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Generalized Bessel numbers and some combinatorial settings
Discrete Mathematics 313 (2013) 2127–2138
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Generalized Bessel numbers and some combinatorial settings✩ Gi-Sang Cheon a,∗ , Ji-Hwan Jung a , Louis W. Shapiro b a
Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea
b
Department of Mathematics, Howard University, Washington, DC 20059, USA
article
abstract
info
Article history: Received 17 October 2012 Received in revised form 1 May 2013 Accepted 2 May 2013 Available online 24 May 2013
Stirling numbers and Bessel numbers have a long history, and both have been generalized in a variety of directions. Here, we present a second level generalization that has both as special cases. This generalization often preserves the inverse relation between the first and second kind, and has simple combinatorial interpretations. We also frame the discussion in terms of the exponential Riordan group. Then the inverse relation is just the group inverse, and factoring inside the group leads to many results connecting the various Stirling and Bessel numbers. © 2013 Elsevier B.V. All rights reserved.
Keywords: Stirling numbers Bessel numbers Partitions G-Permutations Telephone exchange Semi-bipartite Riordan matrices
1. Introduction The Bessel number of the second kind, B(n, k), is the number of ways in which an n-set X can be partitioned into k blocks of size one or two. These Bessel numbers have many properties similar to those of the Stirling numbers of the second kind. An explicit formula for B(n, k) is well known as B(n, k) =
n! 2n−k
0
(n − k)!(2k − n)!
n
≤ k ≤ n, 2 otherwise. if
(1)
If j is the number of singleton blocks in these set partitions, then the Bessel numbers can be displayed in matrix form by n +j means of singleton blocks, the (n, j) entries being the Bessel numbers given by B(n, 2 ) if n + j is even and 0 if n + j is odd: 1 0 1 0 3 0 15
0 1 0 3 0 15 0
0 0 1 0 6 0 45
0 0 0 1 0 10 0
0 0 0 0 1 0 15
0 0 0 0 0 1 0
0 0 0 0 = ez 2 /2 , z . 0 0 1
··· ✩ This work was supported by a National Research Foundation of Korea Grant funded by the Korean Government (NRF-2012-007648).
∗
Corresponding author. E-mail addresses: [email protected] (G.-S. Cheon), [email protected] (J.-H. Jung), [email protected] (L.W. Shapiro).
0012-365X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.disc.2013.05.001
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The ⟨, ⟩ notation is the exponential Riordan group terminology,1 which will be explained briefly at the end of this section. For an alternative approach, see [16]. If we think of a telephone call as a matching or a block of size 2, then we have the telephone exchange problem [10]. The total number of ways in which n subscribers to a telephone exchange can be connected is given by the row sums of the matrix, and the generating function is 1+z+2
z2
+4
2!
z3 3!
+ 10
z4 4!
+ 26
z5 5!
+ · · · = ez +z
2 /2
.
Here, the n subscribers could be vertices, and an edge connecting two vertices would represent a phone call. We would then be thinking about a labeled graph in which every vertex has degree 0 or 1. Then the connected subgraphs are either a single point or an edge. The connected part has the exponential generating function z + z 2 /2!; see [1,3,4,14,15] for the exponential formula. There are many possible variations. For instance, a cell phone could be turned off, so the blocks of size 1 now can be in two states. Then the generating function is e2z +z /2! . Here is a different variation. We now also allow conference calls involving three parties. The connected part generating function is thus z + z 2 /2! + z 3 /3!, and the resulting generating function is 2
ez +z /2!+z /3! . It is natural to allow blocks (or conference calls) of every size giving the connected part generating function z + z 2 /2! + z 3 /3! + · · · = ez − 1. Now (ez − 1)k /k! is the generating function when we want exactly k blocks and n subscribers (labeled vertices, distinguishable elements). If k is the number of blocks, we obtain the matrix ⟨1, ez − 1⟩, the entries being the Stirling numbers of the second kind, S (n, k). Removing the restriction on block size gives the row sums, z starting 1, 1, 2, 5, 15, 51, . . . , the Bell numbers with generating function ee −1 . The dual is given by arranging each block in a circle (or in a line with the largest element at the head of the line). The connected part is 2
3
z+
z2 2!
+ 2!
z3 3!
+ 3!
z4
+ ··· = z +
4!
z2 2
+
z3 3
+
z4 4
+ · · · = ln
1 1−z
.
1 Now, the matrix is 1, ln 1− , the entries being the unsigned Stirling numbers of the first kind, s(n, k). The row sums have z the generating function
e
1 ln 1− z
=
1 1−z
= 1 + z + z2 + · · · =
n≥0
n!
zn n!
.
This generating function summarizes the fact that any element of the symmetric group on n letters can be written uniquely as the product of disjoint cycles. The duality referred to earlier, after inserting minus signs on every other diagonal, can be expressed as
⟨1, ln(1 + z )⟩ 1, ez − 1 = ⟨1, z ⟩ in Riordan group terminology. Bessel numbers and Stirling numbers have been generalized in a variety of directions. One direction in which Bessel numbers have been generalized is by permitting each block to be of size at most s for s ≥ 3. We refer to these numbers as the s-restricted numbers of the second kind, M2s (n, k) [6,7]. There is a second direction in which Stirling numbers of the second kind have been generalized as r-Stirling numbers, Sr (n, k) [5]. Suppose that there are r distinguished elements in separate blocks or cycles. We may consider dividing up an army into squads, where the r lieutenants are in different squads. The Stirling numbers of the first kind sr (n, k) would generalize by having the r distinguished elements in separate cycles. For matchings, we might have r seeded players that we do not want to compete against each other. In this paper, we present a second level generalization that has both the s-restricted numbers and the r-Stirling numbers as special cases. This generalization often preserves the inverse relation between the first kind and the second kind, and has simple combinatorial interpretations. We also frame the discussion in terms of the exponential Riordan group. Then the inverse relation is just the group inverse, and factoring inside the group leads to many results connecting the various Stirling, telephone exchange, and Bessel numbers. Many combinatorial counting problems can be treated systematically using the Riordan group introduced by Shapiro, Getu, Woan, and Woodson [12]. Here, we use various elements from the exponential version of the Riordan group, and the generating functions are exponential generating functions, which we abbreviate as egfs. An exponential Riordan matrix [2], also denoted as an e-Riordan matrix, is an infinite lower triangular matrix L = [ℓn,k ] whose kth column has egf g (z )f (z )k /k!, where g (0) ̸= 0, f (0) = 0 and f ′ (0) ̸= 0. Equivalently, ℓn,k = n![z n ]g (z )f (z )k /k!, where [z n ] is the operator that extracts the coefficient of z n . The matrix L is denoted by ⟨g (z ), f (z )⟩. It is easy to show that, if we multiply L = ⟨g (z ), f (z )⟩ by a column vector h = (h0 , h1 , . . .)T corresponding to the egf h(z ), then the resulting column
1 As is customary in this setting, we index our rows and columns starting with the index 0.
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vector Lh has egf g (z )h(f (z )). This observation is known as the fundamental theorem of a Riordan matrix (FTRM), and we write this as
⟨g (z ), f (z )⟩ h(z ) = g (z )h(f (z )). The FTRM implies that the set of e-Riordan matrices forms a group under the multiplication defined by
⟨g (z ), f (z )⟩ ⟨h(z ), ℓ(z )⟩ = ⟨g (z )h(f (z )), ℓ(f (z ))⟩ . The identity element is ⟨1, z ⟩, the usual identity matrix, and the inverse of ⟨g , f ⟩ is 1 ⟨g (z ), f (z )⟩−1 = , f (z ) , g (f (z ))
(2)
where f is the compositional inverse, i.e., f (f (z )) = f (f (z )) = z. As shown in [12,13], Riordan matrices provide a powerful method for deriving generating functions, proving combinatorial counting results and identities, and inverting identities. For instance, let ⟨g r (z ), f (z )⟩ = [Gr (n, k)], r ≥ 1, where n![z n ]g (z ) = gn and n![z n ]f (z ) = fn . Using ⟨g r (z ), f (z )⟩ = ⟨g (z ), z ⟩ g r −1 (z ), f (z ) , one can derive the identity Gr (n, k) =
n i
i≥k
gn−i Gr −1 (i, k).
(3)
2. Two types of Bessel number generalized In this section, we consider Bessel numbers generalized in two different types, and we will give their combinatorial interpretations. We begin with φs (z ) and ψs (z ) given by
φs (z ) = 1 + z +
z2
+ ··· +
zs
and ψs (z ) = 1 + z +
z2
+ ··· +
zs
.
2! s! 2 s For an indeterminate x and an integer r ≥ 0, we then define two functions Φr ,s (z ) and Ψr ,s (z ) as
Φr ,s (z ) = (φs′ (z ))r (φs (z ))x and Ψr ,s (z ) = (ψs′ (z ))r (ψs (z ))x . Let us now define the (r , s)-Bessel numbers of the second type, Br ,s (n, k), by the identity n dn Φr ,s (z ) (n) = Br ,s (n, k)xk , Φr ,s (0) := dz n z =0
k=0
where x = x(x − 1) · · · (x − k + 1) and x0 = 1. Similarly, the (r , s)-Bessel numbers of the first type, Ar ,s (n, k), are defined by the identity k
Ψr(,ns) (0)
dn Ψr ,s (z )
:=
dz n
=
z =0
n
Ar ,s (n, k)xk .
k=0
If s → ∞, then Br ,s (n, k) and Ar ,s (n, k) are identical to the r-Stirling numbers of the second kind, Sr (n, k), and of the first kind, sr (n, k), respectively; see [5]. Lemma 2.1. Let G1 (z ), G2 (z ) ∈ C[[z ]] such that G1 (0) ̸= 0, G2 (0) = 1, G′2 (0) ̸= 0, and define G(z ) = G1 (z )(G2 (z ))x for an indeterminate x. Then ⟨G1 (z ), G2 (z ) − 1⟩ = [d(n, k)] if and only if G(n) (0) :=
dn G(z ) dz n
= z =0
n
d(n, k)xk .
(4)
k =0
Proof. If (4) holds, the Taylor series expansion of G(z ) at z = 0 gives G(z ) =
(n)
G
(0)
n ≥0
zn n!
=
n n ≥0
k=0
d(n, k)x
k
zn n!
=
k≥0
d(n, k)
n≥k
zn n!
xk .
(5)
Also, we have G(z ) = G1 (z )(G2 (z ))x = G1 (z )(1 + (G2 (z ) − 1))x = G1 (z )
x k≥0
=
k≥0
G1 (z )
k
(G2 (z ) − 1)k
(G2 (z ) − 1) k x . k! k
Equating the coefficients of xk in (5) and (6), we obtain ⟨G1 (z ), G2 (z ) − 1⟩ = [d(n, k)].
(6)
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Conversely, if ⟨G1 (z ), G2 (z ) − 1⟩ = [d(n, k)] = L, by the FTRM we have L[x0 , x1 , . . .]T = ⟨G1 (z ), G2 (z ) − 1⟩ (1 + z )x = G1 (z )(G2 (z ))x = G(z ) G(n) (0)
=
n ≥0
Hence (4) follows from (7).
zn n!
.
(7)
1 . Then Example 1. Let G1 (z ) = G2 (z ) = 1− z
dn dz n Since
1 1−z
x +1 1−z 1
= z =0
n
(x + k) := (x + 1)n .
k=1
, 1−z z = nk nk!! , it follows from (4) that n n n!
(x + 1)n =
k!
k
k=0
xk .
These numbers connecting the rising and falling factorials are the Lah numbers [1,10]. By Lemma 2.1, we have the following theorem. Theorem 2.2. The e-Riordan matrix representations for Br ,s (n, k) and Ar ,s (n, k) are respectively given as follows: (a) [Br ,s (n, k)] = (φs′ (z ))r , φs (z ) − 1 ;
(b) [Ar ,s (n, k)] = (ψs′ (z ))r , ψs (z ) − 1 .
Applying the defining formula of e-Riordan matrices, we obtain the egfs for Br ,s (n, k) and Ar ,s (n, k): ∞
Br ,s (n, k)
n =k
zn n!
=
1
1+z+
k!
z2 2!
r
z s−1
+ ··· +
z+
(s − 1)!
z2 2!
+ ··· +
zs s!
k
,
and ∞
Ar ,s (n, k)
n =k
zn n!
=
1 k!
1 + z + z 2 + · · · + z s−1
r
z+
z2 2
+ ··· +
zs s
k
.
Let r ≥ 1. It can also be readily shown that Br ,s (n, k) = Sr (n, k) if n ≤ k + s − 1, and, from (3), Br ,s (n, k) =
s−1 n
i
i=0
Br −1,s (n − i, k) if n ≥ k + s.
Similarly, for r ≥ 1, we have
s r ( n, k ) s−1 n! Ar ,s (n, k) = Ar −1,s (n − i, k) (n − i)!
if n ≤ k + s − 1, if n ≥ k + s.
i=0
Now, let us turn to the combinatorial interpretations of these two (r , s)-Bessel numbers. The following lemma is addressed in the literature, for example in Aigner [1, p. 112] and Stanley [14, p. 2]. Lemma 2.3. Let gk , h : N0 → C be counting functions defined by h(|X |) =
g1 (|T1 |)g2 (|T2 |) · · · gr (|Tr |),
(T1 ,...,Tr )
where (T1 , . . . , Tr ) runs through all r-tuples with ∪rk=1 Tk = X and Ti ∩ Tj = ∅ (i ̸= j). Then H (z ) = G1 (z )G2 (z ) · · · Gr (z ), where Gk (z ) =
n
n≥0
gk (n) zn! and H (z ) =
n
n≥0
h(n) zn! .
G.-S. Cheon et al. / Discrete Mathematics 313 (2013) 2127–2138
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Theorem 2.4. Let g , f , hr : N0 → C be counting functions with g (0) ̸= 0 and f (0) = 0 defined by
hr (n, k) =
g (|B1 | − 1) · · · g (|Br | − 1)f (|Br +1 |) · · · f (|Br +k |),
(8)
(B1 ,...,Br +k )
where (B1 , . . . , Br +k ) runs through all (r + k)-tuples of (r + k)-partitions of X = [n + r ] such that i ∈ Bi for i = 1, . . . , r. Then the array [hr (n, k)] may be expressed as an e-Riordan matrix as follows:
Gr (z ), F (z ) ,
where G(z ) =
n ≥0
(9) g (n)z n /n! and F (z ) =
n≥1
f (n)z n /n!.
Proof. Since the blocks Br +1 , . . . , Br +k are nonempty, they can be permuted in k! ways. Hence, the result follows from Lemma 2.3. Theorem 2.5. Let r , s ≥ 1. We have combinatorial interpretations for the (r , s)-Bessel numbers of the second and first types respectively. (a) The number Br ,s (n, k) equals the number of (k + r )-partitions of an (n + r )-set X into blocks of size at most s such that r specified elements are in distinct blocks. (b) The number Ar ,s (n, k) equals the number of (k + r )-cyclic partitions of an (n + r )-set X into cycles of size at most s such that r specified elements are in distinct cycles. Proof. Let X = [n + r ], and let the first r elements of X be specified. If we set g (j) = f (j + 1) = 1 for j with 0 ≤ j ≤ s − 1, and g (j) = f (j + 1) = 0 for j with j ≥ s, then it follows from (8) that hr (n, k) counts (k + r )-partitions of X into blocks of size at k
k
z z ′ most s such that the first r elements are in distinct blocks. Since G(z ) = k=1 k! = φs (z ) and F (z ) = k=1 k! = φs (z ) − 1, we have hr (n, k) = Br ,s (n, k) from Theorem 2.2 and (9), which proves (a). The proof of (b) follows similarly, with g (j) = f (j + 1) = j! for j ∈ {0, 1, . . . , s − 1} and g (j) = f (j + 1) = 0 for j ≥ s.
s−1
s
Example 2. Let us consider the set X = {1, 2, 3, 4, 5}, with r = 2 and s = 3. Then B2,3 (3, 0) = 6 counts the number of partitions of X having two blocks with size at most 3, such that 1 and 2 are in distinct blocks as follows:
{{1, 3}, {2, 4, 5}}, {{1, 4}, {2, 3, 5}}, {{1, 5}, {2, 3, 4}}, {{2, 3}, {1, 4, 5}}, {{2, 4}, {1, 3, 5}}, {{2, 5}, {1, 3, 4}}. A2,3 (3, 0) = 12 equals the number of cyclic partitions of X having two cycles of size at most 3, such that 1 and 2 are in distinct cycles as follows:
(1 3)(2 4 5), (1 3)(2 5 4), (1 4)(2 3 5), (1 4)(2 5 3), (1 5)(2 3 4), (1 5)(2 4 3), (2 3)(1 4 5), (2 3)(1 5 4), (2 4)(1 3 5), (2 4)(1 5 3), (2 5)(1 3 4), (2 5)(1 4 3). 3. Combinatorial setting for the inverse relation There are several kinds of generalization for the Bessel numbers of both kinds. But combinatorial interpretations for those of the first kind are less well known than those of the second kind. In this section, we will consider a combinatorial setting for the inverse relations of the (r , s)-Bessel numbers Br ,s (n, k) or Ar ,s (n, k). Since 2! = 2, Br ,2 (n, k) = Ar ,2 (n, k). In general, in order to find the inverse of the matrix [Br ,s (n, k)] (or [Ar ,s (n, k)]), it is necessary to solve an algebraic equation of degree s, so the problem becomes difficult for s > 2. In particular, when s = 2, the numbers Br ,2 (n, k) (or Ar ,2 (n, k)) will be called the r-Bessel numbers of the second kind, which we will denote as Br (n, k). It then follows from Theorem 2.2 that
z2 . [Br ,2 (n, k)] = [Br (n, k)] = (1 + z )r , z + 2
(10)
Thus we obtain an explicit expression for Br (n, k): Br (n, k) =
n −k n! r k 1 z k [z n ](1 + z )r z k 1 + = . n − k! 2 k! j=0 j n − k − j 2 k−j
n!
(11)
The r-Bessel numbers of the first kind, denoted br (n, k), are the elements of the inverse matrix of (10). Using formula (2) √ for the inverse Riordan matrix ⟨g (z ), f (z )⟩−1 , we find f¯ (z ) = −1 + 1 + 2z, which is the solution of y = z + z 2 /2 in z such that f¯ (0) = 0 (here the case s > 2 fails), and consequently, by substituting in g (z ), we obtain
√ r [br (n, k)] = (1 + 2z )− 2 , −1 + 1 + 2z .
(12)
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The br (n, k) are not all positive. In Theorem 3.3, we will give the combinatorial interpretation of |br (n, k)| = (−1)n−k br (n, k) := br (n, k), the numbers of the first kind. The br (n, k) are obtained from the bivariate generating unsigned r-Bessel zn k function Φ (z , u) = n,k≥0 br (n, k) n! u for (12) if we replace z , u by −z , −u, as follows:
√ r [br (n, k)] = (1 − 2z )− 2 , 1 − 1 − 2z .
(13)
As pointed out by a referee, we can now apply Theorem 2.2 of [9]to find the inverse element of the Riordan matrix (10). Using br (n, k) = (−1)n−k br (n, k), we obtain an explicit expression for br (n, k): f ′ (z ) (−1)n−k n! n−k [z ] k! g (z )(f (z )/z )n+1 n −k z −n−1 (−1) n! n−k = [z ](1 + z )1−r 1 + k! 2 n −k (−1)n−k n! 1−r −n − 1 1 = n−k−j k! 2 j n − k − j j =0
b r ( n, k ) =
=
n −k n!
r +j−2
k! j=0
2n − k − j
1 2n−k−j
n−k−j
j
.
(14)
If r = 0, then the r-Bessel numbers of both kinds are reduced to the ordinary Bessel numbers of the second kind, B0 (n, k) = B(n, k), and of the first kind, b0 (n, k) = b(n, k). By setting r = 0 in (11) and (14), we obtain respectively the formula for B(n, k) in (1) and (2n − k − 1)! if 1 ≤ k ≤ n, b(n, k) = 2n−k (n − k)!(k − 1)! (15) 0 if 1 ≤ n < k. Theorem 3.1. The numbers br (n, k) may be expressed in terms of the Bessel numbers b(n, k) of the first kind as follows: b r ( n, k ) =
k n
j
j =0
[r |2]n−j b(j, k),
where [r |2]k = r (r + 2 · 1)(r + 2 · 2) · · · (r + 2(k − 1)) for k ≥ 1 and [r |2]0 = 1. x
Proof. Since (1 − 2z )− 2 =
n
[x|2]n zn! and b0 (n, k) = b(n, k), we have √ √ r r [br (n, k)] = (1 − 2z )− 2 , 1 − 1 − 2z = (1 − 2z )− 2 , z 1, 1 − 1 − 2z n = [r |2]n−k [b(n, k)] .
n ≥0
k
Matrix multiplication finishes this computation.
Theorem 3.2. The numbers br (n, k) satisfy the recurrence relation, as follows: br (n + 1, k) = br (n, k − 1) + (r + 2n − k)br (n, k). Proof. We first note that
(1 − 2z )
br (n + 1, k)
n≥k−1
zn n!
=
{br (n + 1, k) − 2nbr (n, k)}
n≥k−1
zn n!
.
(16)
Using the egf for the br (n, k), we also obtain
(1 − 2z )
n≥k−1
br (n + 1, k)
zn n!
= (1 − 2z )
d dz n≥k
br (n, k) r
= (r − k)(1 − 2z )− 2 =
n≥k−1
(1 −
zn n!
√
1 − 2z )k
k!
r
+ (1 − 2z )− 2
{(r − k)br (n, k) + br (n, k − 1)}
z
n
n!
(1 −
√
1 − 2z )k−1
(k − 1)!
.
By equating the coefficients of z n /n! in the right-hand sides of (16) and (17), the result follows.
(17)
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Now, let us turn to the combinatorial setting of the r-Bessel numbers of the first kind. For this purpose, we introduce the concept of G-permutations on the multiset given by
[[n ]]k := {0, 0, . . . , 0, 1, 1, 2, 2, . . . , n, n}, where k is the number of zeros. We then define a G-permutation to be a permutation of [[n ]]k such that, for each i ≤ n, the elements occurring between the two occurrences of i are larger than i or 0. In particular, a G-permutation of [[n ]]0 = {1, 1, 2, 2, . . . , n, n} is called a Stirling permutation, which has been defined by Gessel and Stanley [8]. A G-permutation of a subset {i1 , i1 , i2 , i2 , . . . , iℓ , iℓ } of [[n ]]0 is called a subpermutation. A G-permutation of [[n ]]k is said to be proper if the k zeros divide the G-permutation into subpermutations. If k = 0, then the concept of a G-permutation and a proper G-permutation on [[n ]]k is identical. For instance, 223 553 is a subpermutation and 22 355 301 447 710 066 is a proper G-permutation of [[7 ]]3 , while 12 201 is not proper. Theorem 3.3. Let r ≥ 1. Then the number br (n, k) equals the number of G-permutations of [[n − k ]]k+r −1 such that the number of zeros between two consecutive occurrences of i ≥ 1 is at most k. In particular, br (n, 0) equals the number of proper Gpermutations of [[n ]]r −1 . Proof. Let us consider G-permutations of [[n − k ]]k+r −1 for a fixed r ≥ 1. We proceed by induction on n ≥ 1 for k = 0, . . . , n. Let n = 1. If k = 0, then the number of G-permutations of [[1 ]]r −1 with no 0s between two 1s is r, as given in 110 . . . 0, 0110 . . . 0, . . . , 0 . . . 011. If k = 1, then 0 . . . 0 is the unique G-permutation of [[0 ]]r . Since br (1, 0) = r and br (1, 1) = 1, the result holds for n = 1. Let n ≥ 2. We observe that G-permutations of [[n + 1 − k ]]k+r −1 such that the number of zeros between the two occurrences of any positive integer is at most k may be obtained in one of the following ways: (i) by including 0 between the two occurrences of the number n + 1 − k in a G-permutation of [[n + 1 − k ]]k+r −2 such that the number of zeros between two occurrences of any positive integer is at most k − 1; (ii) by including the number n − k + 1 twice at any position of a G-permutation of [[n − k ]]k+r −1 such that the number of zeros between some two occurrences of any positive integer is at most k. For case (ii), there are r + 2n − k possible choices, since the length of the G-permutation is (k + r − 1) + 2(n − k) = r + 2n − k − 1. Thus, by Theorem 3.2, we have br (n, k − 1) + (r + 2n − k)br (n, k) = br (n + 1, k). By induction on n, br (n + 1, k) is the number of G-permutations of [[n + 1 − k ]]k+r −1 such that the number of zeros between the two occurrences of any positive integer is at most k. It is obvious that br (n, 0) is the number of proper G-permutations of [[n ]]r −1 . Hence the proof is complete. From Theorem 3.1, the number of proper G-permutations of [[n ]]k is equal to the (k + 1)-Bessel numbers bk+1 (n, 0) = [k + 1|2]n = (2n + k − 1)!!/(k − 1)!!. In particular, b1 (n, 0) is exactly the number of Stirling permutations of {1, 1, 2, 2, . . . , n, n}. Example 3. (i) b2 (3, 2) = 9 counts G-permutations of [[1 ]]3 = {0, 0, 0, 1, 1} such that the number of zeros between 1 and 1 is at most 2: 11000, 01100, 00110, 00011, 10100, 01010, 00101, 10010, 01001. (ii) b2 (2, 0) = 8 counts proper G-permutations of [[2 ]]1 = {0, 1, 1, 2, 2}: 01122, 11022, 11220, 02211, 22011, 22110, 01221, 12210. Corollary 3.4. The number of G-permutations of [[n ]]k equals the Bessel number of the first kind given by b(n + k + 1, k + 1) =
(2n + k)! . 2n n!k!
(18)
Proof. By Theorem 3.3, the number of G-permutations of [[n − k ]]k is equal to b0 (n + 1, k + 1) = b(n + 1, k + 1). By substituting n := n + k into (15) we obtain (18).
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4. Applications of the r-Bessel numbers of the second kind As described in [10, p. 85], a telephone exchange with n subscribers can connect subscribers in pairs only, i.e., conference circuits are not possible. We first observe that the ordinary Bessel numbers B(n, k) of the second kind may be interpreted as the number of ways in which we can select n − k pairs out of n subscribers. In this sense, these numbers are called the telephone exchange numbers and are denoted as T (n, k). These numbers are generalized in Section 5 to the r-telephone exchange numbers. If we let X be the set of n + r subscribers and set s = 2 in Theorem 2.5 we get the following. Theorem 4.1. The number Br (n, k) equals the number of ways in which one can select n − k pairs out of n + r subscribers such that r specified subscribers cannot connect with each other. A general telephone exchange requires a certain amount of internal memory depending on the number of subscribers. n Since k=0 B(n, k) is equal to the total number of possibilities in a telephone exchange with n subscribers, the exchange must have a stable internal state corresponding to each of these possibilities and must have memory capacity given by
Mn := log2
n
B(n, k)
(see [11]).
k=0
If r specified subscribers out of n subscribers cannot connect with each other, it follows from Theorem 4.1 that the exchange must have memory capacity given by
(r )
Mn
:= log2
n −r
Br (n − r , k) .
k=0
The r-Bessel numbers of the second kind can also be viewed as matchings of a complete semi-bipartite graph G = (V1 ∪ V2 , E1 ∪ E2 ) with bipartition V1 , V2 of the vertex set, where E1 is the edge set between V1 and V2 , and E2 the edge set between all the vertices in V2 . A matching of G is the set of pairwise non-adjacent edges, i.e., no two edges share a common vertex. The following theorem is an immediate consequence of Theorem 2.5 by setting s = 2. Theorem 4.2. Let G be a complete semi-bipartite graph with bipartition V1 = {1, 2, . . . , r } and V2 = {r + 1, r + 2, . . . , n + r }. Then the number Br (n, k) equals the number of matchings of G with n − k edges. A perfect matching is a matching which matches all vertices of the graph, i.e., every vertex of the graph is incident to exactly one edge of the matching. Corollary 4.3. Let G be the same graph as in Theorem 4.2. If n + r = 2m, then Br (n, n − m) is equal to the number of perfect matchings of G. Example 4. Let G be a complete bipartite graph with bipartition V1 = {1, 2} and V2 = {3, 4, 5, 6}. Then B2 (4, 1) = 12 is the number of perfect matchings of G, as shown in Fig. 1. 5. The r-telephone exchange numbers Let us define a conference call as one involving more than two people. As discussed in Section 5, the ordinary Bessel numbers B(n, k) of the second kind may be interpreted as the number of ways in which one can select n − k pairs out of n subscribers in a telephone exchange, where conference calls are not allowed. In this section, we are interested in a telephone exchange problem in which conference calls are possible; this concept leads to another generalization of the Bessel numbers B(n, k). For this approach, we begin by defining the r-telephone exchange numbers Tr (n, k) as
T r ( n, k )
n ≥k
zn n!
= erz
(z + z 2 /2)k . k!
(19)
This then immediately gives the e-Riordan matrix expression for Tr (n, k), as follows:
[Tr (n, k)] = erz , z + z 2 /2 ,
(20)
which gives Tr (n, k) =
n! k!
[z
n−k
]e
rz
1+
z k 2
=
n −k n!
r n−k−j
k! j=0 (n − k − j)!
k
1
j
2j
.
We note that 0-telephone exchange numbers are precisely the Bessel numbers B(n, k) of the second kind. More generally, in Theorem 5.2, we will give the combinatorial interpretation of Tr (n, k).
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Fig. 1. Perfect matchings of the semi-bipartite graph for r = 2 and n = 4.
Theorem 5.1. The numbers Tr (n, k) satisfy the recurrence relation Tr (n + 1, k) = rTr (n, k) + Tr (n, k − 1) + nTr (n − 1, k − 1). Proof. It follows from (19) that
Tr (n + 1, k)
n≥k−1
zn n!
=
d dz
n≥k
Tr (n, k)
zn n!
=
d dz
(z + z 2 /2)k e k! rz
(z + z 2 /2)k (z + z 2 /2)k−1 + (1 + z )erz k! (k − 1)! zn = {rTr (n, k) + Tr (n, k − 1) + nTr (n − 1, k − 1)} . n! n≥k−1 = rerz
n
By identifying the coefficients of zn! on both sides of the equation, the result follows.
Theorem 5.2. The number Tr (n, k) equals the number of ways in which one can select n − k pairs out of n + r subscribers such that r specified subscribers cannot connect with each other but any conference call involves one of these r subscribers. Proof. The r special vertices instead of being people could be the devices that enable subscribers to make conference calls. Let X = [n + r ], and let the first r elements of X be specified. If we set g (j) = 1 for all j ≥ 0, f (1) = f (2) = 1 and f (k) = 0 for all k ≥ 3, then it follows from (8) that hr (n, k) counts (k + r )-partitions of [n + r ] such that the first r elements are in distinct 2
blocks and these blocks are of unrestricted size but the other blocks are of size at most 2. Since G(z ) = 1 + z + z2! + · · · = ez 2 and F (z ) = z + z2! , from (9) and (20), we have hr (n, k) = Tr (n, k). From Theorems 4.1 and 5.2, we obtain the following. Corollary 5.3. Let Tr (n, k) be as in Theorem 5.2. Then the number of ways that at least one out of r specified subscribers will be involved in a conference call is Tr (n, k) − Br (n, k). Example 5. There are T2 (3, 0) = 8 ways in which three pairs can be selected such that two specified subscribers, say 1 and 2, cannot connect with each other but they participate in all of the conference calls, as shown in Fig. 2. Now, let Tr (n) =
n ≥0
Tr (n)
zn n!
n
k=0
Tr (n, k). A direct application of the FTRM to [Tr (n, k)] = erz , z + z 2 /2 with h(z ) = ez gives
= e(r +1)z +z
2 /2
,
(21)
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Fig. 2. The cases for T2 (3, 0).
and the equivalent formula is T r ( n) = n!
⌊ n/2⌋ k=0
(r + 1)n−2k 1 . (n − 2k)! k!2k
Further, by differentiating both sides of (21) with respect to z, one can obtain the recurrence relation for Tr (n) as follows: Tr (n + 1) = (r + 1)Tr (n) + nTr (n − 1). Let Hn (x) be the Hermite polynomials defined by
H n ( x)
n ≥0
zn
= e−z
n!
2 +2zx
.
(22)
From (22), we immediately obtain the following theorem. Theorem 5.4. The numbers Tr (n) may be expressed in terms of the Hermite polynomials as follows: Tr (n) =
n
i
√ 2
Hn
r +1
√
i 2
,
√ −1 .
i=
We end this section by giving a connection between Tr (n) and a Hankel matrix H = [hn,k ] whose general term is hn,k = Tr (n + k). The following theorem was found while working on the previously mentioned connection, and the idea for its combinatorial proof was suggested by a referee.
2 Theorem 5.5. Let L := e(r +1)z +z /2 , z = [
n k
Tr (n − k)] and D = diag(0!, 1!, 2!, . . .). Then LDLT is a Hankel matrix of the
form Tr (0) Tr (1) LDLT = [Tr (n + k)] = Tr (2)
.. .
Tr (1) Tr (2) Tr (3)
.. .
Tr (2) Tr (3) Tr (4)
.. .
··· · · · · · · . .. .
Proof. Let [hn,k ] := LDLT . It is enough to prove that hn,k = Tr (n + k), i.e., Tr (n + k) =
n k j! Tr (n − j)Tr (k − j). j≥0
j
j
(23)
We give a combinatorial proof. Since Tr (n + k) = i=0 Tr (n + k, i), by Theorem 5.2, Tr (n + k) is equal to the number of all the possible ways in which one can select telephone calls out of n + k + r subscribers such that r specified subscribers cannot connect with each other but any conference call involves one of these r subscribers. Assume that the first r subscribers are specified. Let us now consider j as a number of telephone calls between n-subset {r + 1, . . . , r + n} and k-subset {n + r + 1, . . . , n + r + k} of the set [n + k + r ]. Then we obtain (23). Hence the result follows.
n+k
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6. Ordered blocks with size at most 2 Let Lr (n, k) be equal to the number of partitions of [n + r ] into k + r ordered blocks of size at most 2 such that the specified r elements are in distinct blocks. Since such partition has n − k blocks with size 2, it follows from (11) that n −k n!
Lr (n, k) = 2n−k Br (n, k) =
k ! j =0
r
k
j
n−k−j
2j .
Theorem 6.1. The e-Riordan matrix expression for Lr (n, k) is [Lr (n, k)] = (1 + 2z )r , z + z 2 .
(24)
Proof. Since [Br (n, k)] = (1 + z )r , z + z 2 /2 , we obtain
[Lr (n, k)] = 2n−k Br (n, k) = ⟨1, 2z ⟩ [Br (n, k)] ⟨1, z /2⟩
= ⟨1, 2z ⟩ (1 + z )r , z + z 2 /2 ⟨1, z /2⟩ = (1 + 2z )r , z + z 2 . The egf for Lr (n, k) is thus
Lr (n, k)
n ≥k
zn n!
= (1 + 2z )r
(z + z 2 )k . k!
(25)
Differentiating both sides of (25) with respect to z, we obtain the recurrence relation for Lr (n, k): Lr (n + 1, k) = 2(r − n + 2k)Lr (n, k) + Lr (n, k − 1). Theorem 6.2. The numbers Lr (n, k) satisfy the equation n
Lr (n, k)4k xk = 2n (2x + r )n .
(26)
k=0
Proof. Since
n n ≥0
n
n≥0
mn xn zn! = (1 + mz )x , it follows from the FTRM that
Lr (n, k)4 x
k k
k=0
zn
= (1 + 2z )r , z + z 2 (1 + 4z )x = (1 + 2z )r {1 + 4(z + z 2 )}x
n!
= (1 + 2z )2x+r =
2n (2x + r )n
n ≥0 n
By equating the coefficients of zn! on both sides, the result follows.
zn n!
.
It is natural to ask for a combinatorial interpretation of the inverse relation of the numbers Lr (n, k). Let us define the (unsigned) number ℓr (n, k) by n (−1)n−k ℓr (n, k)2k (2x + r )k = 4n xn , k=0
which is obtained by inverting (26). That is, (−1)n−k ℓr (n, k) are the elements of the inverse matrix of (24). By a method similar to that used for deriving (12) and (13), it can be readily shown that
[ℓr (n, k)] =
Since [br (n, k)] =
√
r
1 1 − 4z
√ 1 1−2z
r
√ ,
1 − 4z
1−
,1 −
2
√
.
1 − 2z , we have
[ℓr (n, k)] = ⟨1, 2z ⟩ [br (n, k)]n,k≥0 ⟨1, z /2⟩ = [2n−k br (n, k)].
(27)
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Thus we obtain an explicit expression for ℓr (n, k) from (14):
ℓr (n, k) = 2n−k br (n, k) =
n−k n!
r +j−2
k! j=0
2n − k − j
n−k−j
j
2j .
We now obtain the two combinatorial interpretations of the numbers ℓr (n, k). The first result is acquired from Theorem 3.3. Theorem 6.3. Let [[n ]]′k = {0, . . . , 0, 1, 1′ , 2, 2′ , . . . , n, n′ } with k zeros. Then the number ℓr (n, k) equals the number of Gpermutations of [[n − k ]]′k+r −1 such that the number of zeros between the two occurrences of any positive integer is at most k. The egf for ℓr (n, k) may be written in terms of generating functions for the central binomial numbers Bn = 1 Catalan numbers Cn = n+ 1
ℓr (n, k)
n ≥k
zn n!
2n n
= (B(z ))r
2n n
and the
as follows:
(zC (z ))k , k!
where
√
1 B(z ) = √ = Bn z n 1 − 4z n ≥0
and
C (z ) =
1−
1 − 4z 2z
=
Cn z n .
n ≥0
It is well known that the nth Catalan number Cn is equal to the number of ordered trees with n + 1 vertices. Hence the number of labeled ordered trees with n + 1 vertices is (n + 1)!Cn . Theorem 6.4. The number ℓr (n, k) equals the number of labeled ordered (k + r )-forests with n + r vertices such that r specified vertices are in distinct trees. Proof. Let X = [n + r ] be a vertex set, and let the first r elements of X be specified. If we set g (n) = f (n + 1) = (n + 1)!Cn for all n ≥ 0, then it follows from (8) that hr (n, k) counts the number of labeled ordered (k + r )- forests on the vertex 2
set [n + r ] such that the first r vertices are in distinct trees. Since G(z ) = C0 + 2!C1 z + 3!C2 z2! + · · · = (zC (z ))′ and 2
3
F (z ) = C0 z + 2!C1 z2! + 3!C2 z3! + · · · = zC (z ), it follows from (9) and (27) that hr (n, k) = ℓr (n, k).
Acknowledgments The authors would like to thank the editor and referees for their helpful assistance and guidance, which led to many improvements of the paper. The authors also wish to thank Professor Lois Simon for her careful reading. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
M. Aigner, A Course in Enumeration, in: Graduate Texts in Mathematics, Springer, Berlin, 2007. P. Barry, Exponential Riordan arrays and permutation enumeration, J. Integer Seq. (ISSN: 1530-7638) 13 (2010). M. Bóna, A Walk through Combinatorics, second ed., World Scientific, Singapore, 2006. M. Bóna, Introduction to Enumerative Combinatorics, McGraw-Hill Higher Education, Boston, 2007. A.Z. Broder, The r-Stirling numbers, Discrete Math. 49 (1984) 241–259. J.Y. Choi, L. Long, S.-H. Ng, J. Smith, Reciprocity for multi-restricted Stirling numbers, J. Combin. Theory Ser. A 113 (2006) 1050–1060. J.Y. Choi, J.D.H. Smith, On the combinatorics of multi-restricted numbers, Ars Combin. 75 (2005) 45–63. I. Gessel, R.P. Stanley, Stirling polynomials, J. Combin. Theory Ser. A 24 (1978) 24–33. A. Luzón, D. Merlini, M.A. Morón, R. Sprugnoli, Identities induced by Riordan arrays, Linear Algebra Appl. 436 (2012) 631–647. J. Riordan, An Introduction to Combinatorial Analysis, Wiley, New York, 1958. C.E. Shannon, Memory requirements in a telephone exchange, Bell Syst. Tech. J. 29 (1950) 343–349. L.W. Shapiro, S. Getu, W.-J. Woan, L. Woodson, The Riordan group, Discrete Appl. Math. 34 (1991) 229–239. R. Sprugnoli, Riordan arrays and combinatorial sums, Discrete Math. 132 (1994) 267–290. R.P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, Cambridge, 1999. H.S. Wilf, Generating Functionology, second ed., Academic Press, New York, 1994. S.L. Yang, Z.K. Qiao, The Bessel numbers and Bessel matrices, J. Math. Res. Exposition 31 (2011) 627–636.
Contents lists available at SciVerse ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Generalized Bessel numbers and some combinatorial settings✩ Gi-Sang Cheon a,∗ , Ji-Hwan Jung a , Louis W. Shapiro b a
Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea
b
Department of Mathematics, Howard University, Washington, DC 20059, USA
article
abstract
info
Article history: Received 17 October 2012 Received in revised form 1 May 2013 Accepted 2 May 2013 Available online 24 May 2013
Stirling numbers and Bessel numbers have a long history, and both have been generalized in a variety of directions. Here, we present a second level generalization that has both as special cases. This generalization often preserves the inverse relation between the first and second kind, and has simple combinatorial interpretations. We also frame the discussion in terms of the exponential Riordan group. Then the inverse relation is just the group inverse, and factoring inside the group leads to many results connecting the various Stirling and Bessel numbers. © 2013 Elsevier B.V. All rights reserved.
Keywords: Stirling numbers Bessel numbers Partitions G-Permutations Telephone exchange Semi-bipartite Riordan matrices
1. Introduction The Bessel number of the second kind, B(n, k), is the number of ways in which an n-set X can be partitioned into k blocks of size one or two. These Bessel numbers have many properties similar to those of the Stirling numbers of the second kind. An explicit formula for B(n, k) is well known as B(n, k) =
n! 2n−k
0
(n − k)!(2k − n)!
n
≤ k ≤ n, 2 otherwise. if
(1)
If j is the number of singleton blocks in these set partitions, then the Bessel numbers can be displayed in matrix form by n +j means of singleton blocks, the (n, j) entries being the Bessel numbers given by B(n, 2 ) if n + j is even and 0 if n + j is odd: 1 0 1 0 3 0 15
0 1 0 3 0 15 0
0 0 1 0 6 0 45
0 0 0 1 0 10 0
0 0 0 0 1 0 15
0 0 0 0 0 1 0
0 0 0 0 = ez 2 /2 , z . 0 0 1
··· ✩ This work was supported by a National Research Foundation of Korea Grant funded by the Korean Government (NRF-2012-007648).
∗
Corresponding author. E-mail addresses: [email protected] (G.-S. Cheon), [email protected] (J.-H. Jung), [email protected] (L.W. Shapiro).
0012-365X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.disc.2013.05.001
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The ⟨, ⟩ notation is the exponential Riordan group terminology,1 which will be explained briefly at the end of this section. For an alternative approach, see [16]. If we think of a telephone call as a matching or a block of size 2, then we have the telephone exchange problem [10]. The total number of ways in which n subscribers to a telephone exchange can be connected is given by the row sums of the matrix, and the generating function is 1+z+2
z2
+4
2!
z3 3!
+ 10
z4 4!
+ 26
z5 5!
+ · · · = ez +z
2 /2
.
Here, the n subscribers could be vertices, and an edge connecting two vertices would represent a phone call. We would then be thinking about a labeled graph in which every vertex has degree 0 or 1. Then the connected subgraphs are either a single point or an edge. The connected part has the exponential generating function z + z 2 /2!; see [1,3,4,14,15] for the exponential formula. There are many possible variations. For instance, a cell phone could be turned off, so the blocks of size 1 now can be in two states. Then the generating function is e2z +z /2! . Here is a different variation. We now also allow conference calls involving three parties. The connected part generating function is thus z + z 2 /2! + z 3 /3!, and the resulting generating function is 2
ez +z /2!+z /3! . It is natural to allow blocks (or conference calls) of every size giving the connected part generating function z + z 2 /2! + z 3 /3! + · · · = ez − 1. Now (ez − 1)k /k! is the generating function when we want exactly k blocks and n subscribers (labeled vertices, distinguishable elements). If k is the number of blocks, we obtain the matrix ⟨1, ez − 1⟩, the entries being the Stirling numbers of the second kind, S (n, k). Removing the restriction on block size gives the row sums, z starting 1, 1, 2, 5, 15, 51, . . . , the Bell numbers with generating function ee −1 . The dual is given by arranging each block in a circle (or in a line with the largest element at the head of the line). The connected part is 2
3
z+
z2 2!
+ 2!
z3 3!
+ 3!
z4
+ ··· = z +
4!
z2 2
+
z3 3
+
z4 4
+ · · · = ln
1 1−z
.
1 Now, the matrix is 1, ln 1− , the entries being the unsigned Stirling numbers of the first kind, s(n, k). The row sums have z the generating function
e
1 ln 1− z
=
1 1−z
= 1 + z + z2 + · · · =
n≥0
n!
zn n!
.
This generating function summarizes the fact that any element of the symmetric group on n letters can be written uniquely as the product of disjoint cycles. The duality referred to earlier, after inserting minus signs on every other diagonal, can be expressed as
⟨1, ln(1 + z )⟩ 1, ez − 1 = ⟨1, z ⟩ in Riordan group terminology. Bessel numbers and Stirling numbers have been generalized in a variety of directions. One direction in which Bessel numbers have been generalized is by permitting each block to be of size at most s for s ≥ 3. We refer to these numbers as the s-restricted numbers of the second kind, M2s (n, k) [6,7]. There is a second direction in which Stirling numbers of the second kind have been generalized as r-Stirling numbers, Sr (n, k) [5]. Suppose that there are r distinguished elements in separate blocks or cycles. We may consider dividing up an army into squads, where the r lieutenants are in different squads. The Stirling numbers of the first kind sr (n, k) would generalize by having the r distinguished elements in separate cycles. For matchings, we might have r seeded players that we do not want to compete against each other. In this paper, we present a second level generalization that has both the s-restricted numbers and the r-Stirling numbers as special cases. This generalization often preserves the inverse relation between the first kind and the second kind, and has simple combinatorial interpretations. We also frame the discussion in terms of the exponential Riordan group. Then the inverse relation is just the group inverse, and factoring inside the group leads to many results connecting the various Stirling, telephone exchange, and Bessel numbers. Many combinatorial counting problems can be treated systematically using the Riordan group introduced by Shapiro, Getu, Woan, and Woodson [12]. Here, we use various elements from the exponential version of the Riordan group, and the generating functions are exponential generating functions, which we abbreviate as egfs. An exponential Riordan matrix [2], also denoted as an e-Riordan matrix, is an infinite lower triangular matrix L = [ℓn,k ] whose kth column has egf g (z )f (z )k /k!, where g (0) ̸= 0, f (0) = 0 and f ′ (0) ̸= 0. Equivalently, ℓn,k = n![z n ]g (z )f (z )k /k!, where [z n ] is the operator that extracts the coefficient of z n . The matrix L is denoted by ⟨g (z ), f (z )⟩. It is easy to show that, if we multiply L = ⟨g (z ), f (z )⟩ by a column vector h = (h0 , h1 , . . .)T corresponding to the egf h(z ), then the resulting column
1 As is customary in this setting, we index our rows and columns starting with the index 0.
G.-S. Cheon et al. / Discrete Mathematics 313 (2013) 2127–2138
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vector Lh has egf g (z )h(f (z )). This observation is known as the fundamental theorem of a Riordan matrix (FTRM), and we write this as
⟨g (z ), f (z )⟩ h(z ) = g (z )h(f (z )). The FTRM implies that the set of e-Riordan matrices forms a group under the multiplication defined by
⟨g (z ), f (z )⟩ ⟨h(z ), ℓ(z )⟩ = ⟨g (z )h(f (z )), ℓ(f (z ))⟩ . The identity element is ⟨1, z ⟩, the usual identity matrix, and the inverse of ⟨g , f ⟩ is 1 ⟨g (z ), f (z )⟩−1 = , f (z ) , g (f (z ))
(2)
where f is the compositional inverse, i.e., f (f (z )) = f (f (z )) = z. As shown in [12,13], Riordan matrices provide a powerful method for deriving generating functions, proving combinatorial counting results and identities, and inverting identities. For instance, let ⟨g r (z ), f (z )⟩ = [Gr (n, k)], r ≥ 1, where n![z n ]g (z ) = gn and n![z n ]f (z ) = fn . Using ⟨g r (z ), f (z )⟩ = ⟨g (z ), z ⟩ g r −1 (z ), f (z ) , one can derive the identity Gr (n, k) =
n i
i≥k
gn−i Gr −1 (i, k).
(3)
2. Two types of Bessel number generalized In this section, we consider Bessel numbers generalized in two different types, and we will give their combinatorial interpretations. We begin with φs (z ) and ψs (z ) given by
φs (z ) = 1 + z +
z2
+ ··· +
zs
and ψs (z ) = 1 + z +
z2
+ ··· +
zs
.
2! s! 2 s For an indeterminate x and an integer r ≥ 0, we then define two functions Φr ,s (z ) and Ψr ,s (z ) as
Φr ,s (z ) = (φs′ (z ))r (φs (z ))x and Ψr ,s (z ) = (ψs′ (z ))r (ψs (z ))x . Let us now define the (r , s)-Bessel numbers of the second type, Br ,s (n, k), by the identity n dn Φr ,s (z ) (n) = Br ,s (n, k)xk , Φr ,s (0) := dz n z =0
k=0
where x = x(x − 1) · · · (x − k + 1) and x0 = 1. Similarly, the (r , s)-Bessel numbers of the first type, Ar ,s (n, k), are defined by the identity k
Ψr(,ns) (0)
dn Ψr ,s (z )
:=
dz n
=
z =0
n
Ar ,s (n, k)xk .
k=0
If s → ∞, then Br ,s (n, k) and Ar ,s (n, k) are identical to the r-Stirling numbers of the second kind, Sr (n, k), and of the first kind, sr (n, k), respectively; see [5]. Lemma 2.1. Let G1 (z ), G2 (z ) ∈ C[[z ]] such that G1 (0) ̸= 0, G2 (0) = 1, G′2 (0) ̸= 0, and define G(z ) = G1 (z )(G2 (z ))x for an indeterminate x. Then ⟨G1 (z ), G2 (z ) − 1⟩ = [d(n, k)] if and only if G(n) (0) :=
dn G(z ) dz n
= z =0
n
d(n, k)xk .
(4)
k =0
Proof. If (4) holds, the Taylor series expansion of G(z ) at z = 0 gives G(z ) =
(n)
G
(0)
n ≥0
zn n!
=
n n ≥0
k=0
d(n, k)x
k
zn n!
=
k≥0
d(n, k)
n≥k
zn n!
xk .
(5)
Also, we have G(z ) = G1 (z )(G2 (z ))x = G1 (z )(1 + (G2 (z ) − 1))x = G1 (z )
x k≥0
=
k≥0
G1 (z )
k
(G2 (z ) − 1)k
(G2 (z ) − 1) k x . k! k
Equating the coefficients of xk in (5) and (6), we obtain ⟨G1 (z ), G2 (z ) − 1⟩ = [d(n, k)].
(6)
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Conversely, if ⟨G1 (z ), G2 (z ) − 1⟩ = [d(n, k)] = L, by the FTRM we have L[x0 , x1 , . . .]T = ⟨G1 (z ), G2 (z ) − 1⟩ (1 + z )x = G1 (z )(G2 (z ))x = G(z ) G(n) (0)
=
n ≥0
Hence (4) follows from (7).
zn n!
.
(7)
1 . Then Example 1. Let G1 (z ) = G2 (z ) = 1− z
dn dz n Since
1 1−z
x +1 1−z 1
= z =0
n
(x + k) := (x + 1)n .
k=1
, 1−z z = nk nk!! , it follows from (4) that n n n!
(x + 1)n =
k!
k
k=0
xk .
These numbers connecting the rising and falling factorials are the Lah numbers [1,10]. By Lemma 2.1, we have the following theorem. Theorem 2.2. The e-Riordan matrix representations for Br ,s (n, k) and Ar ,s (n, k) are respectively given as follows: (a) [Br ,s (n, k)] = (φs′ (z ))r , φs (z ) − 1 ;
(b) [Ar ,s (n, k)] = (ψs′ (z ))r , ψs (z ) − 1 .
Applying the defining formula of e-Riordan matrices, we obtain the egfs for Br ,s (n, k) and Ar ,s (n, k): ∞
Br ,s (n, k)
n =k
zn n!
=
1
1+z+
k!
z2 2!
r
z s−1
+ ··· +
z+
(s − 1)!
z2 2!
+ ··· +
zs s!
k
,
and ∞
Ar ,s (n, k)
n =k
zn n!
=
1 k!
1 + z + z 2 + · · · + z s−1
r
z+
z2 2
+ ··· +
zs s
k
.
Let r ≥ 1. It can also be readily shown that Br ,s (n, k) = Sr (n, k) if n ≤ k + s − 1, and, from (3), Br ,s (n, k) =
s−1 n
i
i=0
Br −1,s (n − i, k) if n ≥ k + s.
Similarly, for r ≥ 1, we have
s r ( n, k ) s−1 n! Ar ,s (n, k) = Ar −1,s (n − i, k) (n − i)!
if n ≤ k + s − 1, if n ≥ k + s.
i=0
Now, let us turn to the combinatorial interpretations of these two (r , s)-Bessel numbers. The following lemma is addressed in the literature, for example in Aigner [1, p. 112] and Stanley [14, p. 2]. Lemma 2.3. Let gk , h : N0 → C be counting functions defined by h(|X |) =
g1 (|T1 |)g2 (|T2 |) · · · gr (|Tr |),
(T1 ,...,Tr )
where (T1 , . . . , Tr ) runs through all r-tuples with ∪rk=1 Tk = X and Ti ∩ Tj = ∅ (i ̸= j). Then H (z ) = G1 (z )G2 (z ) · · · Gr (z ), where Gk (z ) =
n
n≥0
gk (n) zn! and H (z ) =
n
n≥0
h(n) zn! .
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Theorem 2.4. Let g , f , hr : N0 → C be counting functions with g (0) ̸= 0 and f (0) = 0 defined by
hr (n, k) =
g (|B1 | − 1) · · · g (|Br | − 1)f (|Br +1 |) · · · f (|Br +k |),
(8)
(B1 ,...,Br +k )
where (B1 , . . . , Br +k ) runs through all (r + k)-tuples of (r + k)-partitions of X = [n + r ] such that i ∈ Bi for i = 1, . . . , r. Then the array [hr (n, k)] may be expressed as an e-Riordan matrix as follows:
Gr (z ), F (z ) ,
where G(z ) =
n ≥0
(9) g (n)z n /n! and F (z ) =
n≥1
f (n)z n /n!.
Proof. Since the blocks Br +1 , . . . , Br +k are nonempty, they can be permuted in k! ways. Hence, the result follows from Lemma 2.3. Theorem 2.5. Let r , s ≥ 1. We have combinatorial interpretations for the (r , s)-Bessel numbers of the second and first types respectively. (a) The number Br ,s (n, k) equals the number of (k + r )-partitions of an (n + r )-set X into blocks of size at most s such that r specified elements are in distinct blocks. (b) The number Ar ,s (n, k) equals the number of (k + r )-cyclic partitions of an (n + r )-set X into cycles of size at most s such that r specified elements are in distinct cycles. Proof. Let X = [n + r ], and let the first r elements of X be specified. If we set g (j) = f (j + 1) = 1 for j with 0 ≤ j ≤ s − 1, and g (j) = f (j + 1) = 0 for j with j ≥ s, then it follows from (8) that hr (n, k) counts (k + r )-partitions of X into blocks of size at k
k
z z ′ most s such that the first r elements are in distinct blocks. Since G(z ) = k=1 k! = φs (z ) and F (z ) = k=1 k! = φs (z ) − 1, we have hr (n, k) = Br ,s (n, k) from Theorem 2.2 and (9), which proves (a). The proof of (b) follows similarly, with g (j) = f (j + 1) = j! for j ∈ {0, 1, . . . , s − 1} and g (j) = f (j + 1) = 0 for j ≥ s.
s−1
s
Example 2. Let us consider the set X = {1, 2, 3, 4, 5}, with r = 2 and s = 3. Then B2,3 (3, 0) = 6 counts the number of partitions of X having two blocks with size at most 3, such that 1 and 2 are in distinct blocks as follows:
{{1, 3}, {2, 4, 5}}, {{1, 4}, {2, 3, 5}}, {{1, 5}, {2, 3, 4}}, {{2, 3}, {1, 4, 5}}, {{2, 4}, {1, 3, 5}}, {{2, 5}, {1, 3, 4}}. A2,3 (3, 0) = 12 equals the number of cyclic partitions of X having two cycles of size at most 3, such that 1 and 2 are in distinct cycles as follows:
(1 3)(2 4 5), (1 3)(2 5 4), (1 4)(2 3 5), (1 4)(2 5 3), (1 5)(2 3 4), (1 5)(2 4 3), (2 3)(1 4 5), (2 3)(1 5 4), (2 4)(1 3 5), (2 4)(1 5 3), (2 5)(1 3 4), (2 5)(1 4 3). 3. Combinatorial setting for the inverse relation There are several kinds of generalization for the Bessel numbers of both kinds. But combinatorial interpretations for those of the first kind are less well known than those of the second kind. In this section, we will consider a combinatorial setting for the inverse relations of the (r , s)-Bessel numbers Br ,s (n, k) or Ar ,s (n, k). Since 2! = 2, Br ,2 (n, k) = Ar ,2 (n, k). In general, in order to find the inverse of the matrix [Br ,s (n, k)] (or [Ar ,s (n, k)]), it is necessary to solve an algebraic equation of degree s, so the problem becomes difficult for s > 2. In particular, when s = 2, the numbers Br ,2 (n, k) (or Ar ,2 (n, k)) will be called the r-Bessel numbers of the second kind, which we will denote as Br (n, k). It then follows from Theorem 2.2 that
z2 . [Br ,2 (n, k)] = [Br (n, k)] = (1 + z )r , z + 2
(10)
Thus we obtain an explicit expression for Br (n, k): Br (n, k) =
n −k n! r k 1 z k [z n ](1 + z )r z k 1 + = . n − k! 2 k! j=0 j n − k − j 2 k−j
n!
(11)
The r-Bessel numbers of the first kind, denoted br (n, k), are the elements of the inverse matrix of (10). Using formula (2) √ for the inverse Riordan matrix ⟨g (z ), f (z )⟩−1 , we find f¯ (z ) = −1 + 1 + 2z, which is the solution of y = z + z 2 /2 in z such that f¯ (0) = 0 (here the case s > 2 fails), and consequently, by substituting in g (z ), we obtain
√ r [br (n, k)] = (1 + 2z )− 2 , −1 + 1 + 2z .
(12)
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The br (n, k) are not all positive. In Theorem 3.3, we will give the combinatorial interpretation of |br (n, k)| = (−1)n−k br (n, k) := br (n, k), the numbers of the first kind. The br (n, k) are obtained from the bivariate generating unsigned r-Bessel zn k function Φ (z , u) = n,k≥0 br (n, k) n! u for (12) if we replace z , u by −z , −u, as follows:
√ r [br (n, k)] = (1 − 2z )− 2 , 1 − 1 − 2z .
(13)
As pointed out by a referee, we can now apply Theorem 2.2 of [9]to find the inverse element of the Riordan matrix (10). Using br (n, k) = (−1)n−k br (n, k), we obtain an explicit expression for br (n, k): f ′ (z ) (−1)n−k n! n−k [z ] k! g (z )(f (z )/z )n+1 n −k z −n−1 (−1) n! n−k = [z ](1 + z )1−r 1 + k! 2 n −k (−1)n−k n! 1−r −n − 1 1 = n−k−j k! 2 j n − k − j j =0
b r ( n, k ) =
=
n −k n!
r +j−2
k! j=0
2n − k − j
1 2n−k−j
n−k−j
j
.
(14)
If r = 0, then the r-Bessel numbers of both kinds are reduced to the ordinary Bessel numbers of the second kind, B0 (n, k) = B(n, k), and of the first kind, b0 (n, k) = b(n, k). By setting r = 0 in (11) and (14), we obtain respectively the formula for B(n, k) in (1) and (2n − k − 1)! if 1 ≤ k ≤ n, b(n, k) = 2n−k (n − k)!(k − 1)! (15) 0 if 1 ≤ n < k. Theorem 3.1. The numbers br (n, k) may be expressed in terms of the Bessel numbers b(n, k) of the first kind as follows: b r ( n, k ) =
k n
j
j =0
[r |2]n−j b(j, k),
where [r |2]k = r (r + 2 · 1)(r + 2 · 2) · · · (r + 2(k − 1)) for k ≥ 1 and [r |2]0 = 1. x
Proof. Since (1 − 2z )− 2 =
n
[x|2]n zn! and b0 (n, k) = b(n, k), we have √ √ r r [br (n, k)] = (1 − 2z )− 2 , 1 − 1 − 2z = (1 − 2z )− 2 , z 1, 1 − 1 − 2z n = [r |2]n−k [b(n, k)] .
n ≥0
k
Matrix multiplication finishes this computation.
Theorem 3.2. The numbers br (n, k) satisfy the recurrence relation, as follows: br (n + 1, k) = br (n, k − 1) + (r + 2n − k)br (n, k). Proof. We first note that
(1 − 2z )
br (n + 1, k)
n≥k−1
zn n!
=
{br (n + 1, k) − 2nbr (n, k)}
n≥k−1
zn n!
.
(16)
Using the egf for the br (n, k), we also obtain
(1 − 2z )
n≥k−1
br (n + 1, k)
zn n!
= (1 − 2z )
d dz n≥k
br (n, k) r
= (r − k)(1 − 2z )− 2 =
n≥k−1
(1 −
zn n!
√
1 − 2z )k
k!
r
+ (1 − 2z )− 2
{(r − k)br (n, k) + br (n, k − 1)}
z
n
n!
(1 −
√
1 − 2z )k−1
(k − 1)!
.
By equating the coefficients of z n /n! in the right-hand sides of (16) and (17), the result follows.
(17)
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Now, let us turn to the combinatorial setting of the r-Bessel numbers of the first kind. For this purpose, we introduce the concept of G-permutations on the multiset given by
[[n ]]k := {0, 0, . . . , 0, 1, 1, 2, 2, . . . , n, n}, where k is the number of zeros. We then define a G-permutation to be a permutation of [[n ]]k such that, for each i ≤ n, the elements occurring between the two occurrences of i are larger than i or 0. In particular, a G-permutation of [[n ]]0 = {1, 1, 2, 2, . . . , n, n} is called a Stirling permutation, which has been defined by Gessel and Stanley [8]. A G-permutation of a subset {i1 , i1 , i2 , i2 , . . . , iℓ , iℓ } of [[n ]]0 is called a subpermutation. A G-permutation of [[n ]]k is said to be proper if the k zeros divide the G-permutation into subpermutations. If k = 0, then the concept of a G-permutation and a proper G-permutation on [[n ]]k is identical. For instance, 223 553 is a subpermutation and 22 355 301 447 710 066 is a proper G-permutation of [[7 ]]3 , while 12 201 is not proper. Theorem 3.3. Let r ≥ 1. Then the number br (n, k) equals the number of G-permutations of [[n − k ]]k+r −1 such that the number of zeros between two consecutive occurrences of i ≥ 1 is at most k. In particular, br (n, 0) equals the number of proper Gpermutations of [[n ]]r −1 . Proof. Let us consider G-permutations of [[n − k ]]k+r −1 for a fixed r ≥ 1. We proceed by induction on n ≥ 1 for k = 0, . . . , n. Let n = 1. If k = 0, then the number of G-permutations of [[1 ]]r −1 with no 0s between two 1s is r, as given in 110 . . . 0, 0110 . . . 0, . . . , 0 . . . 011. If k = 1, then 0 . . . 0 is the unique G-permutation of [[0 ]]r . Since br (1, 0) = r and br (1, 1) = 1, the result holds for n = 1. Let n ≥ 2. We observe that G-permutations of [[n + 1 − k ]]k+r −1 such that the number of zeros between the two occurrences of any positive integer is at most k may be obtained in one of the following ways: (i) by including 0 between the two occurrences of the number n + 1 − k in a G-permutation of [[n + 1 − k ]]k+r −2 such that the number of zeros between two occurrences of any positive integer is at most k − 1; (ii) by including the number n − k + 1 twice at any position of a G-permutation of [[n − k ]]k+r −1 such that the number of zeros between some two occurrences of any positive integer is at most k. For case (ii), there are r + 2n − k possible choices, since the length of the G-permutation is (k + r − 1) + 2(n − k) = r + 2n − k − 1. Thus, by Theorem 3.2, we have br (n, k − 1) + (r + 2n − k)br (n, k) = br (n + 1, k). By induction on n, br (n + 1, k) is the number of G-permutations of [[n + 1 − k ]]k+r −1 such that the number of zeros between the two occurrences of any positive integer is at most k. It is obvious that br (n, 0) is the number of proper G-permutations of [[n ]]r −1 . Hence the proof is complete. From Theorem 3.1, the number of proper G-permutations of [[n ]]k is equal to the (k + 1)-Bessel numbers bk+1 (n, 0) = [k + 1|2]n = (2n + k − 1)!!/(k − 1)!!. In particular, b1 (n, 0) is exactly the number of Stirling permutations of {1, 1, 2, 2, . . . , n, n}. Example 3. (i) b2 (3, 2) = 9 counts G-permutations of [[1 ]]3 = {0, 0, 0, 1, 1} such that the number of zeros between 1 and 1 is at most 2: 11000, 01100, 00110, 00011, 10100, 01010, 00101, 10010, 01001. (ii) b2 (2, 0) = 8 counts proper G-permutations of [[2 ]]1 = {0, 1, 1, 2, 2}: 01122, 11022, 11220, 02211, 22011, 22110, 01221, 12210. Corollary 3.4. The number of G-permutations of [[n ]]k equals the Bessel number of the first kind given by b(n + k + 1, k + 1) =
(2n + k)! . 2n n!k!
(18)
Proof. By Theorem 3.3, the number of G-permutations of [[n − k ]]k is equal to b0 (n + 1, k + 1) = b(n + 1, k + 1). By substituting n := n + k into (15) we obtain (18).
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4. Applications of the r-Bessel numbers of the second kind As described in [10, p. 85], a telephone exchange with n subscribers can connect subscribers in pairs only, i.e., conference circuits are not possible. We first observe that the ordinary Bessel numbers B(n, k) of the second kind may be interpreted as the number of ways in which we can select n − k pairs out of n subscribers. In this sense, these numbers are called the telephone exchange numbers and are denoted as T (n, k). These numbers are generalized in Section 5 to the r-telephone exchange numbers. If we let X be the set of n + r subscribers and set s = 2 in Theorem 2.5 we get the following. Theorem 4.1. The number Br (n, k) equals the number of ways in which one can select n − k pairs out of n + r subscribers such that r specified subscribers cannot connect with each other. A general telephone exchange requires a certain amount of internal memory depending on the number of subscribers. n Since k=0 B(n, k) is equal to the total number of possibilities in a telephone exchange with n subscribers, the exchange must have a stable internal state corresponding to each of these possibilities and must have memory capacity given by
Mn := log2
n
B(n, k)
(see [11]).
k=0
If r specified subscribers out of n subscribers cannot connect with each other, it follows from Theorem 4.1 that the exchange must have memory capacity given by
(r )
Mn
:= log2
n −r
Br (n − r , k) .
k=0
The r-Bessel numbers of the second kind can also be viewed as matchings of a complete semi-bipartite graph G = (V1 ∪ V2 , E1 ∪ E2 ) with bipartition V1 , V2 of the vertex set, where E1 is the edge set between V1 and V2 , and E2 the edge set between all the vertices in V2 . A matching of G is the set of pairwise non-adjacent edges, i.e., no two edges share a common vertex. The following theorem is an immediate consequence of Theorem 2.5 by setting s = 2. Theorem 4.2. Let G be a complete semi-bipartite graph with bipartition V1 = {1, 2, . . . , r } and V2 = {r + 1, r + 2, . . . , n + r }. Then the number Br (n, k) equals the number of matchings of G with n − k edges. A perfect matching is a matching which matches all vertices of the graph, i.e., every vertex of the graph is incident to exactly one edge of the matching. Corollary 4.3. Let G be the same graph as in Theorem 4.2. If n + r = 2m, then Br (n, n − m) is equal to the number of perfect matchings of G. Example 4. Let G be a complete bipartite graph with bipartition V1 = {1, 2} and V2 = {3, 4, 5, 6}. Then B2 (4, 1) = 12 is the number of perfect matchings of G, as shown in Fig. 1. 5. The r-telephone exchange numbers Let us define a conference call as one involving more than two people. As discussed in Section 5, the ordinary Bessel numbers B(n, k) of the second kind may be interpreted as the number of ways in which one can select n − k pairs out of n subscribers in a telephone exchange, where conference calls are not allowed. In this section, we are interested in a telephone exchange problem in which conference calls are possible; this concept leads to another generalization of the Bessel numbers B(n, k). For this approach, we begin by defining the r-telephone exchange numbers Tr (n, k) as
T r ( n, k )
n ≥k
zn n!
= erz
(z + z 2 /2)k . k!
(19)
This then immediately gives the e-Riordan matrix expression for Tr (n, k), as follows:
[Tr (n, k)] = erz , z + z 2 /2 ,
(20)
which gives Tr (n, k) =
n! k!
[z
n−k
]e
rz
1+
z k 2
=
n −k n!
r n−k−j
k! j=0 (n − k − j)!
k
1
j
2j
.
We note that 0-telephone exchange numbers are precisely the Bessel numbers B(n, k) of the second kind. More generally, in Theorem 5.2, we will give the combinatorial interpretation of Tr (n, k).
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Fig. 1. Perfect matchings of the semi-bipartite graph for r = 2 and n = 4.
Theorem 5.1. The numbers Tr (n, k) satisfy the recurrence relation Tr (n + 1, k) = rTr (n, k) + Tr (n, k − 1) + nTr (n − 1, k − 1). Proof. It follows from (19) that
Tr (n + 1, k)
n≥k−1
zn n!
=
d dz
n≥k
Tr (n, k)
zn n!
=
d dz
(z + z 2 /2)k e k! rz
(z + z 2 /2)k (z + z 2 /2)k−1 + (1 + z )erz k! (k − 1)! zn = {rTr (n, k) + Tr (n, k − 1) + nTr (n − 1, k − 1)} . n! n≥k−1 = rerz
n
By identifying the coefficients of zn! on both sides of the equation, the result follows.
Theorem 5.2. The number Tr (n, k) equals the number of ways in which one can select n − k pairs out of n + r subscribers such that r specified subscribers cannot connect with each other but any conference call involves one of these r subscribers. Proof. The r special vertices instead of being people could be the devices that enable subscribers to make conference calls. Let X = [n + r ], and let the first r elements of X be specified. If we set g (j) = 1 for all j ≥ 0, f (1) = f (2) = 1 and f (k) = 0 for all k ≥ 3, then it follows from (8) that hr (n, k) counts (k + r )-partitions of [n + r ] such that the first r elements are in distinct 2
blocks and these blocks are of unrestricted size but the other blocks are of size at most 2. Since G(z ) = 1 + z + z2! + · · · = ez 2 and F (z ) = z + z2! , from (9) and (20), we have hr (n, k) = Tr (n, k). From Theorems 4.1 and 5.2, we obtain the following. Corollary 5.3. Let Tr (n, k) be as in Theorem 5.2. Then the number of ways that at least one out of r specified subscribers will be involved in a conference call is Tr (n, k) − Br (n, k). Example 5. There are T2 (3, 0) = 8 ways in which three pairs can be selected such that two specified subscribers, say 1 and 2, cannot connect with each other but they participate in all of the conference calls, as shown in Fig. 2. Now, let Tr (n) =
n ≥0
Tr (n)
zn n!
n
k=0
Tr (n, k). A direct application of the FTRM to [Tr (n, k)] = erz , z + z 2 /2 with h(z ) = ez gives
= e(r +1)z +z
2 /2
,
(21)
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Fig. 2. The cases for T2 (3, 0).
and the equivalent formula is T r ( n) = n!
⌊ n/2⌋ k=0
(r + 1)n−2k 1 . (n − 2k)! k!2k
Further, by differentiating both sides of (21) with respect to z, one can obtain the recurrence relation for Tr (n) as follows: Tr (n + 1) = (r + 1)Tr (n) + nTr (n − 1). Let Hn (x) be the Hermite polynomials defined by
H n ( x)
n ≥0
zn
= e−z
n!
2 +2zx
.
(22)
From (22), we immediately obtain the following theorem. Theorem 5.4. The numbers Tr (n) may be expressed in terms of the Hermite polynomials as follows: Tr (n) =
n
i
√ 2
Hn
r +1
√
i 2
,
√ −1 .
i=
We end this section by giving a connection between Tr (n) and a Hankel matrix H = [hn,k ] whose general term is hn,k = Tr (n + k). The following theorem was found while working on the previously mentioned connection, and the idea for its combinatorial proof was suggested by a referee.
2 Theorem 5.5. Let L := e(r +1)z +z /2 , z = [
n k
Tr (n − k)] and D = diag(0!, 1!, 2!, . . .). Then LDLT is a Hankel matrix of the
form Tr (0) Tr (1) LDLT = [Tr (n + k)] = Tr (2)
.. .
Tr (1) Tr (2) Tr (3)
.. .
Tr (2) Tr (3) Tr (4)
.. .
··· · · · · · · . .. .
Proof. Let [hn,k ] := LDLT . It is enough to prove that hn,k = Tr (n + k), i.e., Tr (n + k) =
n k j! Tr (n − j)Tr (k − j). j≥0
j
j
(23)
We give a combinatorial proof. Since Tr (n + k) = i=0 Tr (n + k, i), by Theorem 5.2, Tr (n + k) is equal to the number of all the possible ways in which one can select telephone calls out of n + k + r subscribers such that r specified subscribers cannot connect with each other but any conference call involves one of these r subscribers. Assume that the first r subscribers are specified. Let us now consider j as a number of telephone calls between n-subset {r + 1, . . . , r + n} and k-subset {n + r + 1, . . . , n + r + k} of the set [n + k + r ]. Then we obtain (23). Hence the result follows.
n+k
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6. Ordered blocks with size at most 2 Let Lr (n, k) be equal to the number of partitions of [n + r ] into k + r ordered blocks of size at most 2 such that the specified r elements are in distinct blocks. Since such partition has n − k blocks with size 2, it follows from (11) that n −k n!
Lr (n, k) = 2n−k Br (n, k) =
k ! j =0
r
k
j
n−k−j
2j .
Theorem 6.1. The e-Riordan matrix expression for Lr (n, k) is [Lr (n, k)] = (1 + 2z )r , z + z 2 .
(24)
Proof. Since [Br (n, k)] = (1 + z )r , z + z 2 /2 , we obtain
[Lr (n, k)] = 2n−k Br (n, k) = ⟨1, 2z ⟩ [Br (n, k)] ⟨1, z /2⟩
= ⟨1, 2z ⟩ (1 + z )r , z + z 2 /2 ⟨1, z /2⟩ = (1 + 2z )r , z + z 2 . The egf for Lr (n, k) is thus
Lr (n, k)
n ≥k
zn n!
= (1 + 2z )r
(z + z 2 )k . k!
(25)
Differentiating both sides of (25) with respect to z, we obtain the recurrence relation for Lr (n, k): Lr (n + 1, k) = 2(r − n + 2k)Lr (n, k) + Lr (n, k − 1). Theorem 6.2. The numbers Lr (n, k) satisfy the equation n
Lr (n, k)4k xk = 2n (2x + r )n .
(26)
k=0
Proof. Since
n n ≥0
n
n≥0
mn xn zn! = (1 + mz )x , it follows from the FTRM that
Lr (n, k)4 x
k k
k=0
zn
= (1 + 2z )r , z + z 2 (1 + 4z )x = (1 + 2z )r {1 + 4(z + z 2 )}x
n!
= (1 + 2z )2x+r =
2n (2x + r )n
n ≥0 n
By equating the coefficients of zn! on both sides, the result follows.
zn n!
.
It is natural to ask for a combinatorial interpretation of the inverse relation of the numbers Lr (n, k). Let us define the (unsigned) number ℓr (n, k) by n (−1)n−k ℓr (n, k)2k (2x + r )k = 4n xn , k=0
which is obtained by inverting (26). That is, (−1)n−k ℓr (n, k) are the elements of the inverse matrix of (24). By a method similar to that used for deriving (12) and (13), it can be readily shown that
[ℓr (n, k)] =
Since [br (n, k)] =
√
r
1 1 − 4z
√ 1 1−2z
r
√ ,
1 − 4z
1−
,1 −
2
√
.
1 − 2z , we have
[ℓr (n, k)] = ⟨1, 2z ⟩ [br (n, k)]n,k≥0 ⟨1, z /2⟩ = [2n−k br (n, k)].
(27)
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Thus we obtain an explicit expression for ℓr (n, k) from (14):
ℓr (n, k) = 2n−k br (n, k) =
n−k n!
r +j−2
k! j=0
2n − k − j
n−k−j
j
2j .
We now obtain the two combinatorial interpretations of the numbers ℓr (n, k). The first result is acquired from Theorem 3.3. Theorem 6.3. Let [[n ]]′k = {0, . . . , 0, 1, 1′ , 2, 2′ , . . . , n, n′ } with k zeros. Then the number ℓr (n, k) equals the number of Gpermutations of [[n − k ]]′k+r −1 such that the number of zeros between the two occurrences of any positive integer is at most k. The egf for ℓr (n, k) may be written in terms of generating functions for the central binomial numbers Bn = 1 Catalan numbers Cn = n+ 1
ℓr (n, k)
n ≥k
zn n!
2n n
= (B(z ))r
2n n
and the
as follows:
(zC (z ))k , k!
where
√
1 B(z ) = √ = Bn z n 1 − 4z n ≥0
and
C (z ) =
1−
1 − 4z 2z
=
Cn z n .
n ≥0
It is well known that the nth Catalan number Cn is equal to the number of ordered trees with n + 1 vertices. Hence the number of labeled ordered trees with n + 1 vertices is (n + 1)!Cn . Theorem 6.4. The number ℓr (n, k) equals the number of labeled ordered (k + r )-forests with n + r vertices such that r specified vertices are in distinct trees. Proof. Let X = [n + r ] be a vertex set, and let the first r elements of X be specified. If we set g (n) = f (n + 1) = (n + 1)!Cn for all n ≥ 0, then it follows from (8) that hr (n, k) counts the number of labeled ordered (k + r )- forests on the vertex 2
set [n + r ] such that the first r vertices are in distinct trees. Since G(z ) = C0 + 2!C1 z + 3!C2 z2! + · · · = (zC (z ))′ and 2
3
F (z ) = C0 z + 2!C1 z2! + 3!C2 z3! + · · · = zC (z ), it follows from (9) and (27) that hr (n, k) = ℓr (n, k).
Acknowledgments The authors would like to thank the editor and referees for their helpful assistance and guidance, which led to many improvements of the paper. The authors also wish to thank Professor Lois Simon for her careful reading. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
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