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On p-adic properties of the Stirling numbers of the first kind
Journal of Number Theory 148 (2015) 73–94
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Journal of Number Theory www.elsevier.com/locate/jnt
On p-adic properties of the Stirling numbers of the first kind Tamás Lengyel Mathematics Department, Occidental College, 1600 Campus Road, Los Angeles, CA 90041, USA
a r t i c l e
i n f o
Article history: Received 12 June 2014 Received in revised form 19 September 2014 Accepted 19 September 2014 Available online 29 October 2014 Communicated by David Goss Keywords: Stirling numbers of the first kind p-Adic properties Congruence relations modulo prime powers Convolution identities for Stirling numbers Generalized harmonic numbers
a b s t r a c t The goal of this paper is to describe s(n, k) mod pe and calculate νp (s(n, k)) for a prime p, fixed integer k 1, and large enough e and n. Some special cases of the form s(apn , k) mod pe and its relation to s(apn+1 , kp) mod pe as well as bounds on νp (s(apn , k)) and its exact values in some special cases are completely determined. We also investigate the properties of νp (s(n, n − k)) and νp (s(apn + b, apn + b − k)) which are significantly different from those of νp (s(n, k)) and νp (s(apn , k)). We use congruential identities for the generalized harmonic numbers, and new congruential and convolution identities for s(n, k). © 2014 Elsevier Inc. All rights reserved.
1. Introduction We note that, unlike for the Stirling number of the second kind, S(n, k), there is no easy way to use an explicit formula for s(n, k) and this fact makes it more difficult to deal with certain characteristics of these numbers from the point of view of congruential and divisibility properties. In fact, the Stirling numbers of the first kind exhibit very different characteristics from the Stirling numbers of the the second kind. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jnt.2014.09.015 0022-314X/© 2014 Elsevier Inc. All rights reserved.
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We will need some basic notation. Let n and k be positive integers, p be a prime, dp (k) and νp (k) denote the sum of digits in the base p representation of k and the highest power of p dividing k, respectively. The latter one is often referred to as the p-adic order of k. For the rational n/k we set νp (n/k) = νp (n) − νp (k). We also use the standard notation for the generalized harmonic numbers:
Hn(m) =
n 1 , im i=1
(1)
with Hn being the harmonic number Hn . As usual, Bn stands for the nth Bernoulli number and νp (B2m ) −1 for m 1 by the von Staudt–Clausen theorem, which implies that the denominator of B2m is squarefree. Let χA denote the indicator variable of the event A, i.e., it is 1 exactly if A occurs. For every fixed integer k 1, we prove that, unlike for Stirling numbers of the second kind, the p-adic order of s(n, k) becomes arbitrarily large for large values of n. We establish this fact in Theorems 1.1 and 1.2 which state the divergence and the asymptotic order of magnitude of νp (s(n, k)) for any fixed k 1. Moreover, only the rate of convergence of the ratio of νp (s(n, k)) to its asymptotic order of magnitude is affected by the choice of k. In Section 2 we study the cases with k = 2 and 3, and determine the exact order νp (s(apn , k)). The ultimate goal is to describe s(n, k) (mod pe ) and then νp (s(n, k)) for large enough e and n. The main results in this direction are summarized in Theorems 1.1–1.3, 1.5, 2.1–2.5, and 4.2. Section 3 deals with νp (s(n, n − k)) for a fixed k 1. The p-adic order of s(n, n − k) does not become arbitrarily large for large values of n except for properly selected subsequences, e.g., if n is of the form apn + b with b = 0 but remains constant if b k + 1 or b = −1 as stated in Theorems 3.2–3.5. Theorems 1.3 and 3.1 present new congruential and convolution identities for s(n, k), respectively. We give a new p-adic relation for the harmonic numbers in Theorem 4.2. Sections 4 and 5 are devoted to the proofs. From now on we rely on two conjectures. Conjecture 1.1. (See [3].) For any odd prime p 5 and integer n 1, we have that νp (Hn ) 3. Note that ν2 (Hn ) = −log2 n 0, which can be proven in a similar fashion to the proof that Hn −Hm with 0 < m < n, is never an integer. Following [3], for every prime p, we define Jp = {n|νp (Hn ) 1}. Remark 1.1. Using J3 , Eswarathasan and Levine (cf. [3]) observed that ν3 (n) 1 for all n 1. In fact, we have that |J3 | = 3 and ν3 (Hn ) = 1 for all three ns in J3 = {2, 7, 22}.
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The above conjecture is extended in [13] for generalized harmonic numbers. Washing(m) (m) (m) ton defines Jp = {n|p divides the numerator of Hn }, i.e, Jp is the set of integers n (m) (1) for which νp (Hn ) 1. Note that Jp = Jp . With some heuristic argument, he states the following (2)
Conjecture 1.2. (See [13].) For m = 2, there is no n > p2 such that νp (Hn ) 1. For (3) m = 3, there is no n > p2 such that νp (Hn ) 1 except possibly for finitely many (m) exceptional primes p. For m 4, there is no n > p such that νp (Hn ) 1. We also use the following two corollaries due to Glaisher. Corollary 1.1. (See Corollary 1, [13].) Let r 1 be odd and suppose p r + 4. Then p−1 r(r + 1) 1 Bp−r−2 p2 ≡− r j 2(r + 2) j=1
mod p3 .
Corollary 1.2. (See Corollary 2, [13].) Let r 2 be even and suppose p r + 3. Then p−1 r 1 Bp−r−1 p ≡ r j r + 1 j=1
mod p2 .
We prove the divergence of νp (s(n, k)) as n → ∞ for all k 1 and establish facts about its asymptotic order of magnitude in the next two theorems. Theorem 1.1. For any prime p and integer k 1, we get that lim νp s(n, k) = ∞.
n→∞
Remark 1.2. By (167) of [11, p. 491], we have that s(n, k) ≡ 0 (mod pr ) for n > krp which also guarantees that limn→∞ νp (s(n, k)) = ∞. In fact, for n > N it yields that νp s(n, k)
N . pk
We improve this inequality in Theorem 1.2. Theorem 1.2. For any prime p and integer k 1, there exists a constant c = c (k, p) > 0 so that for n n0 (k, p) we get νp s(n, k) c n. In fact, we can set c (k, p) to any value less than 1/(1 −p), and we also have the inequality
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n − dp (n) n − νp (n) − (k − 1) logp n − (k + 1) logp n νp s(n, k) p−1 p−1 for n 1. Under Conjectures 1.1 and 1.2, we get that 1 νp (s(n, k)) = . n→∞ n p−1 lim
In this case, for the rate of convergence we obtain that νp (s(n, k)) logp n 1 (k + 1) − n p − 1 n
(1.1)
for any sufficiently large n. By Theorem 1.1 we know that lim νp s apn , k = ∞
n→∞
for any prime p, integers a, n, and k, such that (a, p) = 1. Now we determine s(apn , k) (mod pe ) in (1.4), for any integer e 1, and νp (s(apn , k)) if n is large enough. Motivated by [2], we first determine (x)m = x(x − 1) · · · (x − m + 1) (mod pe ) with m = apn . We note that in [2] only some special cases were considered, e.g., the case with p = 2, e = 2, m = 2n and m = 3 · 2n (with n 5), and the case with p 3, e = 1, and m = pn (with n 2). Theorem 1.3. For any prime p, positive integers a, n, k, and e such that (a, p) = 1, 0 k apn , 3 2e − 1 n, with the notation (x)n = x(x − 1) · · · (x − n + 1),
(1.2) n
for the ordinary generating function (ogf) of the sequence {s(apn , k)}ap k=0 we get that n
(x)apn
ap apn−e n−1 e−1 = s apn , k xk ≡ xap −1 x(p−1)p
mod pe .
(1.3)
k=0
Thus, with the setting g(a, p, n, e, k) =
k − apn−1 , (p − 1)pe−1
if
k − apn−1 is an integer (p − 1)pe−1
and otherwise undefined, and b(a, p, n, e, k) =
n−e
(−1)ap 0,
−g(a,p,n,e,k)
,
apn−e g(a,p,n,e,k)
if
k−apn−1 (p−1)pe−1
otherwise,
is an integer,
(1.4)
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we get that s apn , k ≡ b(a, p, n, e, k)
mod pe .
It follows that νp s apn , k =
⎧ apn−e ⎪ , ν p ⎪ g(a,p,n,e,k) ⎨ ⎪ ⎪ ⎩
if νp
e,
k−apn−1 (p−1)pe−1
is an integer and < e,
apn−e g(a,p,n,e,k)
(1.5)
otherwise.
Remark 1.3. Clearly, if νp (s(apn , k)) is at least as large as e, e.g., as large as (n + 1)/2 then (1.3) and its consequences are of no use. This is the case in Theorem 3.3 as well as Theorem 3.4 and Conjecture 3.1 with regard to νp (s(apn , apn − k)). n−e
n−1
n
Remark 1.4. We note that expanding (1.3) results in (−1)ap xap + · · · + xap e n e (mod p ). In order to have s(ap , k) ≡ 0 (mod p ) we need to consider only values of the form k = apn−1 +j(p−1)pe−1 with 0 j apn−e and thus, pe−1 | k is a must since n > e. Note that here j is playing the role of g(a, p, n, e, k). Somewhat counterintuitively, for any n−e fixed e, we do not get more j values such that νp ap j < e as n increases, as we will see in Theorem 1.5. In special cases of Theorem 1.3, part of (1.5) can be easily calculated, k−pn−1 n e.g., if a = 1 and (p−1)p e−1 is an integer then νp (s(p , k)) = n − e − νp (g(1, p, n, e, k)) provided that it is less than e except if g(1, p, n, e, k) = 0 when νp (s(pn , k)) = 0. This possibility is further explored in Example 1.1. The following theorem gives a generalization of the Jacobstahl–Kazandzidis congruences, cf. [7]. Theorem 1.4. (See Corollary 11.6.22, [4].) Let M and N be such that 0 M N and p prime. We have
pN pM
⎧ N N Bp−3 3 4 ⎪ p N M (N − M )) mod p N M (N − M ) , (1 − ⎪ M 3 ⎨ N M N 4 ≡ (1 + 45N M (N − M )) M mod p N M (N − M ) M , ⎪ N ⎪ ⎩ (−1)M (N −M ) P (N, M ) N mod p4 N M (N − M ) M , M
if p 5, if p = 3, if p = 2,
where P (N, M ) = 1 +6N M (N −M ) −4N M (N −M )(N 2 −N M +M 2 ) +2(N M (N −M ))2 . Theorem 1.3, in combination with Theorem 1.4, helps to determine the exact value of s(apn , k) (mod pe ) when n is sufficiently large. In this case we relate s(apn+1 , kp) (mod pe ) to s(apn , k) (mod pe ). Theorem 1.5. For any prime p, integers a, n, k, and e such that (a, p) = 1, 0 k apn , 3 2e − 1 n, we get that
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(x)apn+1 =
n+1 ap
n
ap s apn+1 , k xk ≡ s apn , k xkp
k=0
mod pe ,
(1.6)
k=0
i.e., s apn+1 , kp ≡ s apn , k
mod pe .
In other words, (x)apn+1 (mod pe ) can be obtained by simply replacing x with xp on the apn right hand side of (x)apn ≡ k=0 s(apn , k)xk (mod pe ). Example 1.1. We consider the case with (a, p, n, e) = (1, 3, 5, 3). The calculations agree with (1.6) of Theorem 1.5: (x)729 ≡
9
s(243, 81 + 18j)x243+54j
j=0
≡ s(243, 81)x243 + s(243, 99)x297 + · · · + s(243, 243)x729 ≡ 26x243 + 9x297 + 18x351 + 3x405 + 9x459 + 18x513 + 24x567 + 9x621 + 18x675 + x729
mod 27.
Note that in the general term s(3n , k)x3k we have k = 3n−1 + 2j3e−1 with 0 j 3n−3 . In our special case, it means k = 81 + 18j with 0 j 9, and j 1 implies that ν3 (s(3n , 3n−1 +2j3e−1 )) = n −e −ν3 (j) < e since ν3 (s(243, 81 +18j)) = 5 −3 −ν3 (j) < 3, in agreement with Remark 1.4. Therefore, the 3-adic orders of the ten consecutive nonzero coefficients of the actual (1.6) are 0, 2, 2, 1, 2, 2, 1, 2, 2, 0. 2. Special cases Theorem 2.1. For any prime p and integer n 1, we have that s(n, 1) = (−1)n−1 (n − 1)!, thus n − dp (n) n n > νp s(n, 1) = − νp (n) − 2 logp n, p−1 p−1 p−1 and s(n, 2) = (−1)n−2 (n − 1)!Hn−1 , hence νp s(n, 2) = νp s(n, 1) + νp (Hn−1 )
n − 3 logp n, p−1
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and under Conjecture 1.1, we get that n + 3 νp s(n, 2) . p−1 Theorem 2.2. For a 1 odd and n 1 we have that ν2 s a2n , 2 = a2n − 2n − d2 (a) − log2 a + χa=1 . Theorem 2.3. For a 1, (a, 3) = 1, and n 1 we have that 1 ν3 s a3n , 2 = a3n − d3 (a) − 2n + ν3 (H3a−1 ) + 1, 2 which is 1 n a3 − d3 (a) − 2n + ν3 (Ha−1 ), 2 for a > 1. In particular, if a = 1 then 1 ν3 s 3n , 2 = 3n + 3 − 2n. 2 We also have the inequalities 1 1 n a3 − d3 (a) − 2n − log3 a ν3 s a3n , 2 a3n − d3 (a) − 2n − log3 a + 2. 2 2 In general, for ν3 (s(n + 1, 2)) we ge that 1 1 n − d3 (n) − log3 n ν3 s(n + 1, 2) n − d3 (n) − log3 n + 3. 2 2 Theorem 2.4. Under Conjecture 1.1, for any prime p 5, a 1, (a, p) = 1, and n 1, we have that νp s apn , 2 =
1 n ap − dp (a) − 2n + 1 + νp (Hap−1 ). p−1
The inequalities νp s apn , 2
1 n ap − dp (a) − 2n + p−1
−logp a, 3,
if a > 1, if a = 1
give lower bounds on the p-adic orders that can be easily calculated. The inequality is strict when a = p2 − p + 1.
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Remark 2.1. We add that the exact values of ν3 (s(a3n , 2)) and ν3 (s(n + 1, 2)) can be found by Theorem 2 of [9]. We also note that, for p 5, νp (Hp−1 ) 2 by Wolstenholme’s theorem and νp (Hp−1 ) = 2 except for the so-called Wolstenholme primes for which νp (Hp−1 ) 3, e.g., p = 16843 and p = 2124679 with νp (Hp−1 ) = 3, and no further examples of such primes are known. Another important conjecture is that νp (Hn ) 4 never occurs, cf. Conjecture 1.1. We define E(i, j) as the elementary symmetric polynomial of degree j in the integers [i] = {1, 2, . . . , i}. In a similar fashion, we define E (i, j) as the elementary symmetric polynomial of degree j in the reciprocals of integers [i] = {1, 2, . . . , i}. For instance, the nth harmonic number Hn is equal to E (n, 1) and E (n, 2) = 1i1
E(S, k) =
i1 i2 · · · ik .
i1
By convention, we set E(i, 0) = E (i, 0) = E(S, 0) = 1. Theorem 2.5. For n 2 we have that ν2 s 2n , 3 = 2n − 3n + 3. For p 3 prime and n 1, we get that pn − 1 − 3n + 2 + νp E (p − 1, 2) , νp s pn , 3 = p−1 −1 which is equal to 12 (3n − 1) − 3n + 2 for p = 3 and pp−1 − 3n + 3 for p 5 provided that νp (Bp−3 ) 0, i.e., if p is not a Wolstenholme prime. In general, for a 1 we have the inequality n
apn − dp (a) − 3n − 2logp a. νp s apn , 3 p−1 Remark 2.2. We note that we found only the sequences {ν2 (s(3·2n , 2))}n1 as http://oeis. org/A050488, {ν2 (s(5 · 2n , 2))}n1 as http://oeis.org/A097809, and {ν2 (s(3 · 2n , 3))}n1 as http://oeis.org/A123203 in [12]. 3. The other end of the range In this section we investigate the case of s(n, n − k) with small values of k. We will use a convolution type identity for Stirling numbers of the first kind that can be derived from (6.46) in [6, p. 272], also cf. (7) in [1]. For any positive integers a, b, and n such that a, b n + 1, we have that
T. Lengyel / Journal of Number Theory 148 (2015) 73–94
s(a, a − k)s(b, b − (n − k)) s(a + b, a + b − n) a+b−2 = . (a + b − 1) a−1 (a + b − n − 1) a+b−n−2 a−k−1 k=0
81
n
(3.1)
We will also use another similar convolution identity (3.2) which seems to be new. It can be derived from (6.47) in [6, p. 272] in a similar manner to (3.1). We leave the details to the reader. Theorem 3.1. For any positive integers a, b, and n such that a, b n + 1, we have that n a + b s(a, a − k)s(b, b − (n − k)) s(a + b, a + b − n) k a+b−2 = . an (a + b − 1) a−1 (a + b − n − 1) a+b−n−2 a−k−1 k=0
(3.2)
We also use the identity from [8, p. 32] s(n, n − k) =
k j=0
d(2k − j, k − j)
n 2k − j
(3.3)
where d(n, k) is the so-called associated Stirling number of the first kind, cf. [5, pp. 256–257], and it is a nonnegative integer. For example, s(n, n − 1) = d(2, 1) n2 + n n d(1, 0) 1 = 2 . We have the following main results for νp (s(apn + b, apn + b − k)) when b 0 in Theorem 3.2, b = 0 in Theorem 3.3 and (the conditional) Theorem 3.4, and b = −1 and −2 in Theorem 3.5. Theorem 3.2. For any prime p, integers a 1 with (a, p) = 1, k so that 2 k + 1 b, and n sufficiently large, we get that νp s apn + b, apn + b − k = νp s(b, b − k) . We note that if b k + 1 then νp (s(apn + b, apn + b − k)) does not depend on n for sufficiently large values of n. This behavior is very different from the one we observed in Theorem 1.2 and its special cases as well as in the case of b = 0. In the latter case we derive a lower bound on the p-adic order in Theorem 3.3. Theorem 3.3. For any prime p, integers a 1 with (a, p) = 1, k 1, and sufficiently large n n0 (p, k) = logp (2k) , we get that νp s apn , apn − k n − logp (2k)
(3.4)
independently of a. For k = 1 the bound is tight when p = 2 and for odd primes the difference between the exact order and the lower bound appears to be constant if k = 1 or k 2 even, according to (the conditional) Theorem 3.4. For odd ks we state Conjecture 3.1.
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Theorem 3.4. For any prime p, integers a 1 with (a, p) = 1, and even k 2 with the condition ∃n1 : n1 > 3 logp k + logp a
such that νp s apn1 , apn1 − k < n1
(3.5)
or k = 1 with n1 = 1 then for n n1 we have that νp s apn+1 , apn+1 − k = νp s apn , apn − k + 1,
(3.6)
νp s apn+1 , apn+1 − k = νp s apn1 , apn1 − k + (n − n1 ).
(3.7)
and hence,
We note that it is easy to check (3.5) for small values of k, for instance, it holds for k = 1 and 2 but not for 3. In fact, we know by [8, Theorem 2.2] that νp (s(apn , apn −k)) n if k and p are both odd. Numerical calculations lead us to believe that (3.5) holds for all even k 2, otherwise we have the following conjecture. Conjecture 3.1. For any prime p, integer a 1 with (a, p) = 1, k 3 odd, and n n1 = n1 (p, k) with some sufficiently large n1 (p, k), we get that νp s apn+1 , apn+1 − k = νp s apn , apn − k + 2,
(3.8)
and hence, νp s apn+1 , apn+1 − k = νp s apn1 , apn1 − k + 2(n − n1 ). We saw in Theorem 3.3 that if b = 0 then limn→∞ νp (s(apn + b, apn + b − k)) = ∞, however, it is still different from the behavior of νp (s(n, k)) since it is not true that limn→∞ νp (s(n, n − k)) = ∞, e.g., we have already seen the case of b k + 1 when limn→∞ νp (s(apn + b, apn + b − k)) = νp (s(b, b − k)) constant. Additional evidence with b = −1 and −2 is provided by Theorem 3.5. For any odd prime p, n > 2k, and n ≡ −1 (mod pk ), we have that s(n, n − k) ≡ 0 (mod p). For p = 2 and n ≡ −1 or −2 (mod 2k+1 ), k 1, we have that s(n, n − k) is odd. 4. Proofs Proof of Theorem 1.1. We proceed with induction on k. Assume that for any prime p and constant B > 0 there exists an n0 (p, j, B) so that for all n n0 (p, j, B) we have νp (s(n, j)) B, and the above statement is true for j = 1, 2, . . . , k − 1. Clearly, the cases with k = 1 and 2 are established by Theorem 2.1.
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We now prove that it also holds for j = k by relying on identity (6) of [1] with m = k + 1:
n−r n m s(n, m) = s(n − i, r)s(i, m − r) r i i=m−r
(4.1) m
for any r : 0 r m. We set m = k + 1 and B = B + νp ( r, 0 < r < m = k + 1. Let
r
) and choose an
n0 (p, k + 1, B) = n0 p, k + 1 − r, B + n0 p, r, B . Now if n n0 (p, k + 1, B) then either n − i n0 (p, r, B ) or i n0 (p, k + 1 − r, B ) or both. In either case we get that on the right side of (4.1) either the factor s(n − i, r) or s(i, k + 1 − r) or both have a p-adic order at least B + νp ( k+1 r ). It follows that νp (s(n, k + 1)) B for all n n0 (p, k + 1, B). 2 Now we determine the asymptotic order of magnitude of νp (s(n, k)). Proof of Theorem 1.2. The ogf (1.2) implies that
s(n, k) = (−1)n−k E(n − 1, n − k) = (−1)n−k
i1 i2 · · · in−k
1i1
= (−1)n−k (n − 1)!
1i1
1 i1 i2 · · · ik−1
with im s being integers. In general, we have that s(n, k) = (−1)n−k (n − 1)!E (n − 1, k − 1).
(4.2)
First we check the cases with k = 1 and 2. We obtain that s(n, 1) = (−1)n−1 (n − 1)! and s(n, 2) = (−1)n (n − 1)!Hn−1 .
(4.3)
For larger values of k, we use the Newton–Girard formulas (−1)k−1 kEk =
k (−1)k−i Si Ek−i
(4.4)
i=1
with Ek = E (n − 1, k) and Sk = Hn−1 . In a similar fashion to the derivation of (4.3), via the application of (4.4) we derive that (k)
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(−1)n (n − 1)! (2) 2 Hn−1 − Hn−1 . 2
s(n, 3) = (−1)n−3 (n − 1)!E (n − 1, 2) =
(4.5)
and s(n, 4) = (−1)n−4 (n − 1)!E (n − 1, 3)
H3 (−1)n (n − 1)! 3 (2) (3) Hn−1 − Hn−1 Hn−1 + n−1 = 3 2 2 by setting k to 2 and 3, respectively. In general, we can express the elementary symmetric polynomials in terms of power sums Ek =
ci1 ,i2 ,...,im Si1 Si2 · · · Sim
1i 1 i2 ···im n−1 m j=1 ij =k;mk
with some rational constants ci1 ,i2 ,...,im by (4.4). It easily follows that νp (Si ) −i logp (n − 1) and hence, νp (Ek ) −k logp (n − 1) + Ok (1) with Ok (1) denoting a constant depending on k. Now n − dp (n) n − νp (n) + νp E (n − 1, k − 1) − (k + 1) logp n νp s(n, k) = p−1 p−1 follows for any large enough n by (4.2). By Conjectures 1.1 and 1.2, we have νp (Si ) 0 except for possibly finitely many exceptional primes p, and thus, except for the exceptional cases, we have that νp(Ek ) n/(p − 1) + Ok (1), and thus, νp (s(n, k)) n/(p − 1) for any sufficiently large n. 2 Example 4.1. We illustrate the changes in ν3 (s(n, 100)) (see Fig. 1). Proof of Theorem 1.3. For the first part of the statement we follow the proof from [2] and derive (1.3) as it was presented for the simpler cases with e = 1 and 2. For the second part we use (1.3). Now we prove (1.3). We proceed in the spirit of Lemmas 16 and 17 of [2]. Let r ∈ + 1 values of k, 0 k m − 1, {0, 1, . . . , pe − 1} and m = apn . There are Mr = m−r−1 pe e such that k is congruent to r modulo p . Clearly, Mr is independent of r and is equal to apn−e . By (1.2), this implies that
(x)apn ≡
e p −1
r=0
n−e
(x − r)ap
mod pe .
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Fig. 1. The functions ν3 (s(n, 100)) and ν3 (s(n, 100))/n with 1 n 5000.
We factor the last product into two factors and get that (x)apn ≡
n−e
(x − r)ap
p|r 0rpe −1
n−e
(x − r)ap
n−e
(x − ip)
=
n−e ap
j=0
≡
with 0 i pe−1 − 1 and
n−e apn−e (−ip)j xap −j j
e−1 n−e ap j=0
(4.6)
pr 1rpe −1
The first factor is the product of terms like (x − ip)ap apn−e
mod pe .
j
n−e
(−ip)j xap
−j
n−e
≡ xap
mod pe .
n−e The last congruence follows by the fact that νp ap j = n − e − νp (j) if 1 j pn−e , and 1 j e − 1 already guarantees e − 1 pn−e if n e + logp (e − 1) which is implied by the assumption that n 2e − 1 3. Then we use the fact that n − e − νp (j) + j e since n 2e − 1 2e − j + νp (j) as j − νp (j) = 1 if j = 1 or p = j = 2, and j − νp (j) j − logp j > 1, otherwise. Since we have pe−1 similar factors with different values of i, the total contribution of n−e e−1 n−1 the first factor of (4.6) is (xap )p = xap . e n−e For the second factor of (4.6), Euler’s theorem yields that it is (xφ(p ) − 1)ap ≡ e−1 n−e (x(p−1)p − 1)ap (mod pe ) with φ(m) denoting the Euler’s totient function. We conclude the proof of (1.3) by putting the two factors together. The other parts of the theorem follow easily by expanding the right hand side of (1.3) and extracting the coefficient of xk for different values of k that actually appear. 2 Proof of Theorem 1.5. We use Theorems 1.3 and 1.4. We note that if g(a, p, n, e, k) is an integer then
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g(a, p, n + 1, e, kp) =
kp − apn k − apn−1 = p = pg(a, p, n, e, k). (p − 1)pe−1 (p − 1)pe−1
We observe that in order for g(a, p, m, e, k) to be an integer, p must divide k if m e 2, otherwise the corresponding term s(apm , k) ≡ 0 (mod pe ). This fact comes in handy as we rewrite
(x)apn+1 =
n+1 ap
s apn+1 , k xk
k=0 n
ap = s apn+1 , kp xkp + k=0
≡
s apn+1 , k xk
pk 1kapn+1 −1
s apn+1 , kp xkp
mod pe .
g(a,p,n+1,e,kp) is an integer 1kapn
We will prove below that last sum is congruent to
s apn , k xkp
mod pe .
g(a,p,n,e,k) is an integer 1kapn
We have two cases. Case 1: p 3 prime. By the definition of function g, Remark 1.4 and Theorem 1.4 we get that if g(a, p, n, e, k) is an integer then g(a, p, n +1, e, kp) is also integer and vice versa, since e 2, and pe−1 | g(a, p, n + 1, e, kp) makes g(a, p, n + 1, e, kp)/p = g(a, p, n, e, k) an integer. Hence
apn+1−e p · apn−e apn−e = ≡ g(a, p, n + 1, e, kp) p · g(a, p, n, e, k) g(a, p, n, e, k)
mod pe
since in Theorem 1.4 we set N = apn−e and M = g(a, p, n, e, k) and observe that min{2 + n − e, 4 + n − e, e} = e if 2e − 1 n. Case 2: p = 2. In this case (1.4) has an extra factor (−1) if g(a, 2, n, e, k) is odd and that compensates for the factor (−1)M (N −M ) = −1 in Theorem 1.4. We also have min{1 + n − e, 4 + n − e, e} = e if 2e − 1 n. 2 Remark 4.1. It immediately follows that limn→∞ νp (s(apn , k)) = ∞ since, as n increases, we clean up the terms in (1.6) with xm , m < apn , for any fixed value e, cf. Example 1.1. Now we turn to the special cases. The identity (4.3) yields that n s apn , 2 = (−1)ap apn − 1 !Hapn −1 ,
(4.7)
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and thus, νp s apn , 2 =
apn − dp (a) − n + νp (Hapn −1 ), p−1
then we try to express the last term in more simple terms. We will use the following theorem due to Eswarathasan and Levine, cf. [3]. Theorem 4.1. (See Theorem 5.2 in [3].) Let p be an odd prime. Then there is a sequence ck ∈ Qp such that, for all m 1, ∞
1 Hpm − Hm = ck p2k m2k , p k=1
where the series converges in p-adic norm. The ck are p-adic integers unless (p − 1)|2k or p|k. In general, νp (ck ) = −1 + νp (1/k) if (p − 1)|2k, and νp (ck ) = νp (1/k) otherwise. Based on the information conveyed in Theorem 4.1 and by adding the case p = 2, we derive Theorem 4.2. For any prime p, integer a 1 with (a, p) = 1, and n 2 we have
νp Hapn −
1 pn−1
Hap
⎧ ⎨ 0, = −n + 4 + 1, ⎩ 2,
if p = 2 if p = 3 if p 5.
Moreover, for a odd and n 1 we have ν2 (Ha2n ) = −(n − 1) + ν2 (H2a ) = −n − log2 a and ν2 (Ha2n −1 ) = −n − log2 a + χa=1 . For p = 3 and, under Conjecture 1.1, for any prime p 5, integer a with (a, p) = 1, and n 1 we have νp (Hapn ) = −(n − 1) + νp (Hap ) and νp (Hapn −1 ) = −(n − 1) + νp (Hap−1 ).
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Proof of Theorem 4.2. First we deal with the p-adic order of the differences and consider the case of p = 2. Note that 1 1 = 4d H4a − H2a = 2 2i − 1 i=1 2a
with ν2 (d) = 0. In fact, since a is odd then by paring up the terms of the above sum with indices i and 2a + 1 − i, 1 i a, we get a, i.e., an odd number of pairs with sums that, after taking common denominators, all have the numerator 4a and odd denominators. This results in a total sum with 2-adic order of 2. In general, for n 3, we have 1 Ha2n − Ha2n−1 = O 2n , 2
(4.8)
now using a different notation: the term O(pm ) stands for any x with νp (x) m and the pairing argument similar to the above case with n = 2 except that now there are a2n−2 pairs. We repeatedly use this fact below 1 Ha2n−1 − Ha2n−2 = O 2n−1 , 2 that is, 1 1 Ha2n − Ha2n−1 = Ha2n − 2 Ha2n−2 + O 2n−2 , 2 2 and finally, 1 1 O 2n = Ha2n − Ha2n−1 = Ha2n − n−2 H4a + O(4) 2 2 1 4d = Ha2n − n−1 H2a + n−2 + O 2−n+5 , 2 2 which already yields that ν2 (Ha2n − 1/2n−1 H2a ) = −n + 4. For odd primes we use Theorem 4.1 with m = apn−1 , n 2, and k = 1. Note that c1 = c1 (n) may depend on n. As above, we repeatedly use 1 Hapn − Hapn−1 = c1 a2 p2n + O p2n+1 , p then with m = apn−2 1 Hapn−1 − Hapn−2 = c1 a2 p2n−2 + O p2n−1 , p that is,
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Hapn −
89
1 Hapn−2 = c1 a2 p2n−3 + O p2n−2 , 2 p
etc. We conclude that Hapn −
1 pn−1
Hap = c1 a2 p−n+6 + O p−n+7 ,
(4.9)
since in the exponent in the right hand side we have 2(n − (n − 2)) − (n − 2) = −n + 6. Note that ν3 (c1 ) = −1. The result on the p-adic order of the differences follows by (4.9). For the other parts, we observe that the case with n = 1 vacuously holds. We also note that for any integer m, 0 m n: Hapn −1 −
1 1 H n−m −1 = Hapn − m Hapn−m . pm ap p
(4.10)
We use this identity with m = n − 1. The case with p = 2 follows by ν2 (1/2n−1 H2a ) = −(n − 1) − log2 2a < −n + 1. For odd primes, we conclude the proof by using (4.9) and Remark 1.1 if p = 3, and (4.9) and Conjecture 1.1 otherwise. 2 Proof of Theorem 2.2. Identity (4.7) combined with Theorem 4.2 yields the result.
2
Now we turn to the case of odd primes. We are ready to present the Proof of Theorem 2.3. Remark 1.1, identities (4.7) and ν3 (Ha3n −1 ) = −(n − 1) + ν3 (H3a−1 ) = −n + ν3 (Ha−1 ) yield the result. The second equation ν3 (H3a−1 ) = −1 +ν3 (Ha−1 ) follows from the special case of Theorem 4.1 with p = 3 and m = a, and ν3 (Ha−1 ) 1. To prove the inequalities, we simply apply Theorem 2 in [9] on calculating (m) ν3 (Hn ). 2 Proof of Theorem 2.4. By identity (4.7) and Theorem 4.2, we get exact p-adic orders, while clearly νp (Hap−1 ) −logp (ap−1) and νp (Hp−1 ) 2 for p 5 by Wolstenholme’s theorem, cf. [3]. If a = p2 − p + 1 then the inequality is strict since 1 + νp (Hap−1 ) 1 + min{2, −1 + νp (Ha−1 )} νp (Ha−1 ) 1 by Theorem 4.1 and a − 1 ∈ Jp (cf. [3]), while −logp a = −1. 2 Proof of Theorem 2.5. We apply identity (4.5), and focus on the term E (apn − 1, 2). m−1 After rearranging its own terms, we can rewrite E (m, 2) as i=1 Hi /(i + 1) if m 2. apn −2 If a > 1 then we use the inequality νp ( i=1 Hi /(i + 1)) −2n − 2logp a, which follows by observing that the term with index i = a pn − 1 where a is the largest power of p not exceeding apn −2 implies that νp (Hi /(i +1)) −2n −2νp (a ). The result follows.
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From now on we assume that a = 1. The case with n = 2 is trivial so we assume that n 3. First we consider the case of p = 2. We prove that ν2 s 2n , 3 = 2n − 1 − n + ν2 E 2n − 1, 2 = 2n − 1 − n − (2n − 4) = 2n − 3n + 3 2n −2 since in the sum i=1 Hi /(i + 1), as we will see, the terms with i = 2n−1 − 1 and i = 3 · 2n−2 − 1 both contribute −2n + 3 to the 2-adic order for a total of −2n + 4, and clearly, the terms with i = 2n−1 − 1 + 2n−3 and i = 3 · 2n−2 − 1 + 2n−3 both contribute −2n + 4, and thus, at least a combined −2n + 5 to the order while all the other terms n−1 have 2-adic orders of at least −2n + 5. In fact, H2n−1 −1 /22 + H3·2n−2 −1 /(3 · 2n−2 ) can be rewritten as 1 1 (3H2n−1 −1 + 2H3·2n−2 −1 ) = (3H3 + 2H5 ) + O 2−2n+7 , 3 · 2n−1 3 · 22n−4 since by (4.10) and similarly to (4.8) we get that 1 H n−2 mod 2n−1 , 2 2 −1 1 ≡ H2n−3 −1 mod 2n−2 , and thus, 2 1 ≡ 2 H2n−3 −1 mod 2n−3 , 2
H2n−1 −1 ≡ H2n−2 −1 H2n−1 −1 etc.,
H2n−1 −1 =
1 2n−3
H22 −1 + O 2−n+7 .
In a similar fashion, we obtain H3·2n−2 −1 ≡
1 H n−3 mod 2n−2 , 2 3·2 −1
etc., H3·2n−2 −1 =
1 H3·2−1 + O 2−n+6 . 2n−3
Putting the relevant terms together, with n 3 we obtain that 3H2n−1 −1 + 2H3·2n−2 −1 =
1 2n−3
(3H3 + 2H5 ) + O 2−n+6 ,
and hence ν2 (E (2n − 1, 2)) = −(n − 1) − (n − 3) = −2n + 4. Now we consider the case with p 3 prime. Identity (4.5) implies that pn − 1 pn − 1 − n + νp E pn − 1, 2 = − 3n + 2 + νp E (p − 1, 2) νp s pn , 3 = p−1 p−1
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since clearly, E({1/pn−1 , 1/(2pn−1 ), . . . , 1/((p−1)pn−1 )}, 2) = 1/p2(n−1) E({1/1, 1/2, . . . , 1/(p − 1)}, 2) determines the p-adic order of the term E (pn − 1, 2) for all n 1. Clearly, ν3 (E (2, 2)) = 0. If p 5 then we can further analyze the last term since E (p − 1, 2) = (2) 2 )/2 by (4.5), hence, its p-adic order is 1 by Corollaries 1.1 and 1.2 if −(Hp−1 − Hp−1 νp (Bp−3 ) 0, i.e., except for the Wolstenholme primes. Therefore, we get that pn − 1 − 3n + 3. νp s pn , 3 = p−1
2
5. More proofs Proof of Theorem 3.2. We use identity (3.1) after substituting apn , b, and k into a, b, and n, respectively. We set
A=
apn + b − 1 apn apn
and derive that k s(apn , apn − l)s(b, b − (k − l)) s apn + b, apn + b − k = A . apn +b−k−1 (apn − l) apn −l l=0
Clearly, for the common factor A we have νp (A) = n. If l = 0 then the p-adic order of the actual term in the right hand side sum, with the factor A included, is νp (s(b, b − k)) as long as 2 k + 1 b < pn with any sufficiently large n. If l 1 then we apply Theorem 3.3, which yields that the p-adic order of the corresponding term is at least n + (n − logp (2k) ) − (n − logp (b − 1) + logp k) > νp (s(b, b − k)) if n is sufficiently large. Therefore, the term with l = 0 determines the p-adic order of the sum. n | We note that if l 1 odd then [8, Theorem 2.2] already implies that ap2 n n s(ap , ap − l), i.e., the corresponding p-adic order is at least n + n − 1 − (n − logp (b − 1)) − logp k n − 1, independently of the actual value of the odd l. 2 Proof of Theorem 3.3. Identity (3.3) already guarantees that νp (s(apn , apn − k)) apn n − logp (2k). 2 n − logp (2k) for 2k < pn since clearly, νp 2k−j Proof of Theorem 3.4. The case with k = 1 is obvious since s(n, n − 1) = n2 and hence, νp (s(apn , apn − 1)) = n − χp=2 if n 1. If k 2 then we prove that νp (s(apn , apn − k)) < n by induction on n n1 . Identity (3.6) will follow. Clearly, the inequality is true in the base case n = n1 by condition (3.5). We assume that it holds for all n : n1 n n and prove that it is also true for n = n + 1.
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We apply (3.2) which has the advantage over (3.1) that the term with index 0 makes no contribution on the right hand side. In (3.2), we substitute apn , a(p − 1)pn , and k for a, b, and n, respectively. We set n+1 apn+1 − 1 n ap ap apn k apn
B= For the common B factor we have
n+1 −1 ap + n + 1 − νp (k). νp (B) = νp apn In fact,
n+1
−1 ap ap − 1 = νp νp apn a although it is irrelevant to what follows and
n+1 apn+1 − 1 −1 apn+1 − k − 1 apn+1 − k apn+1 − k + 1 ap ··· = apn − k + 1 apn − k + 2 apn apn − k apn
thus, νp
n+1
n+1 ap ap −k−1 −1 = n − ν , (k) + ν p p apn − k apn
(5.1)
since here
νp
apn+1 − 1 apn+1 − k apn+1 − k + 1 · · · apn − k + 1 apn − k + 2 apn
= νp
k! (k − 1)!
− n = νp (k) − n.
After some simple manipulations we obtain k s(apn , apn − l)s(a(p − 1)pn , a(p − 1)pn − (k − l)) s apn+1 , apn+1 − k = B l n+1 −k−2 (apn − k − 1) apapn −l−1 l=0 k s(apn , apn − l)s(a(p − 1)pn , a(p − 1)pn − (k − l)) =B l apn+1 −k−1 (apn − l) apn −l l=1
If l = k then the p-adic order of the actual term in the right hand side sum, includ n+1 ing the factor B, is equal to (νp ( ap apn−1 ) + n + 1 − νp (k)) + νp (s(apn , apn − k)) − n+1 ) = νp (s(apn , apn − k)) + 1 by (5.1), and this term is the p-adically domiνp ( ap apn−k−1 −k nating one. Indeed, if 0 < l < k then the p-adic order of the corresponding term is at least n+1 n+1 (νp ( ap apn−1 ) +n +1 −νp (k)) +(n −logp (2l) ) +(n −logp (2(k−l)) ) −νp ( ap apn−k−1 ) −l
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3n + 1 − νp (k) − 2 logp k − (n + logp a) > n + 1 > νp (s(apn , apn − k)) + 1 by Theorem 3.3 and the induction hypothesis. The statements in (3.6) and (3.7) follow. 2 Proof of Theorem 3.5. The next to last identity in [8, p. 31] claims that with h 1, i = 0, 1, . . . , h, and t = 1, 2, . . . , p − 1, s hp + p − 1, h + t + i(p − 1) ≡
h (−1)h−i i
mod p
(5.2)
holds true for the unsigned (or signless) Stirling numbers of the first kind |s(hp + p − 1, h + t + i(p − 1))|. For any odd prime p, we set h = (n − (p − 1))/p, i = h + 1 − (k + 1)/(p − 1) , and t = (h − i + 1)(p − 1) − k in (5.2). Clearly, 0 i h and 1 t p − 1. We obtain that s(n, n −k) ≡ hi (−1)h−i (−1)k ≡ (−1)k (mod p) since n ≡ −1 (mod pk ) yields that, with some positive integer c: n = cpk − 1, h = cpk−1 − 1, and i = cpk−1 − (k + 1)/(p − 1) . k−1 Hence, by Lemma 8 in [10] we have cp j ≡ (−1)j (mod p) for 0 j < pk−1 , which h h h−i h−i implies that i (−1) = h−i (−1) ≡ 1 (mod p) since h − i < pk−1 by 2k < n. If p = 2 then we first prove that, with n ≡ −1 (mod 2k+1 ), s(n, n − k) is odd and then the relation s(N, K) = s(N − 1, K − 1) − (N − 1)s(N − 1, K), with N = M 2k+1 − 1, M 1, and K = M 2k+1 − 1 − k, yields the other part of the statement since N − 1 is even and s(N, K) is odd by the first part; thus, s(M 2k+1 − 2, M 2k+1 − 2 − k) is also odd. To prove the first part, we set h = (r + 1)2k−1 − 1 with some positive odd integer r, i = (r + 1)2k−1 − 1 − k, t = 1, and n = hp + p − 1 = (r + 1)2k − 1 ≡ −1 (mod 2k+1 ) in (5.2), and the result follows. 2 Acknowledgments The author wishes to thank Gregory P. Tollisen for his helpful comments. References [1] T. Agoh, K. Dilcher, Convolution identities for Stirling numbers of the first kind, Integers 10 (2010) 101–109. [2] D. Barsky, J.-P. Bezivin, p-Adic properties of Lengyel’s numbers, J. Integer Seq. 17 (2014), Article 14.7.3. [3] D.W. Boyd, A p-adic study of the partial sums of the harmonic series, Experiment. Math. 3 (1994) 287–302. [4] H. Cohen, Number Theory, vol 2: Analytic and Modern Tools, Graduate Texts in Mathematics, Springer, 2007. [5] L. Comtet, Advanced Combinatorics. The Art of Finite and Infinite Expansions, D. Reidel, Publishing Co., Dordrecht, 1974, enlarged edition. [6] R.L. Graham, D.E. Knuth, O. Patashnik, Concrete Mathematics, Addison–Wesley, Reading, MA, 1994. [7] A. Granville, Arithmetic properties of binomial coefficients. I. Binomial coefficients modulo prime powers, in: Organic Mathematics, Burnaby, BC, 1995, in: CMS Conference Proceedings, vol. 20, Amer. Math. Soc., Providence, RI, 1997, pp. 253–276, electronic version (a dynamic survey): http://www.dms.umontreal.ca/~andrew/Binomial/.
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[8] F.T. Howard, Congruences for the Stirling numbers and associated Stirling numbers, Acta Arith. LV (1990) 29–41. [9] K. Kamano, On the 3-adic valuations of generalized harmonic numbers, Integers 11 (2011) 1–8. [10] T. Lengyel, On the least significant 2-adic and ternary digits of certain Stirling numbers, Integers 13 (2013) 1–10. [11] J. Sándor, B. Crstici, Handbook of Number Theory, vol. II, Kluwer, 2004. [12] N.J.A. Sloane, The On-Line Encyclopedia of Integer Sequences, published electronically at http://oeis.org, 2013. [13] L.C. Washington, p-Adic L-functions and sums of powers, J. Number Theory 69 (1998) 50–61.
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
On p-adic properties of the Stirling numbers of the first kind Tamás Lengyel Mathematics Department, Occidental College, 1600 Campus Road, Los Angeles, CA 90041, USA
a r t i c l e
i n f o
Article history: Received 12 June 2014 Received in revised form 19 September 2014 Accepted 19 September 2014 Available online 29 October 2014 Communicated by David Goss Keywords: Stirling numbers of the first kind p-Adic properties Congruence relations modulo prime powers Convolution identities for Stirling numbers Generalized harmonic numbers
a b s t r a c t The goal of this paper is to describe s(n, k) mod pe and calculate νp (s(n, k)) for a prime p, fixed integer k 1, and large enough e and n. Some special cases of the form s(apn , k) mod pe and its relation to s(apn+1 , kp) mod pe as well as bounds on νp (s(apn , k)) and its exact values in some special cases are completely determined. We also investigate the properties of νp (s(n, n − k)) and νp (s(apn + b, apn + b − k)) which are significantly different from those of νp (s(n, k)) and νp (s(apn , k)). We use congruential identities for the generalized harmonic numbers, and new congruential and convolution identities for s(n, k). © 2014 Elsevier Inc. All rights reserved.
1. Introduction We note that, unlike for the Stirling number of the second kind, S(n, k), there is no easy way to use an explicit formula for s(n, k) and this fact makes it more difficult to deal with certain characteristics of these numbers from the point of view of congruential and divisibility properties. In fact, the Stirling numbers of the first kind exhibit very different characteristics from the Stirling numbers of the the second kind. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jnt.2014.09.015 0022-314X/© 2014 Elsevier Inc. All rights reserved.
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We will need some basic notation. Let n and k be positive integers, p be a prime, dp (k) and νp (k) denote the sum of digits in the base p representation of k and the highest power of p dividing k, respectively. The latter one is often referred to as the p-adic order of k. For the rational n/k we set νp (n/k) = νp (n) − νp (k). We also use the standard notation for the generalized harmonic numbers:
Hn(m) =
n 1 , im i=1
(1)
with Hn being the harmonic number Hn . As usual, Bn stands for the nth Bernoulli number and νp (B2m ) −1 for m 1 by the von Staudt–Clausen theorem, which implies that the denominator of B2m is squarefree. Let χA denote the indicator variable of the event A, i.e., it is 1 exactly if A occurs. For every fixed integer k 1, we prove that, unlike for Stirling numbers of the second kind, the p-adic order of s(n, k) becomes arbitrarily large for large values of n. We establish this fact in Theorems 1.1 and 1.2 which state the divergence and the asymptotic order of magnitude of νp (s(n, k)) for any fixed k 1. Moreover, only the rate of convergence of the ratio of νp (s(n, k)) to its asymptotic order of magnitude is affected by the choice of k. In Section 2 we study the cases with k = 2 and 3, and determine the exact order νp (s(apn , k)). The ultimate goal is to describe s(n, k) (mod pe ) and then νp (s(n, k)) for large enough e and n. The main results in this direction are summarized in Theorems 1.1–1.3, 1.5, 2.1–2.5, and 4.2. Section 3 deals with νp (s(n, n − k)) for a fixed k 1. The p-adic order of s(n, n − k) does not become arbitrarily large for large values of n except for properly selected subsequences, e.g., if n is of the form apn + b with b = 0 but remains constant if b k + 1 or b = −1 as stated in Theorems 3.2–3.5. Theorems 1.3 and 3.1 present new congruential and convolution identities for s(n, k), respectively. We give a new p-adic relation for the harmonic numbers in Theorem 4.2. Sections 4 and 5 are devoted to the proofs. From now on we rely on two conjectures. Conjecture 1.1. (See [3].) For any odd prime p 5 and integer n 1, we have that νp (Hn ) 3. Note that ν2 (Hn ) = −log2 n 0, which can be proven in a similar fashion to the proof that Hn −Hm with 0 < m < n, is never an integer. Following [3], for every prime p, we define Jp = {n|νp (Hn ) 1}. Remark 1.1. Using J3 , Eswarathasan and Levine (cf. [3]) observed that ν3 (n) 1 for all n 1. In fact, we have that |J3 | = 3 and ν3 (Hn ) = 1 for all three ns in J3 = {2, 7, 22}.
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The above conjecture is extended in [13] for generalized harmonic numbers. Washing(m) (m) (m) ton defines Jp = {n|p divides the numerator of Hn }, i.e, Jp is the set of integers n (m) (1) for which νp (Hn ) 1. Note that Jp = Jp . With some heuristic argument, he states the following (2)
Conjecture 1.2. (See [13].) For m = 2, there is no n > p2 such that νp (Hn ) 1. For (3) m = 3, there is no n > p2 such that νp (Hn ) 1 except possibly for finitely many (m) exceptional primes p. For m 4, there is no n > p such that νp (Hn ) 1. We also use the following two corollaries due to Glaisher. Corollary 1.1. (See Corollary 1, [13].) Let r 1 be odd and suppose p r + 4. Then p−1 r(r + 1) 1 Bp−r−2 p2 ≡− r j 2(r + 2) j=1
mod p3 .
Corollary 1.2. (See Corollary 2, [13].) Let r 2 be even and suppose p r + 3. Then p−1 r 1 Bp−r−1 p ≡ r j r + 1 j=1
mod p2 .
We prove the divergence of νp (s(n, k)) as n → ∞ for all k 1 and establish facts about its asymptotic order of magnitude in the next two theorems. Theorem 1.1. For any prime p and integer k 1, we get that lim νp s(n, k) = ∞.
n→∞
Remark 1.2. By (167) of [11, p. 491], we have that s(n, k) ≡ 0 (mod pr ) for n > krp which also guarantees that limn→∞ νp (s(n, k)) = ∞. In fact, for n > N it yields that νp s(n, k)
N . pk
We improve this inequality in Theorem 1.2. Theorem 1.2. For any prime p and integer k 1, there exists a constant c = c (k, p) > 0 so that for n n0 (k, p) we get νp s(n, k) c n. In fact, we can set c (k, p) to any value less than 1/(1 −p), and we also have the inequality
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n − dp (n) n − νp (n) − (k − 1) logp n − (k + 1) logp n νp s(n, k) p−1 p−1 for n 1. Under Conjectures 1.1 and 1.2, we get that 1 νp (s(n, k)) = . n→∞ n p−1 lim
In this case, for the rate of convergence we obtain that νp (s(n, k)) logp n 1 (k + 1) − n p − 1 n
(1.1)
for any sufficiently large n. By Theorem 1.1 we know that lim νp s apn , k = ∞
n→∞
for any prime p, integers a, n, and k, such that (a, p) = 1. Now we determine s(apn , k) (mod pe ) in (1.4), for any integer e 1, and νp (s(apn , k)) if n is large enough. Motivated by [2], we first determine (x)m = x(x − 1) · · · (x − m + 1) (mod pe ) with m = apn . We note that in [2] only some special cases were considered, e.g., the case with p = 2, e = 2, m = 2n and m = 3 · 2n (with n 5), and the case with p 3, e = 1, and m = pn (with n 2). Theorem 1.3. For any prime p, positive integers a, n, k, and e such that (a, p) = 1, 0 k apn , 3 2e − 1 n, with the notation (x)n = x(x − 1) · · · (x − n + 1),
(1.2) n
for the ordinary generating function (ogf) of the sequence {s(apn , k)}ap k=0 we get that n
(x)apn
ap apn−e n−1 e−1 = s apn , k xk ≡ xap −1 x(p−1)p
mod pe .
(1.3)
k=0
Thus, with the setting g(a, p, n, e, k) =
k − apn−1 , (p − 1)pe−1
if
k − apn−1 is an integer (p − 1)pe−1
and otherwise undefined, and b(a, p, n, e, k) =
n−e
(−1)ap 0,
−g(a,p,n,e,k)
,
apn−e g(a,p,n,e,k)
if
k−apn−1 (p−1)pe−1
otherwise,
is an integer,
(1.4)
T. Lengyel / Journal of Number Theory 148 (2015) 73–94
77
we get that s apn , k ≡ b(a, p, n, e, k)
mod pe .
It follows that νp s apn , k =
⎧ apn−e ⎪ , ν p ⎪ g(a,p,n,e,k) ⎨ ⎪ ⎪ ⎩
if νp
e,
k−apn−1 (p−1)pe−1
is an integer and < e,
apn−e g(a,p,n,e,k)
(1.5)
otherwise.
Remark 1.3. Clearly, if νp (s(apn , k)) is at least as large as e, e.g., as large as (n + 1)/2 then (1.3) and its consequences are of no use. This is the case in Theorem 3.3 as well as Theorem 3.4 and Conjecture 3.1 with regard to νp (s(apn , apn − k)). n−e
n−1
n
Remark 1.4. We note that expanding (1.3) results in (−1)ap xap + · · · + xap e n e (mod p ). In order to have s(ap , k) ≡ 0 (mod p ) we need to consider only values of the form k = apn−1 +j(p−1)pe−1 with 0 j apn−e and thus, pe−1 | k is a must since n > e. Note that here j is playing the role of g(a, p, n, e, k). Somewhat counterintuitively, for any n−e fixed e, we do not get more j values such that νp ap j < e as n increases, as we will see in Theorem 1.5. In special cases of Theorem 1.3, part of (1.5) can be easily calculated, k−pn−1 n e.g., if a = 1 and (p−1)p e−1 is an integer then νp (s(p , k)) = n − e − νp (g(1, p, n, e, k)) provided that it is less than e except if g(1, p, n, e, k) = 0 when νp (s(pn , k)) = 0. This possibility is further explored in Example 1.1. The following theorem gives a generalization of the Jacobstahl–Kazandzidis congruences, cf. [7]. Theorem 1.4. (See Corollary 11.6.22, [4].) Let M and N be such that 0 M N and p prime. We have
pN pM
⎧ N N Bp−3 3 4 ⎪ p N M (N − M )) mod p N M (N − M ) , (1 − ⎪ M 3 ⎨ N M N 4 ≡ (1 + 45N M (N − M )) M mod p N M (N − M ) M , ⎪ N ⎪ ⎩ (−1)M (N −M ) P (N, M ) N mod p4 N M (N − M ) M , M
if p 5, if p = 3, if p = 2,
where P (N, M ) = 1 +6N M (N −M ) −4N M (N −M )(N 2 −N M +M 2 ) +2(N M (N −M ))2 . Theorem 1.3, in combination with Theorem 1.4, helps to determine the exact value of s(apn , k) (mod pe ) when n is sufficiently large. In this case we relate s(apn+1 , kp) (mod pe ) to s(apn , k) (mod pe ). Theorem 1.5. For any prime p, integers a, n, k, and e such that (a, p) = 1, 0 k apn , 3 2e − 1 n, we get that
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(x)apn+1 =
n+1 ap
n
ap s apn+1 , k xk ≡ s apn , k xkp
k=0
mod pe ,
(1.6)
k=0
i.e., s apn+1 , kp ≡ s apn , k
mod pe .
In other words, (x)apn+1 (mod pe ) can be obtained by simply replacing x with xp on the apn right hand side of (x)apn ≡ k=0 s(apn , k)xk (mod pe ). Example 1.1. We consider the case with (a, p, n, e) = (1, 3, 5, 3). The calculations agree with (1.6) of Theorem 1.5: (x)729 ≡
9
s(243, 81 + 18j)x243+54j
j=0
≡ s(243, 81)x243 + s(243, 99)x297 + · · · + s(243, 243)x729 ≡ 26x243 + 9x297 + 18x351 + 3x405 + 9x459 + 18x513 + 24x567 + 9x621 + 18x675 + x729
mod 27.
Note that in the general term s(3n , k)x3k we have k = 3n−1 + 2j3e−1 with 0 j 3n−3 . In our special case, it means k = 81 + 18j with 0 j 9, and j 1 implies that ν3 (s(3n , 3n−1 +2j3e−1 )) = n −e −ν3 (j) < e since ν3 (s(243, 81 +18j)) = 5 −3 −ν3 (j) < 3, in agreement with Remark 1.4. Therefore, the 3-adic orders of the ten consecutive nonzero coefficients of the actual (1.6) are 0, 2, 2, 1, 2, 2, 1, 2, 2, 0. 2. Special cases Theorem 2.1. For any prime p and integer n 1, we have that s(n, 1) = (−1)n−1 (n − 1)!, thus n − dp (n) n n > νp s(n, 1) = − νp (n) − 2 logp n, p−1 p−1 p−1 and s(n, 2) = (−1)n−2 (n − 1)!Hn−1 , hence νp s(n, 2) = νp s(n, 1) + νp (Hn−1 )
n − 3 logp n, p−1
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and under Conjecture 1.1, we get that n + 3 νp s(n, 2) . p−1 Theorem 2.2. For a 1 odd and n 1 we have that ν2 s a2n , 2 = a2n − 2n − d2 (a) − log2 a + χa=1 . Theorem 2.3. For a 1, (a, 3) = 1, and n 1 we have that 1 ν3 s a3n , 2 = a3n − d3 (a) − 2n + ν3 (H3a−1 ) + 1, 2 which is 1 n a3 − d3 (a) − 2n + ν3 (Ha−1 ), 2 for a > 1. In particular, if a = 1 then 1 ν3 s 3n , 2 = 3n + 3 − 2n. 2 We also have the inequalities 1 1 n a3 − d3 (a) − 2n − log3 a ν3 s a3n , 2 a3n − d3 (a) − 2n − log3 a + 2. 2 2 In general, for ν3 (s(n + 1, 2)) we ge that 1 1 n − d3 (n) − log3 n ν3 s(n + 1, 2) n − d3 (n) − log3 n + 3. 2 2 Theorem 2.4. Under Conjecture 1.1, for any prime p 5, a 1, (a, p) = 1, and n 1, we have that νp s apn , 2 =
1 n ap − dp (a) − 2n + 1 + νp (Hap−1 ). p−1
The inequalities νp s apn , 2
1 n ap − dp (a) − 2n + p−1
−logp a, 3,
if a > 1, if a = 1
give lower bounds on the p-adic orders that can be easily calculated. The inequality is strict when a = p2 − p + 1.
80
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Remark 2.1. We add that the exact values of ν3 (s(a3n , 2)) and ν3 (s(n + 1, 2)) can be found by Theorem 2 of [9]. We also note that, for p 5, νp (Hp−1 ) 2 by Wolstenholme’s theorem and νp (Hp−1 ) = 2 except for the so-called Wolstenholme primes for which νp (Hp−1 ) 3, e.g., p = 16843 and p = 2124679 with νp (Hp−1 ) = 3, and no further examples of such primes are known. Another important conjecture is that νp (Hn ) 4 never occurs, cf. Conjecture 1.1. We define E(i, j) as the elementary symmetric polynomial of degree j in the integers [i] = {1, 2, . . . , i}. In a similar fashion, we define E (i, j) as the elementary symmetric polynomial of degree j in the reciprocals of integers [i] = {1, 2, . . . , i}. For instance, the nth harmonic number Hn is equal to E (n, 1) and E (n, 2) = 1i1
E(S, k) =
i1 i2 · · · ik .
i1
By convention, we set E(i, 0) = E (i, 0) = E(S, 0) = 1. Theorem 2.5. For n 2 we have that ν2 s 2n , 3 = 2n − 3n + 3. For p 3 prime and n 1, we get that pn − 1 − 3n + 2 + νp E (p − 1, 2) , νp s pn , 3 = p−1 −1 which is equal to 12 (3n − 1) − 3n + 2 for p = 3 and pp−1 − 3n + 3 for p 5 provided that νp (Bp−3 ) 0, i.e., if p is not a Wolstenholme prime. In general, for a 1 we have the inequality n
apn − dp (a) − 3n − 2logp a. νp s apn , 3 p−1 Remark 2.2. We note that we found only the sequences {ν2 (s(3·2n , 2))}n1 as http://oeis. org/A050488, {ν2 (s(5 · 2n , 2))}n1 as http://oeis.org/A097809, and {ν2 (s(3 · 2n , 3))}n1 as http://oeis.org/A123203 in [12]. 3. The other end of the range In this section we investigate the case of s(n, n − k) with small values of k. We will use a convolution type identity for Stirling numbers of the first kind that can be derived from (6.46) in [6, p. 272], also cf. (7) in [1]. For any positive integers a, b, and n such that a, b n + 1, we have that
T. Lengyel / Journal of Number Theory 148 (2015) 73–94
s(a, a − k)s(b, b − (n − k)) s(a + b, a + b − n) a+b−2 = . (a + b − 1) a−1 (a + b − n − 1) a+b−n−2 a−k−1 k=0
81
n
(3.1)
We will also use another similar convolution identity (3.2) which seems to be new. It can be derived from (6.47) in [6, p. 272] in a similar manner to (3.1). We leave the details to the reader. Theorem 3.1. For any positive integers a, b, and n such that a, b n + 1, we have that n a + b s(a, a − k)s(b, b − (n − k)) s(a + b, a + b − n) k a+b−2 = . an (a + b − 1) a−1 (a + b − n − 1) a+b−n−2 a−k−1 k=0
(3.2)
We also use the identity from [8, p. 32] s(n, n − k) =
k j=0
d(2k − j, k − j)
n 2k − j
(3.3)
where d(n, k) is the so-called associated Stirling number of the first kind, cf. [5, pp. 256–257], and it is a nonnegative integer. For example, s(n, n − 1) = d(2, 1) n2 + n n d(1, 0) 1 = 2 . We have the following main results for νp (s(apn + b, apn + b − k)) when b 0 in Theorem 3.2, b = 0 in Theorem 3.3 and (the conditional) Theorem 3.4, and b = −1 and −2 in Theorem 3.5. Theorem 3.2. For any prime p, integers a 1 with (a, p) = 1, k so that 2 k + 1 b, and n sufficiently large, we get that νp s apn + b, apn + b − k = νp s(b, b − k) . We note that if b k + 1 then νp (s(apn + b, apn + b − k)) does not depend on n for sufficiently large values of n. This behavior is very different from the one we observed in Theorem 1.2 and its special cases as well as in the case of b = 0. In the latter case we derive a lower bound on the p-adic order in Theorem 3.3. Theorem 3.3. For any prime p, integers a 1 with (a, p) = 1, k 1, and sufficiently large n n0 (p, k) = logp (2k) , we get that νp s apn , apn − k n − logp (2k)
(3.4)
independently of a. For k = 1 the bound is tight when p = 2 and for odd primes the difference between the exact order and the lower bound appears to be constant if k = 1 or k 2 even, according to (the conditional) Theorem 3.4. For odd ks we state Conjecture 3.1.
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Theorem 3.4. For any prime p, integers a 1 with (a, p) = 1, and even k 2 with the condition ∃n1 : n1 > 3 logp k + logp a
such that νp s apn1 , apn1 − k < n1
(3.5)
or k = 1 with n1 = 1 then for n n1 we have that νp s apn+1 , apn+1 − k = νp s apn , apn − k + 1,
(3.6)
νp s apn+1 , apn+1 − k = νp s apn1 , apn1 − k + (n − n1 ).
(3.7)
and hence,
We note that it is easy to check (3.5) for small values of k, for instance, it holds for k = 1 and 2 but not for 3. In fact, we know by [8, Theorem 2.2] that νp (s(apn , apn −k)) n if k and p are both odd. Numerical calculations lead us to believe that (3.5) holds for all even k 2, otherwise we have the following conjecture. Conjecture 3.1. For any prime p, integer a 1 with (a, p) = 1, k 3 odd, and n n1 = n1 (p, k) with some sufficiently large n1 (p, k), we get that νp s apn+1 , apn+1 − k = νp s apn , apn − k + 2,
(3.8)
and hence, νp s apn+1 , apn+1 − k = νp s apn1 , apn1 − k + 2(n − n1 ). We saw in Theorem 3.3 that if b = 0 then limn→∞ νp (s(apn + b, apn + b − k)) = ∞, however, it is still different from the behavior of νp (s(n, k)) since it is not true that limn→∞ νp (s(n, n − k)) = ∞, e.g., we have already seen the case of b k + 1 when limn→∞ νp (s(apn + b, apn + b − k)) = νp (s(b, b − k)) constant. Additional evidence with b = −1 and −2 is provided by Theorem 3.5. For any odd prime p, n > 2k, and n ≡ −1 (mod pk ), we have that s(n, n − k) ≡ 0 (mod p). For p = 2 and n ≡ −1 or −2 (mod 2k+1 ), k 1, we have that s(n, n − k) is odd. 4. Proofs Proof of Theorem 1.1. We proceed with induction on k. Assume that for any prime p and constant B > 0 there exists an n0 (p, j, B) so that for all n n0 (p, j, B) we have νp (s(n, j)) B, and the above statement is true for j = 1, 2, . . . , k − 1. Clearly, the cases with k = 1 and 2 are established by Theorem 2.1.
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We now prove that it also holds for j = k by relying on identity (6) of [1] with m = k + 1:
n−r n m s(n, m) = s(n − i, r)s(i, m − r) r i i=m−r
(4.1) m
for any r : 0 r m. We set m = k + 1 and B = B + νp ( r, 0 < r < m = k + 1. Let
r
) and choose an
n0 (p, k + 1, B) = n0 p, k + 1 − r, B + n0 p, r, B . Now if n n0 (p, k + 1, B) then either n − i n0 (p, r, B ) or i n0 (p, k + 1 − r, B ) or both. In either case we get that on the right side of (4.1) either the factor s(n − i, r) or s(i, k + 1 − r) or both have a p-adic order at least B + νp ( k+1 r ). It follows that νp (s(n, k + 1)) B for all n n0 (p, k + 1, B). 2 Now we determine the asymptotic order of magnitude of νp (s(n, k)). Proof of Theorem 1.2. The ogf (1.2) implies that
s(n, k) = (−1)n−k E(n − 1, n − k) = (−1)n−k
i1 i2 · · · in−k
1i1
= (−1)n−k (n − 1)!
1i1
1 i1 i2 · · · ik−1
with im s being integers. In general, we have that s(n, k) = (−1)n−k (n − 1)!E (n − 1, k − 1).
(4.2)
First we check the cases with k = 1 and 2. We obtain that s(n, 1) = (−1)n−1 (n − 1)! and s(n, 2) = (−1)n (n − 1)!Hn−1 .
(4.3)
For larger values of k, we use the Newton–Girard formulas (−1)k−1 kEk =
k (−1)k−i Si Ek−i
(4.4)
i=1
with Ek = E (n − 1, k) and Sk = Hn−1 . In a similar fashion to the derivation of (4.3), via the application of (4.4) we derive that (k)
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(−1)n (n − 1)! (2) 2 Hn−1 − Hn−1 . 2
s(n, 3) = (−1)n−3 (n − 1)!E (n − 1, 2) =
(4.5)
and s(n, 4) = (−1)n−4 (n − 1)!E (n − 1, 3)
H3 (−1)n (n − 1)! 3 (2) (3) Hn−1 − Hn−1 Hn−1 + n−1 = 3 2 2 by setting k to 2 and 3, respectively. In general, we can express the elementary symmetric polynomials in terms of power sums Ek =
ci1 ,i2 ,...,im Si1 Si2 · · · Sim
1i 1 i2 ···im n−1 m j=1 ij =k;mk
with some rational constants ci1 ,i2 ,...,im by (4.4). It easily follows that νp (Si ) −i logp (n − 1) and hence, νp (Ek ) −k logp (n − 1) + Ok (1) with Ok (1) denoting a constant depending on k. Now n − dp (n) n − νp (n) + νp E (n − 1, k − 1) − (k + 1) logp n νp s(n, k) = p−1 p−1 follows for any large enough n by (4.2). By Conjectures 1.1 and 1.2, we have νp (Si ) 0 except for possibly finitely many exceptional primes p, and thus, except for the exceptional cases, we have that νp(Ek ) n/(p − 1) + Ok (1), and thus, νp (s(n, k)) n/(p − 1) for any sufficiently large n. 2 Example 4.1. We illustrate the changes in ν3 (s(n, 100)) (see Fig. 1). Proof of Theorem 1.3. For the first part of the statement we follow the proof from [2] and derive (1.3) as it was presented for the simpler cases with e = 1 and 2. For the second part we use (1.3). Now we prove (1.3). We proceed in the spirit of Lemmas 16 and 17 of [2]. Let r ∈ + 1 values of k, 0 k m − 1, {0, 1, . . . , pe − 1} and m = apn . There are Mr = m−r−1 pe e such that k is congruent to r modulo p . Clearly, Mr is independent of r and is equal to apn−e . By (1.2), this implies that
(x)apn ≡
e p −1
r=0
n−e
(x − r)ap
mod pe .
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Fig. 1. The functions ν3 (s(n, 100)) and ν3 (s(n, 100))/n with 1 n 5000.
We factor the last product into two factors and get that (x)apn ≡
n−e
(x − r)ap
p|r 0rpe −1
n−e
(x − r)ap
n−e
(x − ip)
=
n−e ap
j=0
≡
with 0 i pe−1 − 1 and
n−e apn−e (−ip)j xap −j j
e−1 n−e ap j=0
(4.6)
pr 1rpe −1
The first factor is the product of terms like (x − ip)ap apn−e
mod pe .
j
n−e
(−ip)j xap
−j
n−e
≡ xap
mod pe .
n−e The last congruence follows by the fact that νp ap j = n − e − νp (j) if 1 j pn−e , and 1 j e − 1 already guarantees e − 1 pn−e if n e + logp (e − 1) which is implied by the assumption that n 2e − 1 3. Then we use the fact that n − e − νp (j) + j e since n 2e − 1 2e − j + νp (j) as j − νp (j) = 1 if j = 1 or p = j = 2, and j − νp (j) j − logp j > 1, otherwise. Since we have pe−1 similar factors with different values of i, the total contribution of n−e e−1 n−1 the first factor of (4.6) is (xap )p = xap . e n−e For the second factor of (4.6), Euler’s theorem yields that it is (xφ(p ) − 1)ap ≡ e−1 n−e (x(p−1)p − 1)ap (mod pe ) with φ(m) denoting the Euler’s totient function. We conclude the proof of (1.3) by putting the two factors together. The other parts of the theorem follow easily by expanding the right hand side of (1.3) and extracting the coefficient of xk for different values of k that actually appear. 2 Proof of Theorem 1.5. We use Theorems 1.3 and 1.4. We note that if g(a, p, n, e, k) is an integer then
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g(a, p, n + 1, e, kp) =
kp − apn k − apn−1 = p = pg(a, p, n, e, k). (p − 1)pe−1 (p − 1)pe−1
We observe that in order for g(a, p, m, e, k) to be an integer, p must divide k if m e 2, otherwise the corresponding term s(apm , k) ≡ 0 (mod pe ). This fact comes in handy as we rewrite
(x)apn+1 =
n+1 ap
s apn+1 , k xk
k=0 n
ap = s apn+1 , kp xkp + k=0
≡
s apn+1 , k xk
pk 1kapn+1 −1
s apn+1 , kp xkp
mod pe .
g(a,p,n+1,e,kp) is an integer 1kapn
We will prove below that last sum is congruent to
s apn , k xkp
mod pe .
g(a,p,n,e,k) is an integer 1kapn
We have two cases. Case 1: p 3 prime. By the definition of function g, Remark 1.4 and Theorem 1.4 we get that if g(a, p, n, e, k) is an integer then g(a, p, n +1, e, kp) is also integer and vice versa, since e 2, and pe−1 | g(a, p, n + 1, e, kp) makes g(a, p, n + 1, e, kp)/p = g(a, p, n, e, k) an integer. Hence
apn+1−e p · apn−e apn−e = ≡ g(a, p, n + 1, e, kp) p · g(a, p, n, e, k) g(a, p, n, e, k)
mod pe
since in Theorem 1.4 we set N = apn−e and M = g(a, p, n, e, k) and observe that min{2 + n − e, 4 + n − e, e} = e if 2e − 1 n. Case 2: p = 2. In this case (1.4) has an extra factor (−1) if g(a, 2, n, e, k) is odd and that compensates for the factor (−1)M (N −M ) = −1 in Theorem 1.4. We also have min{1 + n − e, 4 + n − e, e} = e if 2e − 1 n. 2 Remark 4.1. It immediately follows that limn→∞ νp (s(apn , k)) = ∞ since, as n increases, we clean up the terms in (1.6) with xm , m < apn , for any fixed value e, cf. Example 1.1. Now we turn to the special cases. The identity (4.3) yields that n s apn , 2 = (−1)ap apn − 1 !Hapn −1 ,
(4.7)
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and thus, νp s apn , 2 =
apn − dp (a) − n + νp (Hapn −1 ), p−1
then we try to express the last term in more simple terms. We will use the following theorem due to Eswarathasan and Levine, cf. [3]. Theorem 4.1. (See Theorem 5.2 in [3].) Let p be an odd prime. Then there is a sequence ck ∈ Qp such that, for all m 1, ∞
1 Hpm − Hm = ck p2k m2k , p k=1
where the series converges in p-adic norm. The ck are p-adic integers unless (p − 1)|2k or p|k. In general, νp (ck ) = −1 + νp (1/k) if (p − 1)|2k, and νp (ck ) = νp (1/k) otherwise. Based on the information conveyed in Theorem 4.1 and by adding the case p = 2, we derive Theorem 4.2. For any prime p, integer a 1 with (a, p) = 1, and n 2 we have
νp Hapn −
1 pn−1
Hap
⎧ ⎨ 0, = −n + 4 + 1, ⎩ 2,
if p = 2 if p = 3 if p 5.
Moreover, for a odd and n 1 we have ν2 (Ha2n ) = −(n − 1) + ν2 (H2a ) = −n − log2 a and ν2 (Ha2n −1 ) = −n − log2 a + χa=1 . For p = 3 and, under Conjecture 1.1, for any prime p 5, integer a with (a, p) = 1, and n 1 we have νp (Hapn ) = −(n − 1) + νp (Hap ) and νp (Hapn −1 ) = −(n − 1) + νp (Hap−1 ).
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Proof of Theorem 4.2. First we deal with the p-adic order of the differences and consider the case of p = 2. Note that 1 1 = 4d H4a − H2a = 2 2i − 1 i=1 2a
with ν2 (d) = 0. In fact, since a is odd then by paring up the terms of the above sum with indices i and 2a + 1 − i, 1 i a, we get a, i.e., an odd number of pairs with sums that, after taking common denominators, all have the numerator 4a and odd denominators. This results in a total sum with 2-adic order of 2. In general, for n 3, we have 1 Ha2n − Ha2n−1 = O 2n , 2
(4.8)
now using a different notation: the term O(pm ) stands for any x with νp (x) m and the pairing argument similar to the above case with n = 2 except that now there are a2n−2 pairs. We repeatedly use this fact below 1 Ha2n−1 − Ha2n−2 = O 2n−1 , 2 that is, 1 1 Ha2n − Ha2n−1 = Ha2n − 2 Ha2n−2 + O 2n−2 , 2 2 and finally, 1 1 O 2n = Ha2n − Ha2n−1 = Ha2n − n−2 H4a + O(4) 2 2 1 4d = Ha2n − n−1 H2a + n−2 + O 2−n+5 , 2 2 which already yields that ν2 (Ha2n − 1/2n−1 H2a ) = −n + 4. For odd primes we use Theorem 4.1 with m = apn−1 , n 2, and k = 1. Note that c1 = c1 (n) may depend on n. As above, we repeatedly use 1 Hapn − Hapn−1 = c1 a2 p2n + O p2n+1 , p then with m = apn−2 1 Hapn−1 − Hapn−2 = c1 a2 p2n−2 + O p2n−1 , p that is,
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Hapn −
89
1 Hapn−2 = c1 a2 p2n−3 + O p2n−2 , 2 p
etc. We conclude that Hapn −
1 pn−1
Hap = c1 a2 p−n+6 + O p−n+7 ,
(4.9)
since in the exponent in the right hand side we have 2(n − (n − 2)) − (n − 2) = −n + 6. Note that ν3 (c1 ) = −1. The result on the p-adic order of the differences follows by (4.9). For the other parts, we observe that the case with n = 1 vacuously holds. We also note that for any integer m, 0 m n: Hapn −1 −
1 1 H n−m −1 = Hapn − m Hapn−m . pm ap p
(4.10)
We use this identity with m = n − 1. The case with p = 2 follows by ν2 (1/2n−1 H2a ) = −(n − 1) − log2 2a < −n + 1. For odd primes, we conclude the proof by using (4.9) and Remark 1.1 if p = 3, and (4.9) and Conjecture 1.1 otherwise. 2 Proof of Theorem 2.2. Identity (4.7) combined with Theorem 4.2 yields the result.
2
Now we turn to the case of odd primes. We are ready to present the Proof of Theorem 2.3. Remark 1.1, identities (4.7) and ν3 (Ha3n −1 ) = −(n − 1) + ν3 (H3a−1 ) = −n + ν3 (Ha−1 ) yield the result. The second equation ν3 (H3a−1 ) = −1 +ν3 (Ha−1 ) follows from the special case of Theorem 4.1 with p = 3 and m = a, and ν3 (Ha−1 ) 1. To prove the inequalities, we simply apply Theorem 2 in [9] on calculating (m) ν3 (Hn ). 2 Proof of Theorem 2.4. By identity (4.7) and Theorem 4.2, we get exact p-adic orders, while clearly νp (Hap−1 ) −logp (ap−1) and νp (Hp−1 ) 2 for p 5 by Wolstenholme’s theorem, cf. [3]. If a = p2 − p + 1 then the inequality is strict since 1 + νp (Hap−1 ) 1 + min{2, −1 + νp (Ha−1 )} νp (Ha−1 ) 1 by Theorem 4.1 and a − 1 ∈ Jp (cf. [3]), while −logp a = −1. 2 Proof of Theorem 2.5. We apply identity (4.5), and focus on the term E (apn − 1, 2). m−1 After rearranging its own terms, we can rewrite E (m, 2) as i=1 Hi /(i + 1) if m 2. apn −2 If a > 1 then we use the inequality νp ( i=1 Hi /(i + 1)) −2n − 2logp a, which follows by observing that the term with index i = a pn − 1 where a is the largest power of p not exceeding apn −2 implies that νp (Hi /(i +1)) −2n −2νp (a ). The result follows.
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From now on we assume that a = 1. The case with n = 2 is trivial so we assume that n 3. First we consider the case of p = 2. We prove that ν2 s 2n , 3 = 2n − 1 − n + ν2 E 2n − 1, 2 = 2n − 1 − n − (2n − 4) = 2n − 3n + 3 2n −2 since in the sum i=1 Hi /(i + 1), as we will see, the terms with i = 2n−1 − 1 and i = 3 · 2n−2 − 1 both contribute −2n + 3 to the 2-adic order for a total of −2n + 4, and clearly, the terms with i = 2n−1 − 1 + 2n−3 and i = 3 · 2n−2 − 1 + 2n−3 both contribute −2n + 4, and thus, at least a combined −2n + 5 to the order while all the other terms n−1 have 2-adic orders of at least −2n + 5. In fact, H2n−1 −1 /22 + H3·2n−2 −1 /(3 · 2n−2 ) can be rewritten as 1 1 (3H2n−1 −1 + 2H3·2n−2 −1 ) = (3H3 + 2H5 ) + O 2−2n+7 , 3 · 2n−1 3 · 22n−4 since by (4.10) and similarly to (4.8) we get that 1 H n−2 mod 2n−1 , 2 2 −1 1 ≡ H2n−3 −1 mod 2n−2 , and thus, 2 1 ≡ 2 H2n−3 −1 mod 2n−3 , 2
H2n−1 −1 ≡ H2n−2 −1 H2n−1 −1 etc.,
H2n−1 −1 =
1 2n−3
H22 −1 + O 2−n+7 .
In a similar fashion, we obtain H3·2n−2 −1 ≡
1 H n−3 mod 2n−2 , 2 3·2 −1
etc., H3·2n−2 −1 =
1 H3·2−1 + O 2−n+6 . 2n−3
Putting the relevant terms together, with n 3 we obtain that 3H2n−1 −1 + 2H3·2n−2 −1 =
1 2n−3
(3H3 + 2H5 ) + O 2−n+6 ,
and hence ν2 (E (2n − 1, 2)) = −(n − 1) − (n − 3) = −2n + 4. Now we consider the case with p 3 prime. Identity (4.5) implies that pn − 1 pn − 1 − n + νp E pn − 1, 2 = − 3n + 2 + νp E (p − 1, 2) νp s pn , 3 = p−1 p−1
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since clearly, E({1/pn−1 , 1/(2pn−1 ), . . . , 1/((p−1)pn−1 )}, 2) = 1/p2(n−1) E({1/1, 1/2, . . . , 1/(p − 1)}, 2) determines the p-adic order of the term E (pn − 1, 2) for all n 1. Clearly, ν3 (E (2, 2)) = 0. If p 5 then we can further analyze the last term since E (p − 1, 2) = (2) 2 )/2 by (4.5), hence, its p-adic order is 1 by Corollaries 1.1 and 1.2 if −(Hp−1 − Hp−1 νp (Bp−3 ) 0, i.e., except for the Wolstenholme primes. Therefore, we get that pn − 1 − 3n + 3. νp s pn , 3 = p−1
2
5. More proofs Proof of Theorem 3.2. We use identity (3.1) after substituting apn , b, and k into a, b, and n, respectively. We set
A=
apn + b − 1 apn apn
and derive that k s(apn , apn − l)s(b, b − (k − l)) s apn + b, apn + b − k = A . apn +b−k−1 (apn − l) apn −l l=0
Clearly, for the common factor A we have νp (A) = n. If l = 0 then the p-adic order of the actual term in the right hand side sum, with the factor A included, is νp (s(b, b − k)) as long as 2 k + 1 b < pn with any sufficiently large n. If l 1 then we apply Theorem 3.3, which yields that the p-adic order of the corresponding term is at least n + (n − logp (2k) ) − (n − logp (b − 1) + logp k) > νp (s(b, b − k)) if n is sufficiently large. Therefore, the term with l = 0 determines the p-adic order of the sum. n | We note that if l 1 odd then [8, Theorem 2.2] already implies that ap2 n n s(ap , ap − l), i.e., the corresponding p-adic order is at least n + n − 1 − (n − logp (b − 1)) − logp k n − 1, independently of the actual value of the odd l. 2 Proof of Theorem 3.3. Identity (3.3) already guarantees that νp (s(apn , apn − k)) apn n − logp (2k). 2 n − logp (2k) for 2k < pn since clearly, νp 2k−j Proof of Theorem 3.4. The case with k = 1 is obvious since s(n, n − 1) = n2 and hence, νp (s(apn , apn − 1)) = n − χp=2 if n 1. If k 2 then we prove that νp (s(apn , apn − k)) < n by induction on n n1 . Identity (3.6) will follow. Clearly, the inequality is true in the base case n = n1 by condition (3.5). We assume that it holds for all n : n1 n n and prove that it is also true for n = n + 1.
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We apply (3.2) which has the advantage over (3.1) that the term with index 0 makes no contribution on the right hand side. In (3.2), we substitute apn , a(p − 1)pn , and k for a, b, and n, respectively. We set n+1 apn+1 − 1 n ap ap apn k apn
B= For the common B factor we have
n+1 −1 ap + n + 1 − νp (k). νp (B) = νp apn In fact,
n+1
−1 ap ap − 1 = νp νp apn a although it is irrelevant to what follows and
n+1 apn+1 − 1 −1 apn+1 − k − 1 apn+1 − k apn+1 − k + 1 ap ··· = apn − k + 1 apn − k + 2 apn apn − k apn
thus, νp
n+1
n+1 ap ap −k−1 −1 = n − ν , (k) + ν p p apn − k apn
(5.1)
since here
νp
apn+1 − 1 apn+1 − k apn+1 − k + 1 · · · apn − k + 1 apn − k + 2 apn
= νp
k! (k − 1)!
− n = νp (k) − n.
After some simple manipulations we obtain k s(apn , apn − l)s(a(p − 1)pn , a(p − 1)pn − (k − l)) s apn+1 , apn+1 − k = B l n+1 −k−2 (apn − k − 1) apapn −l−1 l=0 k s(apn , apn − l)s(a(p − 1)pn , a(p − 1)pn − (k − l)) =B l apn+1 −k−1 (apn − l) apn −l l=1
If l = k then the p-adic order of the actual term in the right hand side sum, includ n+1 ing the factor B, is equal to (νp ( ap apn−1 ) + n + 1 − νp (k)) + νp (s(apn , apn − k)) − n+1 ) = νp (s(apn , apn − k)) + 1 by (5.1), and this term is the p-adically domiνp ( ap apn−k−1 −k nating one. Indeed, if 0 < l < k then the p-adic order of the corresponding term is at least n+1 n+1 (νp ( ap apn−1 ) +n +1 −νp (k)) +(n −logp (2l) ) +(n −logp (2(k−l)) ) −νp ( ap apn−k−1 ) −l
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3n + 1 − νp (k) − 2 logp k − (n + logp a) > n + 1 > νp (s(apn , apn − k)) + 1 by Theorem 3.3 and the induction hypothesis. The statements in (3.6) and (3.7) follow. 2 Proof of Theorem 3.5. The next to last identity in [8, p. 31] claims that with h 1, i = 0, 1, . . . , h, and t = 1, 2, . . . , p − 1, s hp + p − 1, h + t + i(p − 1) ≡
h (−1)h−i i
mod p
(5.2)
holds true for the unsigned (or signless) Stirling numbers of the first kind |s(hp + p − 1, h + t + i(p − 1))|. For any odd prime p, we set h = (n − (p − 1))/p, i = h + 1 − (k + 1)/(p − 1) , and t = (h − i + 1)(p − 1) − k in (5.2). Clearly, 0 i h and 1 t p − 1. We obtain that s(n, n −k) ≡ hi (−1)h−i (−1)k ≡ (−1)k (mod p) since n ≡ −1 (mod pk ) yields that, with some positive integer c: n = cpk − 1, h = cpk−1 − 1, and i = cpk−1 − (k + 1)/(p − 1) . k−1 Hence, by Lemma 8 in [10] we have cp j ≡ (−1)j (mod p) for 0 j < pk−1 , which h h h−i h−i implies that i (−1) = h−i (−1) ≡ 1 (mod p) since h − i < pk−1 by 2k < n. If p = 2 then we first prove that, with n ≡ −1 (mod 2k+1 ), s(n, n − k) is odd and then the relation s(N, K) = s(N − 1, K − 1) − (N − 1)s(N − 1, K), with N = M 2k+1 − 1, M 1, and K = M 2k+1 − 1 − k, yields the other part of the statement since N − 1 is even and s(N, K) is odd by the first part; thus, s(M 2k+1 − 2, M 2k+1 − 2 − k) is also odd. To prove the first part, we set h = (r + 1)2k−1 − 1 with some positive odd integer r, i = (r + 1)2k−1 − 1 − k, t = 1, and n = hp + p − 1 = (r + 1)2k − 1 ≡ −1 (mod 2k+1 ) in (5.2), and the result follows. 2 Acknowledgments The author wishes to thank Gregory P. Tollisen for his helpful comments. References [1] T. Agoh, K. Dilcher, Convolution identities for Stirling numbers of the first kind, Integers 10 (2010) 101–109. [2] D. Barsky, J.-P. Bezivin, p-Adic properties of Lengyel’s numbers, J. Integer Seq. 17 (2014), Article 14.7.3. [3] D.W. Boyd, A p-adic study of the partial sums of the harmonic series, Experiment. Math. 3 (1994) 287–302. [4] H. Cohen, Number Theory, vol 2: Analytic and Modern Tools, Graduate Texts in Mathematics, Springer, 2007. [5] L. Comtet, Advanced Combinatorics. The Art of Finite and Infinite Expansions, D. Reidel, Publishing Co., Dordrecht, 1974, enlarged edition. [6] R.L. Graham, D.E. Knuth, O. Patashnik, Concrete Mathematics, Addison–Wesley, Reading, MA, 1994. [7] A. Granville, Arithmetic properties of binomial coefficients. I. Binomial coefficients modulo prime powers, in: Organic Mathematics, Burnaby, BC, 1995, in: CMS Conference Proceedings, vol. 20, Amer. Math. Soc., Providence, RI, 1997, pp. 253–276, electronic version (a dynamic survey): http://www.dms.umontreal.ca/~andrew/Binomial/.
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[8] F.T. Howard, Congruences for the Stirling numbers and associated Stirling numbers, Acta Arith. LV (1990) 29–41. [9] K. Kamano, On the 3-adic valuations of generalized harmonic numbers, Integers 11 (2011) 1–8. [10] T. Lengyel, On the least significant 2-adic and ternary digits of certain Stirling numbers, Integers 13 (2013) 1–10. [11] J. Sándor, B. Crstici, Handbook of Number Theory, vol. II, Kluwer, 2004. [12] N.J.A. Sloane, The On-Line Encyclopedia of Integer Sequences, published electronically at http://oeis.org, 2013. [13] L.C. Washington, p-Adic L-functions and sums of powers, J. Number Theory 69 (1998) 50–61.