there are many different ways to generalize associativity from binary to

This type of associativity implies a general associativity law that will be explained in Section

Left-right-pushing was introduced by Dörnte [

We would like to complete Table

Number of nonisomorphic associative mono-

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||

1 | 1 | 3 | 7 | 19 | 47 | 130 | 343 | 951 | |

2 | 1 | 5 | 24 | 188 | |||||

3 | 1 | 6 | |||||||

Since it seems to be very difficult to find general formulas for the rows even for small values of

These results might be interesting for some aspects of nonlinear coding theory. An example of an application is given in Section

When denoting a (multi)product of several elements of a semigroup it is not necessary to write brackets in order to indicate the order of calculation since the value of the product only depends on the elements and their total order. This is called “general associativity law,” and it may be generalized for multiproducts in associative mono-

Let us fix some terms. If

every element of the algebra is a term,

every product with terms as entries (the

Furthermore, a

We can formulate now a basic result that was remarked by Dörnte [

In an associative mono-

By induction on the number of subterms of a given term we prove that every term can be transformed into a right-tower-term. By induction we may assume that the left-most subterm that is not a single element is a right-tower-term; thus we have

Here we consider associative mono-

We define the following types of algebras on the set

type

type

type

type

type

type

In case of odd

The types

The main result of this article will be as follows.

Let

Let

The general idea of the proof is to distinguish cases according to the values of certain products. In each case

There are four possibilities for the values of the products

The case

If

Indeed, whenever two products have the same number of zero entries, each zero can be replaced by the product

If

Using the assumptions and the associativity of

By Lemma

If

Prove by induction that

If

For

For

If

If

For all

If we had instead of

By a similar argument (transposition of 0 and 1) one could change the conditions

If

For all

If

We need some preparations before realizing that for all types of entries of a product the values are fixed by the premises. Note that we may not use the remark, since the algebra restricted on the main diagonal is the identical combination.

For

Assume that the assertion is not true, that is, that

The proof is by induction. The case

This proves Claim

Indeed, if

This proves Claim

Suppose that

This completes the proof of the lemma.

Now we continue with the proof of the main theorem. As described in the beginning of this subsection and by the notice between Lemma

We have seen that, for any fixed

If we consider the case of 4 or more elements, the numbers in the column of Table

If there is a 0 or an 1 entry, then the product is 0.

Otherwise the product is 0 or 1.

There are at least

The question arises whether or not a classification of the associative mono-

As an application of Theorem

To be precise we want to assign a single control bit

which we apply recursively on substrings, replacing these substrings by the value of the mapping. For reasons of efficiency it might be advantageous not to choose the substrings in canonical order, rather in a randomlike order. This may be the case if it is cheaper to find and calculate two identical substrings than to calculate two different substrings. In order to have for each possible word a well-defined control bit

From Theorem

The situation changes for an alphabet with

Although the control character produced by this semigroup contains hardly any information about the number of the letters 2 and 0 in the string it provides a lot of information about letter 1, namely the parity of its occurrence.

The author would like to thank A. Romanowska, P. Burmeister, U. Faigle, B. Fuchs, and B. Peis for some hints and comments.