arXiv:1008.1573v1 [math.NT] 9 Aug 2010
Zhi-Wei Sun and Don Zagier
Abstract. In this paper we derive congruences expressing Bell numbers and derangement numbers in terms of each other modulo any prime.
1. Introduction
Let Bn denote the nth Bell number, defined as the number of partitions
of a set of cardinality n (with B0 = 1). In 1933 Touchard [T] proved that
for any prime p we have
Bp+n ≡ Bn+ Bn+1 (mod p) for all n = 0, 1, 2, . . . . (1)
Thus it is natural to look at the numbers Bn (mod p) for n < p. In [S], the
first author discovered experimentally that for a fixed positive integer m the sumPp−1
n=0Bn/(−m)
nmodulo a prime p not dividing m is independent
of the prime p, a typical case being
p−1
X
n=0
Bn
(−8)n ≡ −1853 (mod p) for all primes p 6= 2 .
In this note we will prove this fact and give some related results.
Our theorem involves another combinatorial quantity, the derangement number Dn, defined either as the number of fixed-point-free permutations
of a set of cardinality n (with D0 = 1) or by the explicit formula
Dn n! = n X k=0 (−1)k k! (n = 0, 1, 2, . . . ) . (2)
2010 Mathematics Subject Classification. Primary 11B75; Secondary 05A15, 05A18, 11A07.
Keywords. Bell numbers, derangement numbers, congruences.
The first author is supported by the National Naturaal Science Foundation (grant 10871087) and the Overseas Cooperation Fund (grant 10928101) of China.
Theorem 1. For every positive integer m and any prime p not dividing m we have X 0<k<p Bk (−m)k ≡ (−1) m−1 Dm−1 (mod p) . (3) Using P 0<m<p(−m) n−k ≡ −δ n,k (mod p) for k, n ∈ {1, . . . , p − 1}, we
immediately obtain a dual formula for Bn(n < p) in terms of D0, . . . , Dp−2.
Corollary. Let p be any prime. Then for all n = 1, . . . , p − 1 we have
Bn ≡ p−1
X
m=1
(−1)mDm−1(−m)n (mod p) .
For the reader’s convenience we give a small table of values of Bnand Dn.
n 0 1 2 3 4 5 6 7 8
Bn 1 1 2 5 15 52 203 877 4140
Dn 1 0 1 2 9 44 265 1854 14833
We will show Theorem 1 in the next section and derive an extension of Theorem 1 in Section 3.
2. Proof of Theorem 1
We first observe that it suffices to prove (3) for 0 < m < p, since both sides are periodic in m (mod p) with period p. For the left-hand side this is obvious and for the right-hand side it follows from (2), which gives the expression (−1)nD
n = P
∞
r=0(−1)rn(n − 1) · · · (n − r + 1) for Dn (mod p)
as a terminating infinite series of polynomials in n.
We will prove (3) for 0 < m < p by induction on m. Denote by Sm the
sum on the left-hand side of (3), where we consider the prime p as fixed and omit it from the notation. Since Dn = nDn−1+ (−1)n for n = 1, 2, 3, . . .
(obvious from (2)), we have to prove the two formulas
S1 ≡ 1 (mod p) , m Sm ≡ S1 − Sm+1 (mod p) . (4)
Recall that the Bell numbers can be given by the generating function
∞ X n=0 Bn xn n! = e ex −1 (5)
equivalent to the well-known closed formula Bn = 1 e ∞ X r=0 rn r! . Since the function y = eex
−1 satisfies y′
= exy, this also gives the recursive
definition B0 = 1 , Bn+1 = n X k=0 n k Bk for all n ≥ 0 . (6)
This recursion is the key ingredient in proving (4).
For the first formula in (4) we use (6) with n = p − 1 to obtain S1 = p−1 X k=1 (−1)kB k ≡ p−1 X k=1 p − 1 k Bk = Bp − B0 (mod p) ,
so it suffices to prove that Bp ≡ 2 (mod p). This is a special case of
Touchard’s congruence (1), but can also be seen by writing (5) in the form
∞ X n=0 Bn xn n! = e x + X 1<r<p ex− 1r r! + xp p! + O x p+1 to get Bp/p! = 1/p! + (p-integral) + 1/p! .
Now using Fermat’s little theorem we have −m Sm ≡ p−2 X n=0 (−m)p−1−n n X k=0 n k Bk ≡ p−2 X k=0 (−1)kBk p−k−2 X r=0 p − k − 1 r mp−k−1−r (r = n − k) ≡ p−2 X k=0 (−1)kBk (m + 1)p−1−k− 1 ≡ Sm+1 − S1 (mod p)
for 1 ≤ m ≤ p − 2. This completes the proof of (4) and the theorem. 3. An extension of Theorem 1
Recall that for nonnegative integers n and k the Stirling number S(n, k) of the second kind is the number of ways to partition a set of n elements into k groups. Obviously
Bn = n
X
k=0
The Touchard polynomial Tn(x) of degree n is given by Tn(x) = n X k=0 S(n, k)xk. (7)
Note that Tn(1) = Bn. Similar to the recursion for Bell numbers, we
have the recursion
Tn+1(x) = x n X k=0 n k Tk(x). (8)
Let p be a prime and let Zp be the ring of p-adic integers. For two
polynomials P (x), Q(x) ∈ Zp[x], by P (x) ≡ Q(x) (mod p) we mean that
the corresponding coefficients of P (x) and Q(x) are congruent modulo p. Our next theorem is a further generalization of Theorem 1.
Theorem 2. For every positive integer m, we have
(−x)m X 0<n<p Tn(x) (−m)n ≡ −x p m−1 X l=0 (m − 1)! l! (−x) l (mod p) (9)
for any prime p not dividing m.
As a consequence, if x is a p-adic integer not divisible by p, then X 0<n<p Tn(x) (−m)n ≡ 1 (−x)m−1 m−1 X k=0 (m − 1)! k! (−x) k (mod p). (10) In particular, X 0<n<p Tn(x) (−2)n ≡ x − 1 x (mod p) for p 6= 2, X 0<n<p Tn(x) (−3)n ≡ x2− 2x + 2 x2 (mod p) for p 6= 3 , X 0<n<p Tn(x) (−4)n ≡ x3− 3x2+ 6x − 6 x3 (mod p) for p 6= 2 .
Although we can show Theorem 2 via a slight modification of the proof of Theorem 1, below we prove Theorem 2 by a new approach.
Proof of Theorem 2. Observe that p−1 X n=1 Tn(x) (−m)n = p−1 X n=1 Pn k=1S(n, k)x k (−m)n = p−1 X k=1 xk p−1 X n=1 S(n, k) (−m)n. It is known that S(n, k) = 1 k! k X j=0 k j (−1)k−jjn for all n, k = 0, 1, 2, . . . . Thus p−1 X n=1 Tn(x) (−m)n = p−1 X k=1 xk k! p−1 X n=1 k X j=1 k j (−1)k−j − j m n . For each j ∈ {1, . . . , p − 1}, if p | m + j then
p−1 X n=1 −j m n ≡ p−1 X n=1 1 ≡ −1 (mod p) , if p ∤ m + j then p−1 X n=1 − j m n = p−1 X n=0 − j m n − 1 = (−j/m) p− 1 −j/m − 1 − 1 ≡ 0 (mod p) with the help of Fermat’s little theorem.
Let r denote the least positive residue of −m modulo p. By the above,
p−1 X n=1 Tn(x) (−m)n = p−1 X k=1 xk k! p−1 X j=1 k j (−1)k−j p−1 X n=1 − j m n ≡ p−1 X k=1 xk k! k r (−1)k−r(−1) = (−1)r+1 r! p−1 X k=r (−x)k (k − r)! (mod p). Therefore (−x)m p−1 X n=1 Tn(x) (−m)n ≡ (−1)r+1 r! m−1 X l=m+r−p (−x)p+l (p + l − m − r)! (mod p). So it remains to show that
(m − 1)!
l! ≡
(−1)r+1/(r!(p + l − m − r)!) (mod p) if m + r − p 6 l < m,
If 0 6 l < m + r − p, then we have (m − 1)!/l! ≡ 0 (mod p) since l < m + r − p < m and m + r ≡ 0 (mod p).
Now suppose that m + r − p 6 l < m. Then (−1)r+1 (p + l − m)! p + l − m r ≡ (−1) r+1Qm−l−1 s=1 (p − s) (p − 1)! l − m r ≡ (−1)m−l−1(m − l − 1)!m − l + r − 1 m − l − 1
(by Wilson’s theorem) ≡ (−1)m−l−1(m − l − 1)! −l − 1 m − l − 1 = (m − 1)! l! (mod p) (as p | m + r). In view of the above, we have completed the proof of Theorem 2. Acknowledgments. The joint work was done during the authors’ visit to the National Center for Theoretical Sciences (Hsinchu, Taiwan) during August 1–8, 2010. Both authors are indebted to Prof. Winnie Wen-Ching Li for the kind invitation and the center for the financial support.
References
[S] Z. W. Sun, A conjecture on Bell numbers, a message to Number Theory List, http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1007&L=nmbrthry&T=0&P=1066. [T] J. Touchard, Propri´et´es arithm´etiques de certains nombres recurrents, Ann. Soc.
Sci. Bruxelles 53A (1933), 21–31.
Department of Mathematics, Nanjing University, Nanjing 210093, Peo-ple’s Republic of China
E-mail address: zwsun@nju.edu.cn Homepage: http://math.nju.edu.cn/∼zwsun
Max-Planck-Institut f¨ur Mathematik, Bonn 53111, Germany E-mail address: don.zagier@mpim-bonn.mpg.de
