1. Introduction

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 [1, 2]. However, there are some very difficult combinatorial problems that have not been solved until now. One of these problems is to determine the number of semigroups with elements, where is a positive integer. An asymptotic formula for the number of labelled semigroups with elements was found in [3]. But a different problem consists in counting up to isomorphism. Let us reformulate this problem into the language of universal algebra. An -ary operation on a set is a function . In this article the ground set will always be finite. is called mono- -ary algebra (with elements). Two mono--ary algebras and are isomorphic if there is a bijection from to , so that for all

This is the standard notion of isomorphism for algebras, (cf. [4])

Hence we have asked how many associative mono-2-ary algebras with elements exist up to isomorphism. We may consider the more general problem of determining the number of associative mono--ary algebras with elements, where and are positive integers. But what does it mean for an algebra with a single operation to be associative?

In fact, there are many different ways to generalize associativity from binary to -ary operations. A well-known example is superassociativity, introduced by Menger [5]. Another example is the associativity of diagonal algebras [6]. This paper will treat only a natural kind of associativity which could be called left-right-pushing. Let be a mono--ary algebra. It is called associative (in the left-right sense) if for all and

Note that a list of the type denotes an empty list.

This type of associativity implies a general associativity law that will be explained in Section 2.

Left-right-pushing was introduced by Dörnte [7] who generalized the notion of groups to -groups where the binary operation is replaced by an -ary. These -groups (or polyadic groups) were investigated further by Post [8]. Later the concept was widened to polyadic semigroups by Zupnik [9] or associatives as Gluskin and Shvarts [10] called them. These are other names for the associative mono--ary algebras we study here. Whilst Zupnik, Gluskin, and Shvarts investigated the representation of certain polyadic semigroups by (associative) binary and unary operations, we are interested in counting polyadic semigroups with no restrictions on their structure, only depending on their order and arity.

We would like to complete Table 1 which lists the numbers of associative mono--ary algebras with elements (up to isomorphism) for some small values of and . The entries of the table were computed by the author using brute force algorithms

Table 1: Number of nonisomorphic associative mono--ary algebras with elements.

Since it seems to be very difficult to find general formulas for the rows even for small values of , one might try to find formulas for the columns for small values of . Surprisingly we will find a simple formula for the case . Moreover, we explicitly classify all associative mono--ary algebras with 2 elements using purely elementary methods. Note that we even need not make use of the methods of Zupnik [9] to prove our result. The occurring algebras will be introduced in Section 3.1. In Section 3.2 we prove that every associative mono--ary algebra with 2 elements is isomorphic to one of these algebras which will be the main subject of this article. Similar results for the columns with could not be found, but in Section 4 we will see that the entries of the columns with are at least exponentially increasing.

These results might be interesting for some aspects of nonlinear coding theory. An example of an application is given in Section 5.

2. The General Associativity Law

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--ary algebras.

Let us fix some terms. If is a mono--ary algebra, then a product is a formal expression with some entries . A term is defined recursively by the following:

Furthermore, a right-tower-term is a term where for each occurring product all entries are single elements of the algebra except for the last entry.

We can formulate now a basic result that was remarked by Dörnte [7]. Because of its importance for this article the idea of a formal proof is added.

Theorem 2.1. In an associative mono--ary algebra the value of any term does not depend on the structure of the subterms, only on the total order of the entries.

Proof. 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 where are single elements and are terms. Applying associativity times we may shift the left-most inner product to the right and obtain By using the induction hypothesis again the right subterm can be transformed into a right-tower-term, and we are done.

3. The 2-Element-Column

Here we consider associative mono--ary algebras with 2 elements. This main section is organized as follows: In Section 3.1 the main result is presented whereas Section 3.2 is devoted to its proof.

3.1. The Occurring Algebras

We define the following types of algebras on the set with single -ary operation :

type 0:

type A:

type L:

type R:

type G0:

type G1:

It is easy to see that these algebras are associative. (For type G0 and type G1 distinguish cases according to the parity of the number of zeros in the inner and outer brackets when verifying the associativity laws.)

In case of odd these types are pairwise nonisomorphic. The same holds in case of even except for type G0 and type G1 which are then isomorphic. In the latter case we also call type G0 and type G1 simply type G. In the trivial case type L, type R and type G0 are isomorphic to type A.

The types G0 and G1 are the only polyadic groups with 2 elements which can be seen easily [7]. However, the analogue classification of polyadic semigroups is more complicated because of the lack of invertibility.

The main result of this article will be as follows.

Theorem 3.1. Let be an associative mono--ary algebra with and . Then is isomorphic to one of the types defined above.

3.2. Proof of Theorem 3.1

Let be an associative mono--ary algebra, . In order to simplify the notation we denote the products only with brackets, that is,

The general idea of the proof is to distinguish cases according to the values of certain products. In each case will be determined only by these values.

There are four possibilities for the values of the products and . The case 1, 0 is treated in Lemma 3.3, and the case 0, 0 in Lemmas 3.4 and 3.5. Obviously, by exchanging 0 and 1, the case 1, 1 leads to isomorphic algebras as in the case 0, 0.

The case and is more complicated. Here we need to consider the products , , , and , too. The subcases 1, 0 and 0, 1, respectively, 0, 0 for and the subcases 0, 1 and 1, 0, respectively, 1, 1 for are examined by Lemma 3.6, respectively, Lemma 3.7. Note that these cases are not excluding each other and may be contradictory which has no effect on our proof. The remaining case is Lemma 3.8.

Lemma 3.2. If in , then the value of a product depends only on the number of zeros in the product and not on the order of the elements.

Proof. Indeed, whenever two products have the same number of zero entries, each zero can be replaced by the product and the new expressions contain both the same number of ones and thus have the same value because of the general associativity law.

Lemma 3.3. If and , then is odd and is isomorphic to type G1.

Proof. Using the assumptions and the associativity of we obtain Thus (in order to avoid the contradiction ) we conclude Assume that is even. Then, using (3.9) we have the contradiction So is odd. We will now establish the following claim.Claim 1. and for .We prove the claim by induction. The first assertion is clear for . Assume that the first assertion has been proved for all and that the second assertion has been proved for all and that . Then by assumption Let us fist consider the case Then in (3.11) we obtain the contradiction . Thus we have Now assume that the first and the second assertions have been proved for all and that . Then by (3.13) we conclude Let us first consider the case Then we obtain the contradiction Therefore we have This proves the claim.
By Lemma 3.2 it follows immediately from Claim 1 that a product is 0 if and only if the number of its zero entries is even. Thus is of type G1.

Lemma 3.4. If and and then is isomorphic to type 0.

Proof. Prove by induction that The induction step is Use Lemma 3.2 again and conclude that every product is 0.

Lemma 3.5. If and and then is even and is isomorphic to type G.

Proof. In order to avoid the contradiction we have Claim 1. for .
For this is (3.23). Assume the claim has been proved for . Then
that is, the claim is true for .Claim 2. for .
For this is (3.24). Assume that the claim has been proved for . Then
that is, the claim is true for .
If is odd, then the two claims contradict each other. So is even and according to Lemma 3.2, the premises, and Claims 1 and 2 a product is 0 if and only if it contains an even number of zeros; that is, is of type G.

Lemma 3.6. If and and then is isomorphic to type R.

Proof. For all we have Thus is isomorphic to type R.

If we had instead of and the alternative conditions and in Lemma 3.6, we also would obtain type R since exchanging 0 and 1 does not affect this type. On the other hand, if we exchanged the order of the entries of all products, that is, if we had and (or and ), would be isomorphic to type L.

By a similar argument (transposition of 0 and 1) one could change the conditions and in the next lemma into and .

Lemma 3.7. If and and then is isomorphic to type A.

Proof. For all we have and for arbitrary Thus is isomorphic to type A.

Lemma 3.8. If and and then is odd and is isomorphic to type G0.

Proof. 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 the premises
are contradictory. So assume that . We will now establish three claims.Claim 1.
Assume that the assertion is not true, that is, that
Then we obtain the contradiction This proves Claim 1.Claim 2. for
The proof is by induction. The case is given by a premise. Now assume that and that the assertion is true for . Then by assumption and Claim 1
and thus the assertion is true for .
This proves Claim 2.Claim 3. is odd.
Indeed, if was even, we would have
by premise and by Claim 2 which is a contradiction.
This proves Claim 3. We are now ready to prepare the main argument in the proof of Lemma 3.8. If we write , the dots at the beginning and at the end denote an arbitrary number of arbitrary entries, so that the total number of entries is . Let be even. By Claim 2 we obtain According to (3.41) we may shift even numbers of zeros (resp., ones) as ones (resp., zeros) to the left without changing the value of a product, so that every product can be written in an equivalent normal form: For example, the product (0111011) can be transformed in three steps into the normal form: The next claims determine the values of these normal forms.Claim 4. for odd and for even.Claim 5. for odd and for even. By symmetry arguments it is sufficient to prove Claim 4. We use the same kind of double-step-induction as in the proof of Claim 1.
Suppose that