Semigroup Forum Vol. 75 (2007) 221–236
c 2007 Springer
DOI: 10.1007/s00233-007-0732-8
RESEARCH ARTICLE
Combinatorial Results for the Symmetric Inverse Semigroup A. Laradji and A. Umar Communicated by Thomas E. Hall
Abstract In this note we obtain and discuss formulae for the number of partial oneone transformations (of an n -element set) of height (equivalently, width) r and having exactly k fixed points. Moreover, we consider a generalization of the ‘Hat Problem’ or ‘Probl`eme des Rencontres’. Keywords: Semigroup, nilpotent, permutation, derangement, full transformation, partial transformation, partial one-one transformation, Laguerre polynomial, ordinary generating function, exponential generating function. AMS Classification: 20M18, 20M20, 05A10, 05A15.
1. Introduction Let Xn = {1, 2, . . . n} be a finite n -element set, and let Tn and Pn be the full transformation semigroup and the partial transformation semigroup of Xn , respectively. Another closely related semigroup to Tn and Pn is In , the semigroup of partial one-one transformations of Xn , also known as the finite symmetric inverse semigroup on Xn . This paper investigates certain combinatorial properties of In . As remarked by Higgins [7], combinatorial properties of Tn and some of its subsemigroups have been studied over a long period and many interesting and delightful results have emerged (see for example [2], [8], [9], [10], [11], [12]). Garba [4] obtained some corresponding results in the semigroup Pn and recently, the authors in [15], [16] investigated combinatorial properties of certain subsemigroups of Pn and found analogous results. However, all these developments concerning combinatorial properties of Tn and Pn , and some of their subsemigroups have not been matched by corresponding results for the semigroup In or some of its subsemigroups. The results in this paper complement the combinatorial results for In in [5], [1], [13] and more recently, in [3]. At the end of this introductory section we gather some known combinatorial results about In that we shall need in later sections. In Section 2 we establish many combinatorial results for In , the main result being Proposition 2.9 which gives a recurrence formula for an,k , the number of partial one-one maps α (of Xn ) having exactly k fixed points. Finally, in Section 3 it is observed that our main result from Section 2 could be interpreted as a further generalization of the ‘Hat
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Problem’ or ‘Probl`eme des Rencontres’ where we show (in Proposition 3.2) that the limit ratio of the number of partial derangements to the total number of partial permutations is, as in the classical case of full permutations, 1/e. For a given (partial) mapping or transformation α: Y ⊆ X −→ X we denote its set of fixed points by F (α) = {x ∈ Y : xα = x}, its domain Y by Dom α and its image set by Im α. If Domα = X then α is called a full or total mapping, otherwise it is strictly partial. The height of α is |Im α| and width of α is |Dom α|. We list some known combinatorial results (about In ) that we shall need later. Throughout, we let rn be the cardinality of In and an be the number of all partial one-one maps (of Xn ) having no fixed points. Result 1.1. 2 n r! . r Result 1.2.
[5, p. 383] Let J(n, r) = |{α ∈ In : |Im α| = r}|. Then J(n, r) =
2 n n [1, Proposition 2.1] rn = r! . Moreover, for n ≥ 2 r r=0
rn = 2nrn−1 − (n − 1)2 rn−2 with r0 = 1 and r1 = 2. Result 1.3.
[1, Theorem 2.3] Let r(x) =
rn xn . Then r(x) converges for n!
n≥0
|x| < 1 to the function ex/(1−x) /(1 − x) . 2. Combinatorial results As in [15] or [14] we define f (n, r, k) as f (n, r, k) = |{α ∈ In : |Im α| = r (= |Dom α|) ∧ |F (α)| = k}|.
(2.1)
Then it is not difficult to see that f (n, r, k) =
n k
f (n − k, r − k, 0).
(2.2)
Thus to compute f (n, r, k) it is sufficient to compute f (n, r, 0) . However, note that f (n, r, 0) is the number of partial one-one maps of width r and without fixed points. For simplicity of notation we set f (n, r, 0) = f (n, r) . Howie and Giraldes (in a private communication) obtained a recurrence for f (n, n−1, k) on Tn , however, characterizing f (n, r, k) on Tn still remains an open problem. It is
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also worth noting that nilpotent partial one-one maps have no fixed points [13], but the converse is not true. In fact α defined by 1α = 2 and 2α = 1 is clearly without fixed points but α2 is the (partial) identity map on {1, 2} . It is clear that f (n, n) is the number of derangements (permutations without any fixed points) of an n -element set, and it is well-known that (see for example [2] or [17]) f (n, n) = (n − 1){f (n − 1, n − 1) + f (n − 2, n − 2)} = nf (n − 1, n − 1) + (−1)n = n!
n k=0
(2.3a)
(−1)k k!
(2.3b)
with f (n, 0) = 1 . However, f (n, r) may also be expressed as n f (n, r) = c(n, r) r
(2.4)
where c(n, r) is the number of partial one-one maps of width r , without fixed points and having a fixed domain, say {x1 , x2 , . . . , xr } ⊆ Xn . It is now easy to see that c(n, 0) = 1 and c(n, n) = f (n, n). More generally, we have Proposition 2.1. Let c(n, r) be the number of partial one-one maps α: {x1 , x2 , . . . , xr } → Xn having no fixed points. Then r (−1)m n−m c(n, r) = r! , r−m m! m=0
(0 ≤ r ≤ n).
Proof. First observe that the number of all partial one-one maps α: {x1 , x2 , . . . , xr } → Xn is n(n − 1) · · · (n − r + 1) = (n)r , the falling factorial of n of degree r . Now by the Principle of Inclusion-Exclusion we have r r c(n, r) = (n − m)r−m (−1)m m m=0 r
r! (n − m)! (r − m)!m! (n − r)! m=0 r (−1)m n − m = r! , r−m m!
=
(−1)m
m=0
as required.
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Laradji and Umar The following two lemmas are now immediate
Lemma 2.2.
For 1 ≤ r < n , c(n, r) = rc(n − 1, r − 1) + c(n − 1, r).
Lemma 2.3. r ≤ n).
(n − r)f (n, r) = n(n − r)f (n − 1, r − 1) + nf (n − 1, r), (1 ≤
Theorem 2.4. we have
Let f (n, r, k) as defined in (2.1). Then for n ≥ r ≥ k ≥ 0 ,
r−k n − k − m (−1)m n! f (n, r, k) = . k!(n − r)! m=0 r − k − m m!
Proof. First note that the result for f (n, r) follows directly from Proposition 2.1 and (2.4). Hence the result for f (n, r, k) now follows directly from (2.2) and the result for f (n, r) . Using Result 1.1, Theorem 2.4 and by making a suitable change of variable we obtain the following identity which may be new: Corollary 2.5. n n−m (−1)m
1=
m=0 t=0
Lemma 2.6.
m!t!
.
f (n + 1, n + 1) = f (n, n − 1) (n ≥ 1).
Proof. f (n, n − 1) = n!
n−1
(n − m)
m=0
(−1)m m!
= n(n!)
n−1
n−1 (−1)m−1 (−1)m + n! m! (m − 1)! m=0 m=1
= n(n!)
n n−1 (−1)m−1 (−1)m − (−1)n n + n! m! (m − 1)! m=0 m=1
= nf (n, n) +
n−1 (−1)m−1 (−1)n−1 · n! + n! (n − 1)! (m − 1)! m=1
= nf (n, n) + n!
n−1
(−1)m m! m=0
= n[f (n, n) + f (n − 1, n − 1)] = f (n + 1, n + 1)
(by (2.3a)).
(by (2.3b))
(by (2.3b))
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Now let an,k be the number of partial one-one maps with exactly k fixed points where 0 ≤ k ≤ n . Then it is easy to see that an,k = =
n k n k
an−k,0
(2.5a)
n−m 1 (−1)m n − m (n − k)! , j m! j! m=0 j=0 n−k
and it is also clear that an = an,0 =
n
f (n, r)
(2.5b)
(2.6)
r=0
is the number of partial one-one maps (of Xn ) having no fixed points. Thus, we have Proposition 2.7.
Let an be as defined in (2.6). Then n n−m 1 (−1)m n − m an = n! . j m! j=0 j! m=0
Proof. By using (2.6), Theorem 2.4 and algebraic manipulations successively, we have an
r n! (−1)m n−m = f (n, r) = (n − r)! m=0 r − m m! r=0 r=0 n n n! (−1)m n−m = (n − r)! r − m m! m=0 r=m n
n
n n−m 1 (−1)m n − m = n! j m! j=0 j! m=0 as required. Our next objective is to establish a recurrence for an . For that, recall (0) from [21, p. 101] that the n -th degree Laguerre polynomial Ln (x) = Ln (x) is given by n (−x)m n Ln (x) = (2.7) m m! m=0 and when x = −1 , it satisfies the recurrence pλp = 2pλp−1 − (p − 1)λp−2
(2.8)
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with λ0 = 1, λ1 = 2 and λp = Lp (−1) . Thus we now have n an (−1)m = λn−m n! m! m=0
(2.9a)
which by inverse relation (see [17, p. 44]) gives n n n!λn = ak = rn . k
(2.9b)
k=0
Next consider n (−1)k k=0
k!
(n − k)λn−k = n
n (−1)k k=0
k!
λn−k −
n (−1)k k k=0
k!
λn−k
(−1)k−1 an + λ(n−1)−(k−1) (n − 1)! (k − 1)! n
=
(by (2.9a))
k=1
(−1)k an + λn−1−k (n − 1)! k! n−1
=
k=0
an an−1 = + . (n − 1)! (n − 1)!
(2.10)
On the other hand, using (2.8) we see that n (−1)k k=0
k!
(n − k)λn−k =
n−2 k=0
=
n−2 k=0
+
= 2
(−1)k (−1)n−1 (n − k)λn−k + λ1 k! (n − 1)! (−1)k {2(n − k)λn−k−1 − (n − k − 1)λn−k−2 } k!
2(−1)n−1 (n − 1)!
n−1 k=0
−
n−2 k=0
= 2
n−1 k=0
−
n−2 k=0
(−1)k 2(−1)n−1 (n − k)λn−k−1 − λ0 k! (n − 1)! 2(−1)n−1 (−1)k ((n − 1) − k)λn−k−2 + k! (n − 1)! (−1)k ((n − 1) − (k − 1))λ(n−1)−k k! (−1)k ((n − 2) − (k − 1))λ(n−2)−k . k!
(2.11)
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Laradji and Umar However, using (2.9a) and (2.10) we see that n (−1)k k=0
k!
(n − (k − 1))λn−k =
n (−1)k k=0
=
k!
(n − k + 1)λn−k
an−1 an an + + (n − 1)! (n − 1)! n!
and hence (2.11) becomes an−1 an−2 an−1 + + (n − 2)! (n − 2)! (n − 1)! an−2 an−3 an−2 − + + (n − 3)! (n − 3)! (n − 2)!
an an−1 + = 2 (n − 1)! (n − 1)!
which simplifies to an = (2n − 1)an−1 − (n − 1)(n − 3)an−2 − (n − 1)(n − 2)an−3 . Thus we have proved the following result. Theorem 2.8.
Let an be as defined in (2.6). Then for all n ≥ 3 ,
an = (2n − 1)an−1 − (n − 1)(n − 3)an−2 − (n − 1)(n − 2)an−3 , with a0 = 1. More generally we have Proposition 2.9.
Let an,k be as defined in (2.5a). Then for all n ≥ k ≥ 0 ,
(n−k)an,k = n(2n−2k−1)an−1,k −n(n−1)(n−k−3)an−2,k −n(n−1)(n−2)an−3,k . Proof. The result follows directly from Theorem 2.8 and (2.5a) after some simplifications. Remark 2.10. The triangular array of numbers c(n, r) is [19, A060475], however the triangular arrays of numbers f (n, r, k) (0 ≤ k ≤ r ≤ n) and the sequences an,k (0 ≤ k < n) are not yet listed in [19]. For some selected values of f (n, r, k) (k = 0, 1, 2) and an,k , see Tables 1–4.
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Table 1: f (n, r, 0) n\r 0 1 2 3 4 5 6 7
0 1 1 1 1 1 1 1 1
1
2
3
4
5
6
7
0 2 6 12 20 30 42
1 9 42 130 315 651
2 44 320 1420 4690
9 265 2715 16275
44 1854 25494
265 14833
1854
Σf (n, r, 0) 1 1 4 18 108 780 6600 63840
Table 2: f (n, r, 1) n\r 0 1 2 3 4 5 6 7
0 0 0 0 0 0 0 0 0
1
2
3
4
5
6
7
1 2 3 4 5 6 7
0 6 24 60 120 210
3 36 210 780 2205
8 220 1920 9940
45 1590 19005
264 12978
1855
Σf (n, r, 1) 0 1 2 12 72 540 4680 46200
Table 3: f (n, r, 2) n\r 0 1 2 3 4 5 6 7
0 0 0 0 0 0 0 0 0
1
2
3
4
5
6
7
0 0 0 0 0 0 0
1 3 6 10 15 21
0 12 60 180 420
6 90 630 2730
20 660 6720
135 5565
924
Σf (n, r, 2) 0 0 1 3 24 180 1620 16380
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Laradji and Umar Table 4: an,k n\k 0 1 2 3 4 5 6 7
0
1
2
3
4
5
6
7
1 1 4 18 108 780 6600 63840
1 2 12 72 540 4680 46200
1 3 24 180 1620 16380
1 4 40 360 3780
1 5 60 630
1 6 84
1 7
1
Σan,k = rn 1 2 7 34 209 1546 13327 130922
The following divisibility property of an is worth recording: Proposition 2.11. Proof. that
For n ≥ 1 , lcm(1, 2, . . . , n) divides an and an,1 .
We first prove that for all n ≥ 1 , n|an . Recall (from Theorem 2.4)
r n! (−1)m n−m f (n, r) = (n − r)! m=0 r − m m! =
=
r n(n − 1) · · · (n − r + 1)(n − m)! (−1)m r! r!(r − m)!(n − r)! m! m=0
r−1 (r − 1)!(−1)m n(n − 1) · · · (n − r + 1) n − m r−m (r − 1)! m! m=0 n + (−1)r , r
but (r − 1)!|(n − 1)(n − 2) · · · (n − r + 1) and m!|(r − 1)! for m ≤ r − 1. Hence n f (n, r) = nA(n, r) + (−1)r , r n for some integer A(n, r). Therefore, an = r=0 f (n, r) is divisible by n , since n n (−1)r = 0. r r=0
It remains to prove that n|an+k for all k ≥ 0 . To do this, we note that since an+3 , an+4 , . . . can all be expressed as linear combinations of an , an+1 and an+2 , we need only show that n|an+1 and n|an+2 . However, by Theorem 2.8, n|(an+1 − an ) and so n|an+1 . Also, from the recurrence an+2 = (2n + 3)an+1 − (n + 1)(n − 1)an − (n + 1)nan−1
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we obtain n|an+2 . Finally, since lcm(1, 2, . . . , n)|n.lcm(1, 2, . . . , n − 1) and an,1 = nan−1 , it follows that lcm(1, 2, . . . , n − 1, n)|an,1 . 3. A generalization of the Hat Problem The ‘Hat Problem’ also known as ‘Probl`eme des Rencontres’ is often presented as follows: At a restaurant n people check their hats in, and when they leave their hats are returned in a random order. In how many ways can it happen that no one receives his own hat back, and further what is the probability (for large n ) of such an event E ? The surprising answer which is now folklore is P (E) → e−1 (as n → ∞), and the convergence is very rapid. There are now several equivalent formulations to the ‘Hat Problem’, one of which is the ‘Derangement Problem’. Recall that a permutation σ of Xn is a derangement if σ(x) = x for all x in Xn . It is now not difficult to see that in the ‘Hat Problem’, the number of ways it can happen that no one receives his own hat back is equal to the number of derangements of Xn . As pointed out in [20, p. 85], the ‘Derangement Problem’ was first solved by Montmort in probabilistic terms (in 1708), and later independently investigated by Euler. In fact Montmort, Euler and Laplace considered a more general situation, see Table 5 below. Table 5: Summary Investigated by/in
No. of persons No. of persons No. of persons checking in leaving receiving own hat n n 0
Monmort, Euler & Laplace
n
n
k (≤ n)
Hanson, Seyffarth & Weston
n
r -fixed persons
k (≤ r ≤ n)
this article
n
r -varied persons
k (≤ r ≤ n)
this article
n
any number from k to n
k (≤ n)
Observe that the combinatorial questions of the last two problems in Table 5 have been answered by Theorem 2.4 and (2.5b), respectively. To answer the probabilistic question of the last problem (in Table 5) we first let f (x) be the exponential generating function for an . Then using (2.9a) we see that
n xn (−1)m n = x λn−m n! m! n≥0 n≥0 m=0 (−x)m (−1)m = = xn λn−m xn−m λn−m m! m!
f (x) =
an
m≥0
n≥m
m≥0
n≥m
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Laradji and Umar (−x)m = λn xn . m!
m≥0
n≥0
However, the ordinary generating function for λn (from [18, p. 219]) is
λn xn =
n≥0
ex/(1−x) . 1−x
Thus
2 x/(1−x) (−x)m ex /(1−x) −x e f (x) = λn xn = e · = . m! 1−x 1−x m≥0
(3.1)
n≥0
More generally, we have Proposition 3.1. Let fk (x) be the exponential generating function for an,k = 2 xk ex /(1−x) n an−k . Then fk (x) = . k k!(1 − x) Proof. lhs
= fk (x) =
n k
n!
n≥k
=
(n − k)!
=
n−k
(−1)m
m=0
n≥k t
an−k · xn
xt+k (−1)m
t≥0 m=0
λn−k−m n!m!
λt−m k!m!
n k
(t = n − k)
t 1 (−1)m λt−m = xt xk k! m! t≥0 m=0 1 at t k = (using (2.9a)) x x k! t! t≥0
2
=
xk xk ex /(1−x) f (x) = = rhs, k! k!(1 − x)
as required. Next we establish:
xn
(using (2.9a))
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Proposition 3.2. Let rn and an be the number of partial one-one maps (of Xn ) and partial one-one maps (of Xn ) without fixed points, respectively. Then lim an /rn = e−1 . n→∞
Proof. We are going to prove this result in stages. First, we show by induction that an ≥ nan−1 . Note that a1 = 1 ≥ 1 · 1 = 1 · a0 , is true. Now suppose that ak ≥ kak−1 for all 1 ≤ k < n. Then by Theorem 2.8 and the induction hypothesis successively, we have an = (2n − 1)an−1 − (n − 1)(n − 3)an−2 − (n − 1)(n − 2)an−3 (n − 1)(n − 3) (n − 1)(n − 2) ≥ (2n − 1)an−1 − an−1 − an−2 n−1 n−2 = (2n − 1)an−1 − (n − 3)an−1 − (n − 1)an−2 ≥ (n + 2)an−1 − an−1 = (n + 1)an−1 ≥ nan−1 ,
(3.2)
as required. an Next we show that ≥ e−1 . By iteration on (3.2) we deduce an ≥ rn n! an−k and so (n − k)! an an an an + + + ··· + 0! 1! 2! n! n n n n ≥ an + an−1 + an−2 + · · · + an n n−1 n−2 0 = rn ,
ean ≥
as required. an an+1 We continue the proof by establishing that > , for n ≥ 4 . Again, rn rn+1 we use induction. First notice that a1 a2 a2 a3 a4 < but > > . r1 r2 r2 r3 r4 So suppose that
an−2 an−1 an > > . rn−2 rn−1 rn
Then using Theorem 2.8, the induction hypothesis and Result 1.2 successively, we have an+1 (2n + 1)an − n(n − 2)an−1 − n(n − 1)an−2 = rn+1 rn+1 (2n + 1)an − n(n − 2) arnn · rn−1 − n(n − 1) arnn · rn−2 < rn+1 an (2n + 1)rn − n(n − 2)rn−1 − n(n − 1)rn−2 < rn rn+1
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(2n + 2)rn − n2 rn−1 rn − 2nrn−1 + n(n − 1)rn−2 − rn+1 rn+1 2 an (n − 1) rn−2 − n(n − 1)rn−2 an (n − 1)rn−2 = 1+ = 1− rn rn+1 rn rn+1 an < , rn
an = rn
as required. Now put, Fn (x) =
n ak k=0
k!
xk and Gn (x) =
n rk k=0
k!
xk .
Then for all x in [0, 1) , we have using Result 1.3 and (3.1) Fn (x) → e−x as n → ∞. Gn (x) Moreover, for all n ≥ 6 , Fn (x)/Gn (x) ≥ this note that rn Fn − an Gn = (a0 rn − an r0 ) + +
an for each x in [0, 1) . To see rn
x (a1 rn − an r1 ) + · · · 1!
xn−1 (an−1 rn − an rn−1 ) (n − 1)!
≥ 0 as
ai an ≥ for all i ≤ n and n ≥ 6 , from above. ri rn To complete the proof we now establish the statement:
an ≤ e−1 . rn an We achieve this by first noting that is a convergent sequence since it is rn positive and eventually decreasing. Moreover, from above we see that lim
n→∞
lim
n→∞
i.e., i.e., i.e.,
Fn (x) ≥ Gn (x) e−x ≥
lim e−x
x→1−
e−1
an rn an lim n→∞ rn an ≥ lim n→∞ rn an ≥ lim , n→∞ rn lim
n→∞
thereby completing the proof of Proposition 3.2.
(x ∈ [0, 1))
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Now suppose we randomly pick a map α from In and let Ek be the event that such an α has exactly k fixed points, what is the probability of such an event occurring? The corresponding ‘Generalized Hat Problem’ is: Suppose n people check in their hats in a restaurant, and when some (possibly none or all) of them leave, their hats are returned in a random order. In how many ways can it happen that only k (0 ≤ k ≤ n) of them received their own hats back, and further what is the probability of such an event occurring for large n ? As noted earlier the combinatorial question has been answered. Moreover, the probabilistic question has been answered only for k = 0 , by Proposition 3.2. The next proposition provides an answer to this question for an arbitrary k . Let rn be the number of partial one-one maps of Xn , and 1 an,k be as defined in (2.5a). Then, for a fixed k , lim = e−1 . n→∞ rn k!
Proposition 3.3. let an,k
First observe that from [21, Theorem 8.23], we have √ 1 1 −1/4 2 n √ √ λn = L(0) 1 + O (−1) = e n . n 2 πe n
Proof.
Hence
√ 2n1/4 πe √ λn → 1 as n → ∞, e2 n
that is, for a fixed k (0 ≤ k ≤ n) we have √
λn−k (n − k)−1/4 e2 n−k √ ∼ as n → ∞ λn n−1/4 e2 n 1 √ = , √ 1/4 (1 − k/n) e2 n−2 n−k but (1 − k/n)1/4 → 1 and e2
√
√ n−2 n−k
√
= e2
4k √ n+2 n−k
→ 1 as n → ∞.
Thus λn−k /λn → 1 as n → ∞ . Now, using (2.9b) and Proposition 3.3 we have n an−k k an,k an−k rn−k an−k (n − k)!λn−k n n = = · = · k k rn rn rn−k rn rn−k n!λn =
1 an−k λn−k 1 · → (e−1 )(1) as n → ∞, k! rn−k λn k!
as required. Remark 3.4. Putting k = 0 in Proposition 3.3, we recover Proposition 3.2. Moreover, the probability (for large n ) that no more than two persons receive
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their own hats back is more than 0.91 . It is worth mentioning that the probabilities that no one receives his own hat back and that exactly one person does are equal. Acknowledgment The authors would like to gratefully acknowledge support from the King Fahd University of Petroleum and Minerals. References [1] Borwein, D., S. Rankin and L. Renner, Enumeration of injective partial transformations, Discrete Math. 73 (1989), 291–296. [2] Comtet, L., “Advanced Combinatorics: The Art of Finite and Infinite Expansions”, D. Reidel Publishing Company, Dordrecht, Holland, 1974. [3] Ganyushkin, O. and V. Mazorchuk, Some remarks on the combinatorics of IS n , Semigroup Forum 70 (2005), 391–405. [4] Garba, G. U., Idempotents in partial transformation semigroups, Proc. Roy. Soc. Edinburgh 116 (1990), 359–366. [5] Gomes, G. M. S. and J. M. Howie, Nilpotents in finite symmetric inverse semigroups, Proc. Edinburgh Math. Soc. 30 (1987), 383–395. [6] Hanson, D., K. Seyffarth and J. H. Weston, Matchings, derangements, rencontres, Math. Mag. 56 (1983), 224–229. [7] Higgins, P. M., Combinatorial results for semigroups of order-preserving mappings, Math. Proc. Camb. Phil. Soc. 113 (1993), 281–296. [8] Higgins, P. M. and E. J. Williams, Random functions on a finite set, Ars. Combinatoria 26 (1988), 93–102. [9] Howie, J. M., Combinatorial and probabilistic results in transformation semigroups, pp. 200–206 in “Words, Languages and Combinatorics II”, World Scientific Publishing, River Edge, NJ, 1994. [10] Howie, J. M., “Fundamentals of Semigroup Theory”, Clarendon Press, Oxford, 1995. [11] Howie, J. M., E. L. Lusk and R. B. McFadden, Combinatorial results relating to products of idempotents in finite full transformation semigroups, Proc. Roy Soc. Edinburgh 115 (1990), 289–299. [12] Howie, J. M., E. F. Robertson and B. M. Schein, A combinatorial property of finite full transformation semigroups, Proc. Roy. Soc. Edinburgh 109 (1988), 319–328.
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[13] Laradji, A. and A. Umar, On the number of nilpotents in the partial symmetric semigroup, Commun. in Algebra 32 (2004), 3017–3023. [14] Laradji, A. and A. Umar, On certain finite semigroups of order-decreasing transformations I, Semigroup Forum 69 (2004), 184–200. [15] Laradji, A. and A. Umar, Combinatorial results for semigroups of orderdecreasing partial transformations, Journal of Integer Sequences 7 (2004), 04.3.8. [16] Laradji, A. and A. Umar, Combinatorial results for semigroups of orderpreserving partial transformations, Journal of Algebra 278 (2004), 342–359. [17] Liu, C. L., “Introduction to Combinatorial Mathematics”, McGraw Hill Company, New York, 1968. [18] Riordan, J., “Combinatorial Identities”, John Wiley and Sons, New York, 1968. [19] Sloane, N. J. A., “The On-Line Encyclopedia of Integer Sequences”, available at http://www.research.att.com/ ∼njas/sequences/. [20] Stanley, R. P., “Enumerative Combinatorics”, Vol. I, Cambridge University Press, 1997. [21] S¨zego, G., “Orthogonal Polynomials”, Amer. Math. Soc. Colloquium Publications Vol. XXIII, Amer. Math. Soc., Providence, RI, 1975.
Department of Mathematical Sciences King Fahd University of Petroleum and Minerals Dhahran 31261 Saudi Arabia
[email protected] [email protected]
Received March 18, 2006 and in final form May 17, 2007 Online publication August 27, 2007
