The cardinal of various monoids of transformations that preserve a uniform partition

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1 The cardinal of various monoids of transformations that preserve a uniform partition Vítor H. Fernandes 1 and Teresa M. Quinteiro 2 July 30, 2010 Abstract In this paper we give formulas for the number of elements of the monoids OR m n of all full transformations on a finite chain with mn elements that preserve a uniform m-partition and preserve or reverse the orientation and for its submonoids OD m n of all order-preserving or order-reversing elements, OP m n of all orientationpreserving elements, O m n of all order-preserving elements, O + m n of all extensive order-preserving elements and O m n of all co-extensive order-preserving elements Mathematics subject classification: 20M20, 05A10. Keywords: order-preserving/reversing, orientation-preserving/reversing, extensive, equivalence-preserving, transformations. Introduction and preliminaries For n N, let X n be a finite chain with n elements, say X n = {1 < 2 < < n}. Following the standard notations, we denote by PT n the monoid (under composition of all partial transformations on X n and by T n and I n its submonoids of all full transformations and of all injective partial transformations, respectively. A transformation α PT n is said to be extensive (resp., co-extensive if x xα (resp., xα x, for all x Dom(α. We denote by T + n (resp., T n the submonoid of T n of all extensive (resp., co-extensive transformations. A transformation α PT n is said to be order-preserving (resp., order-reversing if x y implies xα yα (resp., yα xα, for all x, y Dom(α. We denote by PO n the submonoid of PT n of all order-preserving partial transformations. As usual, we denote by O n the monoid PO n T n of all full transformations that preserve the order. This monoid has been extensively studied since the sixties (e.g. see [2, 1, 20, 34, 7, 3, 31, 9]. In particular, in 1971, Howie [21] showed that the cardinal of O n is ( 2 and later, jointly with Gomes, in [18] they proved that PO n = n ( n ( n+i 1 i i + 1. See also Laradji and Umar papers [27] and [28]. Next, denote by O n + (resp., by On the monoid T n + O n (resp., Tn O n of all extensive (resp., co-extensive order-preserving full transformations. The monoids O n + and On are isomorphic and it is well-known that the pseudovariety of J-trivial monoids, which are the syntactic monoids of piecewise testable languages (see e.g. [30], is generated by the family {O + n n N}. Moreover, the cardinal of O n + (or On is the n th -Catalan number, i.e. O n + = n+1( 1 2n n (see [32]. Regarding the injective counterpart of O n, i.e. the inverse monoid POI n = PO n I n of all injective orderpreserving partial transformations, we have POI n = ( 2n n. This result was first presented by Garba in [17] (see also [7]. 1 The author gratefully acknowledges support of FCT and PIDDAC, within the projects ISFL and PTDC/MAT/69514/2006 of CAUL. 2 The author gratefully acknowledges support of ISEL and of FCT and PIDDAC, within the projects ISFL and PTDC/MAT/69514/2006 of CAUL. 1

2 Now, being POD n the submonoid of PT n of all partial transformations that preserve or reverse the order, OD n = POD n T n and PODI n = POD n I n (the full and partial injective counterparts of POD n, respectively, Fernandes et al. [10, 11] proved that POD n = n ( n ( i 2 ( n+i 1 i n + 1, OD n = 2 ( 2 n and PODI n = 2 ( 2n n n 2 1. Wider classes of monoids are obtained when we consider transformations that either preserve or reverse the orientation. Let a = (a 1, a 2,..., a t be a sequence of t, t 0, elements from the chain X n. We say that a is cyclic (resp., anti-cyclic if there exists no more than one index i {1,..., t} such that a i > a i+1 (resp., a i < a i+1, where a t+1 denotes a 1. Let α T n and suppose that Dom(α = {a 1,..., a t }, with t 0 and a 1 < < a t. We say that α is orientation-preserving (resp., orientation-reversing if the sequence of its images (a 1 α, a 2 α,..., a t α is cyclic (resp., anti-cyclic. This notions were introduced by McAlister in [29] and independently Catarino and Higgins in [6]. Denote by POP n (resp., POR n the submonoid of PT n of all orientation-preserving (resp., orientationpreserving or orientation-reversing transformations. The cardinalities of POP n and POR n were calculated by Fernandes et al. [12] and are 1 + (2 n 1n + n k=2 k( n 22 k n k and 1 + (2 n 1n + 2 ( n 22 2 n 2 + n k=3 2k( n 22 k n k, respectively. As usual, OP n denotes the monoid POP n T n of all full transformations that preserve the orientation, OR n denotes the monoid POR n T n of all full transformations that preserve or reserve the orientation and POPI n and PORI n denote the submonoids of POP n and POR n, respectively, whose elements are the injective transformations. McAlister in [29], and independently Catarino and Higgins in [6], proved that OP n = n ( 2 n(n 1 and ORn = n ( 2n n n 2 2 (n2 2n n. The monoids OP n and OR n were also studied by Arthur and Rušcuk in [5]. Regarding their injective counterparts, in [8], Fernandes established that POPI n = 1 + n ( 2n 2 n and, in [10], Fernandes et al. showed that PORIn = 1 + n ( 2n n n 2 2 (n2 2n + 3. All these results are summarized in [13]. Now, let X be a set and denote by T (X the monoid (under composition of all full transformations on X. Let ρ be an equivalence relation on X and denote by T ρ (X the submonoid of T (X of all transformations that preserve the equivalence relation ρ, i.e. T ρ (X = {α T (X (aα, bα ρ, for all (a, b ρ}. This monoid was studied by Huisheng in [23] who determined its regular elements and described its Green s relations. Let m, n N. Of particular interest is the submonoid T m n = T ρ (X mn of T mn, with ρ the equivalence relation on X mn defined by ρ = (A 1 A 1 (A 2 A 2 (A m A m, where A i = {(i 1n + 1,..., in}, for i {1,..., m}. Notice that the ρ-classes A i, with 1 i m, form a uniform m-partition of X mn. Regarding the rank of T m n, first, Huisheng [22] proved that it is at most 6 and, later on, Araújo and Schneider [4] improved this result by showing that, for X mn 3, the rank of T m n is precisely 4. Finally, denote by OR m n the submonoid of T m n of all orientation-preserving or orientation-reversing transformations, i.e. OR m n = T m n OR mn. Similarly, let OD m n = T m n OD mn, OP m n = T m n OP mn and O m n = T m n O mn. Consider also the submonoids O + m n = O m n T + mn and O m n = O m n T mn of O m n whose elements are the extensive transformations and the co-extensive transformations, respectively. Example 0.1 Consider the following transformations of T 12 : ( ( α 1 = ; α = ; ( ( α 3 = ; α = ; ( ( α 5 = ; α = ; ( ( α 7 = ; α = Then, we have: α 1 T 3 4, but α 1 OR 3 4 ; α 2 OR 3 4, but α 2 OP 3 4 ; α 3 OD 3 4, but α 3 O 3 4 ; α 4 OP 3 4, but α 4 O 3 4 ; α 5 O 3 4, but α 5 O and α 5 O3 4 ; α 6 O ; α 7 O3 4 ; and, finally, α 8 T

3 Notice that, as On and O n +, the monoids Om n and O+ m n are isomorphic [15]. Recall that in [25] Kunze proved that the monoid O n is a quotient of a bilateral semidirect product of its subsemigroups On and O n +. This result was generalized by the authors [15] by showing that O m n also is a quotient of a bilateral semidirect product of its subsemigroups Om n and O+ m n. See also [26, 14]. In [24] Huisheng and Dingyu described the regular elements and the Green s relations of O m n. On the other hand, the ranks of the monoids O m n, O m n + and O m n were calculated by the authors in [15]. Regarding OP m n, a description of the regular elements and a characterization of the Green s relations were given by Sun et al. in [33]. Its rank was determined by the authors in [16], who also computed in the same paper the ranks of the monoids OD m n and OR m n. In this paper we calculate the cardinals of the monoids OR m n, OP m n, OD m n, O m n, O + m n and O m n. In order to achieve this objective, we use a wreath product description of T m n, due to Araújo and Schneider [4], that we recall in Section 1. 1 Wreath products of transformation semigroups In [4] Araújo and Schneider proved that the rank of T m n is 4, by using the concept of wreath product of transformation semigroups. This approach will also be very useful in this paper. Next, we recall some facts from [4, 15, 16]. First, we define the wreath product T n T m of T n and T m as being the monoid with underlying set Tn m T m and multiplication defined by (α 1,..., α m ; β(α 1,..., α m; β = (α 1 α 1β,..., α mα mβ ; ββ, for all (α 1,..., α m ; β, (α 1,..., α m; β Tn m T m. Now, let α T m n and let β = α/ρ T m be the quotient map of α by ρ, i.e. for all j {1,..., m}, we have A j α A jβ. For each j {1,..., m}, define α j T n by kα j = ((j 1n + kα (jβ 1n, for all k {1,..., n}. Let α = (α 1, α 2,..., α m ; β Tn m T m. With these notations, the function ψ : T m n T n T m, α α, is an isomorphism (see [4, Lemma 2.1]. Observe that, from this fact, we can immediately conclude that the cardinal of T m n is n nm m m. ( Example 1.1 Consider the transformation α = T ( ( ( ( Then, being β =, α =, α = and α =, we have α = (α 1, α 2, α 3 ; β. Next, consider O m n = {(α 1,..., α m ; β O m n O m jβ = (j + 1β implies nα j 1α j+1, for all j {1,..., m 1}}. Notice that, if (α 1,..., α m ; β O m n and 1 i < j m are such that iβ = jβ, then nα i 1α j. Proposition 1.2 [15] The set O m n is a submonoid of T n T m (and of O n O m isomorphic to O m n. and On the other hand, being we have: O + m n = {(α 1,..., α m ; β O m 1 n O m n = {(α 1,..., α m ; β O n O m 1 n O + n O + m jβ = (j + 1β implies nα j 1α j+1 and jβ = j implies α j O + n, for all j {1,..., m 1}} O m (j 1β = jβ implies nα j 1 1α j and jβ = j implies α j O n, for all j {2,..., m}}, Proposition 1.3 [15] The set O + m n [resp. O m n] is a submonoid of T n T m (and of O n O m isomorphic to O + m n [resp. O m n ]. 3

4 A description of OP m n in terms of wreath products is more elaborate. In fact, considering addition modulo m (in particular, m + 1 = 1, we have: Proposition 1.4 [16] A (m + 1-tuple (α 1, α 2,..., α m ; β of Tn m satisfies one of the following conditions: 1. β is a non-constant transformation of OP m, for all i {1,..., m}, α i O n and, for all j {1,..., m}, jβ = (j + 1β implies nα j 1α j+1 ; 2. β is a constant transformation, for all i {1,..., m}, α i O n and there exists at most one index j {1,..., m} such that nα j > 1α j+1 ; 3. β is a constant transformation, T m belongs to OP m n ψ if and only if it there exists one index i {1,..., m} such that α i OP n \ O n and, for all j {1,..., m} \ {i}, α j O n and, for all j {1,..., m}, nα j 1α j+1. Let α OP m n. We say that α is of type i if αψ satisfies the condition i. of the previous proposition, for i {1, 2, 3}. 2 The cardinals In this section we use the previous bijections to obtain formulas for the number of elements of the monoids O m n, O + m n, O m n, OD m n, OP m n and OR m n. In order to count the elements of O m n, on one hand, for each transformation β O m, we determine the number of sequences (α 1,..., α m On m such that (α 1,..., α m ; β O m n and, on the other hand, we notice that this last number just depends of the kernel of β (and not of β itself. With this purpose, let β O m. Suppose that Im(β = {b 1 < b 2 < < b t }, for some 1 t m, and define k i = b i β 1, for i = 1,..., t. Being β an order-preserving transformation, the sequence (k 1,..., k t determines the kernel of β: we have {k k i 1 + 1,..., k k i }β = {b i }, for i = 1,..., t (considering k k i = 1, with i = 1. We define the kernel type of β as being the sequence (k 1,..., k t. Notice that 1 k i m, for i = 1,..., t, and k 1 + k k t = m. Now, recall that the number of non-decreasing sequences of length k from a chain with n elements (which is the same as the number of k-combinations with repetition from a set with n elements is ( ( n+k 1 k = n+k 1 (see [19], for example. Next, notice that, as a sequence (α 1,..., α k On k satisfies the condition nα j 1α j+1, for all 1 j k 1, if and only if the concatenation sequence of the images of the transformations α 1,..., α k (by this order is still a non-decreasing sequence, then we have ( n+k such sequences. Since (α 1,..., α m ; β O m n if and only if, for all 1 i t, α k1 + +k i 1 +1,..., α k1 + +k i are k i orderpreserving transformations such that the concatenation sequence of their images (by this order is still a nondecreasing sequence, then we have t ( ki n+ elements in Om n whose (m + 1-component is β. Finally, now it is also clear that if β and β are two elements of O m with the same kernel type then (α 1,..., α m ; β O m n if and only if (α 1,..., α m ; β O m n. Thus, as the number of transformations β O m with kernel type of length t (1 t m coincides with the number of t-combinations (without repetition from a set with m elements, it follows: Theorem 2.1 O m n = 1 k 1,...,k t m k 1 + +k t=m 1 t m ( ki n+ ( m t t. 4

5 The table below gives us an idea of the size of the monoid O m n. m \ n In view of Theorem ( 2.1, finding the cardinal of OD m n is not difficult. Indeed, consider the reflexion 1 2 mn 1 mn permutation h =. Observe that h OD mn mn m n and, given α T m n, we have α OD m n if and only if α O m n or hα O m n. On the other hand, as clearly O m n = ho m n and O m n ho m n = {α O m n Im(α = 1} = mn, it follows immediately that: Theorem 2.2 OD m n = 2 O m n mn = 2 1 k 1,...,k t m k 1 + +k t=m 1 t m ( m t t ( ki n+ mn. Next, we describe a process to count the number of elements of O m n +. First, recall that the cardinal of O n + is the n th -Catalan number, i.e. O n + = n+1( 1 2n n. See [32]. It is also useful to consider the following numbers: θ(n, i = {α O n + 1α = i}, for 1 i n. Clearly, we have O n + = n θ(n, i. Moreover, for 2 i n 1, we have θ(n, i = θ(n, i θ(n 1, i 1. In fact, {α O n + 1α = i} = {α O n + 1α = i < 2α} {α O n + 1α = 2α = i} and it is easy to show that the function which maps each transformation β {α O n + 1α = i < 2α} into the transformation ( n {α O n + 1α = i + 1} i + 1 2β... nβ and the function which maps each transformation β {α O + 1α = i 1} into the transformation ( n 1 n {α O n + 1α = 2α = i} i i 2β (n 2β + 1 (n 1β + 1 are bijections. Thus θ(n, i = {α O n + 1α = i < 2α} + {α O n + 1α = 2α = i} = {α O n + 1α = i + 1} + {α O + 1α = i 1} = θ(n, i θ(n 1, i 1. Also, it is not hard to prove that θ(n, 2 = θ(n, 1 = θ(n 1, i = O+. Now, we can prove: Lemma 2.3 For all 1 i n, θ(n, i = i ( 2n i 1 ( n = i 2n i 1 n. n i Proof. We prove the lemma by induction on n. For n = 1, it is clear that θ(1, 1 = 1 = 1 ( Let n 2 and suppose that the formula is valid for n 1. Next, we prove the formula for n by induction on i. For i = 1, as observed above, we have θ(n, 1 = O + ( = 1 2n 2 ( n. For i = 2, we have θ(n, 2 = θ(n, 1 = 1 2n 2 n = 2 (2n 2! n (!(! 2n 2 = 2 (2n 3! n (!(n 2! = 2 ( 2n 3 n. Now, suppose that the formula is valid for i 1, with 3 i n. Then, using both induction hypothesis on i and on n in the second equality, we have θ(n, i = θ(n, i 1 θ(n 1, i 2 = i 1 n (2n i! (2n i 1! (2n i!, as required. i 1 n (!(n i+1! i 2 (n 2!(n i+1! = i(n i+1 n(2n i (!(n i+1! = i n ( 2n i 1 ( 2n i i 2 ( 2n i 1 n 2 = 5

6 Recall that (α 1,..., α m ; β O + m n if and only if β O m, + α m O n +, α 1,..., α m 1 O n and, for all j {1,..., m 1}, jβ = (j + 1β implies nα j 1α j+1 and jβ = j implies α j O n +. Let β O m. + As for the monoid O m n, we aim to count the number of sequences (α 1,..., α m On m such that (α 1,..., α m ; β O + m n. Let (k 1,..., k t be the kernel type of β. Let K i = {k 1 + +k i 1 +1,..., k 1 + +k i }, for i = 1,..., t. Then, β fixes a point in K i if and only if it fixes k k i, for i = 1,..., t. It follows that (α 1,..., α m ; β O + m n if and only if, for all 1 i t: 1. If β does not fix a point in K i, then α k1 + +k i 1 +1,..., α k1 + +k i are k i order-preserving transformations such that the concatenation sequence of their images (by this order is still a non-decreasing sequence (in this case, we have ( k i n+ subsequences (αk1 + +k i 1 +1,..., α k1 + +k i allowed; 2. If β fixes a point in K i, then α k1 + +k i 1 +1,..., α k1 + +k i 1 are k i 1 order-preserving transformations such that the concatenation sequence of their images (by this order is still a non-decreasing sequence, nα k1 + +k i 1 1α k1 + +k i and α k1 + +k i O n + (in this case, we have n ( (ki 1n+j 1 j=1 j 1 θ(n, j subsequences (α k1 + +k i 1 +1,..., α k1 + +k i allowed. Define for all 1 i t. Thus, we have: { ( ki n+, if (k1 + + k i β k k i d(β, i = j ( 2n j 1 ( (ki 1n+j 1, if (k1 + + k i β = k k i, n j=1 n Proposition 2.4 O + m n = β O m + t d(β, i. j 1 Next, we obtain a formula for O + m n which does not depend of β O+ m. Let β be an element of O + m with kernel type (k 1,..., k t. Define s β = (s 1,..., s t {0, 1} t 1 {1} by s i = 1 if and only if (k k i β = k k i, for all 1 i t 1. Let 1 t, k 1,..., k t m be such that k k t = m and let (s 1,..., s t {0, 1} t 1 {1}. Let k = (k 1,..., k t and s = (s 1,..., s t. Define (k, s = {β O + m β has kernel type k and s β = s}. In order to get a formula for (k, s, we count the number of distinct restrictions to unions of partition classes of the kernel of transformations β of O + m with kernel type k and s β = s corresponding to maximal subsequences of consecutive zeros of s. Let β be an element of O + m with kernel type k and s β = s. First, notice that, given i {1,..., t}, if s i = 1 then K i β = {k k i } and if s i = 0 then the (unique element of K i β belongs to K j, for some i < j t. Next, let i {1,..., t} and r {1,..., t i} be such that s j = 0, for all j {i,..., i + r 1}, s i+r = 1 and, if i > 1, s i 1 = 1 (i.e. (s i,..., s i+r 1 is a maximal subsequence of consecutive zeros of s. Then (K i K i+r 2 K i+r 1 β K i+1 K i+r 1 (K i+r \ {k k i+r }. Let l j = K i+j (K i K i+r 1 β, for 1 j r. Hence, we have l 1,..., l r 1 0, l r 1, l l r = r and 0 l l j j, for all 1 j r 1. On the other hand, given l 1,..., l r such that l 1,..., l r 1 0, l r 1, l 1 + +l r = r and 0 l 1 + +l j ( ki+2 ( ki+r 1 l r 1 ( ki+r 1 l r = ( ki+r 1 l r j, for all 1 j r 1, we have precisely ( k i+1 l 1 l 2 restrictions to K i K i+r 1 of transformations β of O m, + with kernel type k and s β r 1 j=1 ( ki+j l j distinct = s, such that 6

7 l j = K i+j (K i K i+r 1 β, for 1 j r. It follow that the number of distinct restrictions to K i K i+r 1 of transformations β of O + m with kernel type k and s β = s is l 1 + +l r=r 0 l 1 + +l j j, 1 j r 1 l 1,...,l r 1 0, l r 1 ( ki+r 1 l r r 1 ( ki+j j=1 l j. Now, let p be the number of distinct maximal subsequences of consecutive zeros of s. Clearly, if p = 0 then (k, s = 1. Hence, suppose that p 1 and let 1 u 1 < v 1 < u 2 < v 2 < < u p < v p t be such that {j {1,..., t} s j = 0} = p {u i,..., v i 1} (i.e. (s ui,..., s vi 1, with 1 i p, are the p distinct maximal subsequences of consecutive zeros of s. Then, being r i = v i u i, for 1 i p, we have (k, s = p l 1 + +l ri =r i 0 l 1 + +l j j 1 j r i 1 l 1,...,l ri 1 0, l ri 1 ( kui +r i 1 l ri ri 1 ( kui +j l j=1 j Finally, notice that, if β and β two elements of O m + with kernel type k = (k 1,..., k t such that s β = s β, then d(β, i = d(β, i, for all 1 i t. Thus, defining Λ(k, s = t d(β, i, where β is any transformation of O m + with kernel type k and s β = s, we have: Theorem 2.5 O + m n = O m n = k=(k 1,...,k t s {0,1} t 1 {1} 1 k 1,...,k t m k 1 + +k t=m 1 t m (k, sλ(k, s. The next table gives us an idea of the size of the monoid O + m n (or O m n. m \ n Despite the unpleasant appearance, the previous formula allows us to calculate the cardinal of O + m n, even for larger m and n. For instance, we have O = In order to count( the number of elements of the monoid OP m n, we begin by recalling that, for k N, 1 2 k 1 k being g k the k-cycle OP 2 3 k 1 k, each element α OP k admits a factorization α = g j k γ, with 0 j k 1 and γ O k, which is unique unless α is constant [6]. Next, consider the permutations (of {1,..., mn} ( 1 2 mn 1 mn g = g mn = OP 2 3 mn 1 mn. 7

8 and f = g n = ( 1 n n + 1 mn n mn n + 1 mn n + 1 2n 2n + 1 mn 1 n OP m n. Being α an element of OP m n \ O m n of type 1 or 2 (see Proposition 1.4 and j {1,..., m 1} such that (jnα > (jn + 1α, as (jn + 1α (mnα 1α (jnα, it is clear that f j α O m n. Thus, each element α of OP m n of type 1 or 2 admits a factorization α = f j γ, with 0 j m 1 and γ O m n, which is unique unless α is constant. Notice that, this uniqueness follows immediately from Catarino and Higgins s result mentioned above. Therefore we have precisely m( O m n mn non-constant transformations of OP m n of types 1 and 2 and mn constant transformations (which are elements of type 2 of OP m n. Now, let α be a transformation of OP m n of type 3. As α is not constant, it can be factorized in a unique way as g r γ, for some r {0,..., mn 1} \ {jn 0 j m 1} and some non-constant order-preserving transformation γ from {1,..., mn} to A i, for some 1 i m. Since only elements of OP m n of type 3 have factorizations of this form and the number of non-constant and non-decreasing sequences of length mn from a chain with n elements is equal to ( ( (mn+ mn+ n, we have precisely m(mn m n elements of type 3 in OP m n. Thus OP m n = m O m n + m 2 (n 1 ( mn+ mn(mn 1 and so we obtain: Theorem 2.6 OP m n = m 1 k 1,...,k t m k 1 + +k t=m 1 t m ( m t t ( ki n+ + m 2 (n 1 ( mn+ mn(mn 1. It follows a table that gives us an idea of the size of the monoid OP m n. m \ n We finish this paper computing the cardinal of the monoid OR m n. Notice that, as for OD m n and O m n, we have a similar relationship between OR m n and OP m n. In fact, α OR m n if and only if α OP m n or hα OP m n. Hence, since OP m n = hop m n and OP m n hop m n = {α OP m n Im(α 2}, we obtain OR m n = 2 OP m n {α OP m n Im(α = 2} mn. It remains to calculate the number of elements of A = {α OP m n Im(α = 2}. First, we count the number of elements of A of types 2 and 3. Let α be such a transformation. Then, there exists k {1,..., m} such that Im(α A k. Clearly, in this case, the number of distinct kernels allowed for α coincides with the number of distinct kernels allowed for transformations of OP mn of rank 2, which is ( mn 2 (see [6]. On the hand, it is easy to check that we have m ( n 2 distinct images for α. Furthermore, for each such possible kernel and image, we have two distinct transformations of A. Hence, the total number of elements of A of types 2 and 3 is precisely 2m ( ( n mn 2 2. Finally, we determine the number of elements of A of type 1. Let α A be of type 1 and suppose that αψ = (α 1,..., α m ; β. Then β must have rank 2 and so, as β OP m, we have 2 ( m 2 2 distinct possibilities for β (see [6]. Moreover, for each 1 i m, α i must be a constant transformation of O n and, for 1 i, j m, if iβ = jβ then α i = α j. Thus, for a fixed β, since β as rank 2, we have precisely n 2 sequences (α 1,..., α m ; β allowed. Hence, A has 2n 2( m 2 2 distinct elements of type 1. Therefore, OR m n = 2 OP m n 2m ( ( n mn 2 2 2n 2 ( m 2 ( mn 2 2n 2 ( m 2 2 mn(2mn 1 and so we get: 2m ( n 2 2 mn = 2m Om n + 2m 2 (n 1 ( mn+ 8

9 Theorem 2.7 OR m n = 2m 1 k 1,...,k t m k 1 + +k t=m 1 t m ( m t t ( ki n+ + +2m 2 (n 1 ( mn+ ( 2m n mn 2( 2 2n 2 ( m 2 2 mn(2mn 1. References [1] A.Ya. Aĭzenštat, Homomorphisms of semigroups of endomorphisms of ordered sets, Uch. Zap. Leningr. Gos. Pedagog. Inst. 238 (1962, (Russian. [2] A.Ya. Aĭzenštat, The defining relations of the endomorphism semigroup of a finite linearly ordered set, Sb. Math. 3 (1962, (Russian. [3] J. Almeida and M.V. Volkov, The gap between partial and full, Internat. J. Algebra Comput. 8 (1998, [4] J. Araújo and C. Schneider, The rank of the endomorphism monoid of a uniform partition, Semigroup Forum 78 (2009, [5] R.E. Arthur and N. Ruškuc, Presentations for two extensions of the monoid of order-preserving mappings on a finite chain, Southeast Asian Bull. Math. 24 (2000, 1 7. [6] P.M. Catarino and P.M. Higgins, The monoid of orientation-preserving mappings on a chain, Semigroup Forum 58 (1999, [7] V.H. Fernandes, Semigroups of order-preserving mappings on a finite chain: a new class of divisors, Semigroup Forum 54 (1997, [8] V.H. Fernandes, The monoid of all injective orientation-preserving partial transformations on a finite chain, Comm. Algebra 28 (2000, [9] V.H. Fernandes, Semigroups of order-preserving mappings on a finite chain: another class of divisors, Izv. Vyssh. Uchebn. Zaved. Mat. 3 (478 (2002, (Russian. [10] V.H. Fernandes, G.M.S. Gomes and M.M. Jesus, Presentations for some monoids of injective partial transformations on a finite chain, Southeast Asian Bull. Math. 28 (2004, [11] V.H. Fernandes, G.M.S. Gomes and M.M. Jesus, Congruences on monoids of order-preserving or orderreversing transformations on a finite chain, Glasgow Math. J. 47 (2005, [12] V.H. Fernandes, G.M.S. Gomes and M.M. Jesus, Congruences on monoids of transformation preserving the orientation on a finite chain, J. Algebra 321 (2009, [13] V.H. Fernandes, G.M.S. Gomes and M.M. Jesus, The cardinal and the idempotent number of various monoids of transformations on a finite chain, Bull. Malays. Math. Sci. Soc., to appear. [14] V.H. Fernandes and T.M. Quinteiro, Bilateral semidirect product decompositions of transformation monoids, DM-FCTUNL pre-print series 10/2009. [15] V.H. Fernandes and T.M. Quinteiro, On the monoids of transformation that preserve the order and a uniform partition, Comm. Algebra, to appear. [16] V.H. Fernandes and T.M. Quinteiro, On the ranks of certain monoids of transformation that preserve a uniform partition. 9

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