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On partitions, surjections, and Stirling
numbers
Peter Hilton, Jean Pedersen and J¨urgen Stigter
1 Introduction
The connection between P (m, n), the number of partitions of a set containing m elements as a disjoint union of n non-empty subsets; S(m, n), the number of sur- jections of a set of m elements onto a set of n-elements; and St(m, n), the Stirling number of the second kind, given by^1
St(m, n) =
n!
∑^ n r=
(−1)n−r
( n r
) rm, (1.1)
has long been known (see, eg, [B, C, L, T]). Indeed, their mutual relation is given by 1 n!
S(m, n) = P (m, n) = St(m, n), (1.2)
the first equality being very elementary and the second somewhat less immediate.
Our primary object in this paper is to provide an explicit formula for St(m, n), and hence, by (1.2), for P (m, n) and S(m, n), in the case that m > n.
(^1) Other definitions are given in the litterature; for example, St(m, n) is characterized by the equation
xm^ =
∑^ m n=
n!St(m, n)
( x n
) .
Received by the editors November 1993 Communicated by A. Warrinier
Bull. Belg. Math. Soc. 1 (1994),713–
714 P. Hilton, J. Pedersen and J. Stigter
The form our formula takes is the following. We write γk (m) = St(m, m − k), k ≥ 1;
then we represent γk (m) as a linear combination of binomial coefficients
( m r
) ,
k + 1 ≤ r ≤ 2 k. The weight ahk attached to the binomial coefficient
( m k + h + 1
)
is a positive integer independent of m, which we call the Stirling factor of type (h, k). We do not calculate these Stirling factors explicitly, except for low values of k (see Figure 3) and the cases h = 0, 1 , 2 , k − 2 , k − 1 (Theorems 4.1, 4.3, 4.4). However we give a recurrence relation expressing ahk , h ≥ 1, as a linear combination of Stirling factors ah− 1 ,j , h ≤ j ≤ k − 1, the weight attached to the Stirling factor
ah− 1 ,j being the binomial coefficient
( k + h j + h
)
. It thus seems natural to represent
the Stirling factors in a triangle too, to illustrate the interaction with the Pascal Triangle (see Figure 4). The recurrence relation, together with the initial condition a 0 k = 1 for all k ≥ 1, of course determines ahk.
We believe that our formula, given by Theorem 4.1, should lead to useful esti- mates of P (m, n), since the Stirling factors are always positive integers and should be readily estimated. We believe further that the Stirling factors ahk should have interesting number-theoretical properties and interrelations. One such relation is displayed in Lemma 4.5. We hope to return to these aspects in a subsequent paper.
We draw attention to two further features of this paper. In Section 3 we prove a generalized version of a simple, but attractive, combinatorial property of binomial coefficients, often known as the Christmas Stocking Theorem; the reason for this choice of name is made quite clear by Figure 1. The generalization (which plays a key role in our proof of our main result, Theorem 4.1) also admits a geometric representation, which we try to convey in Figure 2. Second, we combine Theorem 4.1 with the classical recurrence relation
P (m, n) = nP (m − 1 , n) + P (m − 1 , n − 1) (1.3)
to obtain certain bilinear relations connecting binomial coefficients and Stirling fac- tors (see, for example, Theorem 5.1).
It is, we think interesting to mention that this paper arose out of the (success- ful!) attempt to reconstruct an elementary proof of Theorem 2.6(b), a result which had been discovered by the third-named author many years ago. Of course, this elementary proof had to proceed without invoking (1.2), since the plan was to use it in a proof that P (m, n) = St(m, n).
2 Classical results
All the results in this section may be found in any standard work on combina- torics (see, eg [B, C, L, T]); we collect them here for the convenience of the reader. Throughout this section m and n are non-negative integers.
716 P. Hilton, J. Pedersen and J. Stigter
Thus we may prove Theorem 2.6 by induction on n, starting with n = 0. To prove (a) we assume that, for a given n ≥ 1, St(m, n − 1) = 0 for all m < n − 1. Thus, expanding (r + 1)m−^1 by the binomial theorem, we conclude that
n∑− 1
r=
(−1)n−^1 −r
( n − 1 r
) (r + 1)m−^1 = 0, 1 ≤ m < n. (2.5)
Set s = r + 1, so that
∑^ n s=
(−1)n−s
( n − 1 s − 1
) sm−^1 = 0, 1 ≤ m < n. (2.6)
Since
( n s
)
n s
( n − 1 s − 1
) , we infer from (2.6) that
∑^ n s=
(−1)n−s
( n s
) sm^ = 0, 1 ≤ m < n.
We may adjoin s = 0, since m ≥ 1, and replace s by r, to obtain
∑^ n r=
(−1)n−r
( n r
) rm^ = 0, 1 ≤ m < n. (2.7)
Since we already know, by (2.4), that St(0, n) = 0, n ≥ 1, it follows that (2.7) establishes the inductive step in the proof of (a).
The proof of (b) is now very similar. Thus we assume that, for a given n ≥ 1, St(n − 1 , n − 1) = 1. In view of (a), this implies that
n∑− 1
r=
(−1)n−^1 −r
( n − 1 r
) (r + 1)n−^1 = (n − 1)!
Set s = r + 1 and exploit again the identity
( n s
)
n s
( n − 1 s − 1
) to infer that
∑^ n s=
(−1)n−s
( n s
) sn^ = n!
We may adjoin s = 0, since n ≥ 1, and thereby complete the inductive step in the proof of (b).
One immediate consequence of Theorem 2.4 which we may include here is
Theorem 2.7. St(n + 1, n) =
( n + 1 2
) .
Proof. It is plain that P (n + 1, n) =
( n + 1 2
) , since a partition of a set containing
(n + 1) elements into n non-empty subsets is equivalent to a choice of one subset containing 2 elements.
Of course, Theorem 2.7 could also be proved using a variant of our proof of Theorem 2.6.
On partitions, surjections, and Stirling numbers 717
3 A combinatorial lemma
In this section we prove a combinatorial result which may be interpreted ‘geometri- cally’ within the Pascal Triangle. We will use this result in an essential way in the proof of our main theorem. We start with the familiar result ∑^ n r=
r(r + 1)... (r + s − 1) =
n(n + 1)... (n + s) s + 1
Now consider A =
∑^ n r=s
r(r − 1)... (r − s + 1), 1 ≤ s ≤ n. By replacing the variable r
by R = r − s + 1 (and then writing r for R) we see that
A =
n−∑s+
r=
(r + s − 1)(r + s − 2)... r =
(n − s + 1)(n − s + 2)... (n + 1) s + 1
by (3.1). Thus
∑^ n r=s
r(r−1)... (r−s+1) =
(n + 1)(n)... (n − s + 2)(n − s + 1) s + 1
, 1 ≤ s ≤ n. (3.2)
We now enunciate the lemma
Lemma 3.1.
∑^ n r=s
( r j
) ( r − j s − j
)
( n + 1 s + 1
) ( s j
) , j ≤ s ≤ n.
Remark. If j = s, then Lemma 3.1 asserts that
∑^ n r=s
( r s
)
( n + 1 s + 1
)
. (3.3)
This is sometimes known as the Christmas Stocking Theorem and admits a very elementary proof. Its name (suggested, we believe, by our late colleague Dave Logo- thetti) is easily understood when one locates the binomial coefficients entering into the statement in the Pascal Triangle (see Figure 1).
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1
The case s = 2, n = 5 of the Christmas Stocking Theorem (3.3) Figure 1.
On partitions, surjections, and Stirling numbers 719
Theorem 4.1. We may express γk (m), 1 ≤ k ≤ m − 1 , as a linear combination
γk(m) =
k∑− 1
h=
ahk
( m k + h + 1
)
where the Stirling factors ahk are positive integers, independent of m, given induc- tively by the rule
a 0 k = 1, ahk =
k∑−h
j=
( k + h j
) ah− 1 ,k−j , 1 ≤ h ≤ k − 1. (4.2)
Proof. We find it convenient, in the proof, to work with
gk(n) = γk (n + k) = P (n + k, n).
We first prove a lemma.
Lemma 4.2.
gk+1(n) =
( n + k + 1 k + 2
)
∑^ n r=
∑^ k j=
( r + k j
) gk−j+1 (r − 1).
Proof of Lemma 4.2. Let^2 S =
n∑− 1
r=
(−1)n−r−^1
(r + 1)n+k r!(n − r − 1)!
. Setting s = r + 1 and
adjoining s = 0, we see that
S =
∑^ n s=
(−1)n−s^
sn+k+ s!(n − s)!
= gk+1(n).
On the other hand, expanding (r + 1)n+k^ by the binomial theorem, and applying Theorem 2.6, we have
S = gk+1(n − 1) + (n + k)gk(n − 1) +
( n + k 2
) gk− 1 (n − 1) +...
( n + k k
) g 1 (n − 1) +
( n + k k + 1
) .
Thus
gk+1(n) − gk+1(n − 1) =
( n + k k + 1
)
∑^ k j=
( n + k j
) gk−j+1 (n − 1). (4.3)
Adding up (4.3) for n = 1, 2 ,... , N, noting that gk (0) = 0 for all k ≥ 1, and finally replacing N by n, we infer that
gk+1(n) =
∑^ n r=
( r + k k + 1
)
∑^ n r=
∑^ k j=
( r + k j
) gk−j+1(r − 1). (4.4)
(^2) Compare the proof of Theorem 2.6.
720 P. Hilton, J. Pedersen and J. Stigter
But
∑^ n r=
( r + k k + 1
)
n∑+k
r=k+
( r k + 1
)
( n + k + 1 k + 2
) , by (3.3), and the lemma is
proved.
We return to the proof of Theorem 4.1. We reformulate the theorem in terms of gk (n) as asserting that, for k ≥ 1 ,
gk (n) =
k∑− 1
h=
ahk
( n + k k + h + 1
) , (4.5)
where the weights ahk are given by (4.2); and we prove the theorem in this form by induction on k.
We note first that (4.5) holds if k = 1 by Theorem 2.7. The inductive hypothesis allows us to write
gk−j+1(r − 1) =
k∑−j
`=
a`,k−j+
( r + k − j k − j + ` + 2
) , 1 ≤ j ≤ k,
with a`,k−j+1 independent of r, whence, by Lemma 4.2,
gk+1(n) =
( n + k + 1 k + 2
)
∑^ n r=
∑^ k j=
k∑−j
`=
( r + k j
) ( r + k − j k − j + ` + 2
) a`,k−j+1. (4.6)
We now calculate
∑^ n r=
( r + k j
) ( r + k − j k − j + ` + 2
) , using Lemma 3.1. For
∑^ n
r=
( r + k j
) ( r + k − j k − j + ` + 2
)
∑^ n r=`+
( r + k j
) ( r + k − j k − j + ` + 2
) ,
by removing zero terms
=
n∑+k
r=k+`+
( r j
) ( r − j k − j + ` + 2
)
( n + k + 1 k + ` + 3
) ( k + ` + 2 j
) , by Lemma 3.1.
Thus, by (4.6),
gk+1(n) =
( n + k + 1 k + 2
)
∑^ k j=
k∑−j
`=
( n + k + 1 k + ` + 3
) ( k + ` + 2 j
) a`,k−j+1. (4.7)
Formula (4.7) shows that, as required,
gk+1(n) =
∑^ k
h=
ah,k+
( n + k + 1 k + h + 2
) , (4.8)
722 P. Hilton, J. Pedersen and J. Stigter
the entries in question running consecutively from j = h to j = k −1, and the weight
of the entry ah− 1 ,j being the binomial coefficient
( k + h j + h
)
. (See Figure 4).
( 0 0
)
( 1 0
)( 1 1
)
( 2 0
)( 2 1
)( 2 2
)
( n 0
)( n 1
)( n 2
)
...
( n k + 1
)( n k + 2
) · · ·
( n 2 k
)
︸ ︷︷ ︸ Pascal Triangle
( n n − 1
)( n n
)
P (n, n − k) = St(n, n − k) = γk(n) =
k∑− 1
h=
ahk
( n k + h + 1
) , 1 ≤ k ≤ n − 1
a 01 a 02 a 12 a 03 a 13 a 23 .. .
a 0 h a 1 h a 2 h · · · ah− 1 ,h .. . ah− 1 ,k− 1
a 0 k a 1 k a 2 k · · · ah− 1 ,k ahk · · · ak− 1 ,k ↑
Stirling Factor Triangle
ahk =
k∑−h
j=
( k + h j
) ah− 1 ,k−j =
k∑− 1
j=h
( k + h j + h
) ah− 1 ,j , 1 ≤ h ≤ k − 1 , a 0 k = 1
The interaction of the Pascal Triangle and the Stirling Factor Triangle Figure 4
In general, it does not appear profitable to seek explicit formulae for the Stir- ling factors ahk, although there are clearly some interesting divisibility properties underlying their definitions. However, the extreme cases h = 1, h = k − 1 are easily calculated. Thus
On partitions, surjections, and Stirling numbers 723
Theorem 4.3. a 1 k = 2k+1^ − (k + 3); ak− 1 ,k =
(2k)! 2 k^ k!
Proof. By (4.2),
a 1 k =
k∑− 1
j=
( k + 1 j
)
k∑+
j=
( k + 1 j
) − 1 − (k + 1) − 1 = 2k+1^ − k − 3.
Again, by (4.2), ak− 1 ,k = (2k − 1)ak− 2 ,k− 1 , k ≥ 2. Since a 01 = 1, this yields
ak− 1 ,k = (2k − 1)(2k − 3)... 3 =
(2k)! 2 k^ k!
this formula holds, of course, also for k = 1.
We may use Theorems 4.1 and 4.3 to calculate a 2 k and ak− 2 ,k. Thus
Theorem 4.4.
a 2 k =
3 k+2^ − (k + 5)2k+1^ +
(k^2 + 7k + 13); ak− 2 ,k =
k − 1 3
(2k)! 2 k^ k!
Proof. By (4.9), a 2 k =
k∑− 1
j=
( k + 2 j + 2
) a 1 j =
k∑− 1
j=
( k + 2 j + 2
) (2j+1^ − j − 3), by Theorem
4.3. We complete the calculation of a 2 k by an elementary argument, using the identities
(1 + x)k+2^ =
k∑+
j=
( k + 2 j
) xj^ , (k + 2)(1 + x)k+1^ =
k∑+
j=
j
( k + 2 j
) xj−^1.
To calculate ak− 2 ,k, we exploit Lemma 4.5 below. For if a akk−−^21 ,k,k = k− 3 1 , then Theorem
4.3 immediately yields the given value of ak− 2 ,k.
Lemma 4.5. a akk−−^21 ,k,k = k− 3 1 , k ≥ 2.
Proof of Lemma. We argue by induction on k. If k = 2, the result certainly holds. Suppose that a akk−−^21 ,k,k = k− 3 1 , for some k ≥ 2. Now by (4.9)
ak− 1 ,k+1 =
( 2 k 2
) ak− 2 ,k− 1 + 2k ak− 2 ,k
= k(2k − 1)ak− 2 ,k− 1 + 2k ak− 2 ,k
= k ak− 1 ,k +
2 k(k − 1) 3
ak− 1 ,k, using the proof of Theorem 4.
and the inductive hypothesis
=
k 3
(2k + 1)ak− 1 ,k
k 3
ak,k+1, again by the proof of Theorem 4.3.
This establishes the inductive hypothesis, the lemma, and, with it, Theorem 4.4.
On partitions, surjections, and Stirling numbers 725
References
[B] Brualdi, Richard A., Introductory Combinatorics, North Holland (1977).
[C] Comtet, Louis, Analyse combinatoire, Tome Premier, Presses Universitaires de France (1970).
[L] Ledermann, Walter, (ed.), Handbook of Applicable Mathematics, Volume 5, Part B, John Wiley, Chichester (1985).
[T] Tucker, Alan, Applied Combinatorics, 2nd^ edition, John Wiley, New York (1984).
Peter Hilton Department of Mathematical Sciences SUNY Binghamton Binghamton, New York 13902-6000, U.S.A.
Jean Pedersen Department of Mathematics Santa Clara University Santa Clara, California 95053, U.S.A.
J¨urgen Stigter Faculty of Technical Mathematics and Informatics Delft University of Technology 2628 BL Delft, Netherlands
