On the p -adic properties of Stirling numbers of the first kind
aa r X i v : . [ m a t h . N T ] M a r ON THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THEFIRST KIND SHAOFANG HONG AND MIN QIU ∗ Abstract.
Let n, k and a be positive integers. The Stirling numbers of the firstkind, denoted by s ( n, k ), count the number of permutations of n elements with k disjoint cycles. Let p be a prime. In recent years, Lengyel, Komatsu and Young,Leonetti and Sanna, Adelberg, Hong and Qiu made some progress in the study ofthe p -adic valuations of s ( n, k ). In this paper, by using Washington’s congruenceon the generalized harmonic number and the n -th Bernoulli number B n and theproperties of m -th Stirling numbers of the first kind obtained recently by the authors,we arrive at an exact expression or a lower bound of v p ( s ( ap, k )) with a and k being integers such that 1 ≤ a ≤ p − ≤ k ≤ ap . This infers that for anyregular prime p ≥ a and k with 5 ≤ a ≤ p − a − ≤ k ≤ ap −
1, one has v p ( H ( ap − , k )) < − log ( ap − p with H ( ap − , k )being the k -th elementary symmetric function of 1 , , ..., ap − . This gives a partialsupport to a conjecture of Leonetti and Sanna raised in 2017. We also present resultson v p ( s ( ap n , ap n − k )) from which one can derive that under certain condition, forany prime p ≥
5, any odd number k ≥ n , if( a, p ) = 1, then v p ( s ( ap n +1 , ap n +1 − k )) = v p ( s ( ap n , ap n − k )) + 2. It confirmspartially Lengyel’s conjecture proposed in 2015. Introduction
Let n and k be positive integers such that k ≤ n . The Stirling number of the firstkind , denoted by s ( n, k ), counts the number of permutations of n elements with k disjointcycles. One can also characterize s ( n, k ) by( x ) n = x ( x + 1)( x + 2) · · · ( x + n −
1) = n X k =0 s ( n, k ) x k . The
Stirling number of the second kind S ( n, k ) is defined as the number of ways topartition a set of n elements into exactly k nonempty subsets. Thus S ( n, k ) = 1 k ! k X i =0 ( − i (cid:18) ki (cid:19) ( k − i ) n . There are many authors who are interested in the divisibility properties of Stirling num-bers of the second kind, see, for example, [2, 11, 17, 18, 21, 23, 35, 36]. But very fewseems to be known about the divisibility properties of s ( n, k ).The Stirling number of the first kind s ( n, k ) is closely related to the k -th elementarysymmetric function H ( n, k ) of 1 , / , ..., /n by the following identity Mathematics Subject Classification.
Primary 11B73, 11A07.
Key words and phrases. p -adic valuation, p -adic analysis, Stirling numbers of the first kind, the m -thStirling numbers of the first kind, Bernoulli numbers, elementary symmetric function. ∗ M. Qiu is the corresponding author. S.F. Hong was supported partially by National ScienceFoundation of China Grant s ( n + 1 , k + 1) = n ! H ( n, k ) , (see Lemma 1.1 in [20]), where H ( n,
0) := 1 and H ( n, k ) := X ≤ i < ···
1) = H n is the n -th harmonic number. Theisinger [25] andNagell [24] proved that H n is an integer only for n = 1. Erd˝os and Niven [7] provedthat H ( n, k ) is an integer only for finitely many positive integers n and k , and laterChen and Tang [4] showed that H (1 ,
1) and H (3 ,
2) are the only integral values. Seealso [8, 10, 22, 30, 34] for some further studies on this topic. In 1862, Wolstenholme [33]proved that for any prime p ≥ H p − is divisible by p . Eswarathasanand Levine [6] conjectured that the set of positive integers n such that the numerator of H n is divisible by p is finite for any given prime p . Boyd [3] confirmed this conjecturefor all primes p ≤
547 but 83, 127 and 397. He also conjectured that v p ( H n ) ≤ n ≥ p ≥
5. See [14, 28] for more results on the p -adic properties of H n .Let p be a prime and n be a positive integer. Lengyel [19] showed that there exists aconstant c ′ = c ′ ( k, p ) > n ≥ n ( k, p ), one has v p ( s ( n, k )) ≥ c ′ n . Thisimplies that the p -adic valuation of H ( n, k ) has a lower bound. Consequently, Leonettiand Sanna [20] conjectured that there exists a positive constant c = c ( k, p ) such that v p ( H ( n, k )) < − c log n (1.1)for all large n and confirmed this conjecture for some special cases. Moreover, Ko-matsu and Young [16] used the theory of p -adic Newton polygon to show that if k is a nonnegative integer and n is of the form n = kp r + m with 0 ≤ m < p r , then v p ( s ( n + 1 , k + 1)) = v p ( n !) − v p ( k !) − kr . Using the study of the higher order Bernoullinumbers B ( l ) n , Adelberg [2] investigated some p -adic properties of Stirling numbers of bothkinds. Qiu and Hong [27] presented a detailed 2-adic analysis and obtained an exact for-mula for v ( s (2 n , k )) with n and k being positive integers such that 1 ≤ k ≤ n . In [26],Qiu, Feng and Hong provide a 3-adic analysis and arrive at a formula for v ( s ( a n , k ))with k being an integer such that 1 ≤ k ≤ a n , where a ∈ { , } .In this paper, we are mainly concerned with the p -adic valuations of the Stirlingnumbers of the first kind. Actually, we yield an exact expression or a lower bound of v p ( s ( ap, i )) with p ≥ a and i being integers such that 1 ≤ a ≤ p − ≤ i ≤ ap . Throughout let p ≥ n be a positive integer. Let a bean integer with 1 ≤ a ≤ p −
1. Obviously, we have v p ( s ( ap, v p (( ap − a − v p ( s ( ap, ap − v p (cid:0)(cid:18) ap (cid:19)(cid:1) = 1 , v p ( s ( ap, ap )) = v p (1) = 0 . N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 3 For any integer i with 2 ≤ i ≤ ap −
2, one can always write i = ap − k for some integer k with 2 ≤ k ≤ ap − y , ⌊ y ⌋ and ⌈ y ⌉ stand for the largest integer nomore than y and the smallest integer no less than y , respectively. For any integer k andprime p , let h k i denote the integer such that 0 ≤ h k i ≤ p − k ≡ h k i (mod p − ǫ k := 0 if k is even and ǫ k := 1 if k is odd. The n -th Bernoulli number B n is definedby the Maclaurin series as xe x − ∞ X n =0 B n x n n ! . By the von Staudt-Clausen theorem (see, for instance, [13]), we know that if n is even,then B n + X ( p − | n p ∈ Z , where the sum is over all primes p such that ( p − | n . The von Staudt-Clausentheorem tells us that v p ( B l ) ≥ l with 2 ≤ l ≤ p −
3. In particular,if v p ( B l ) = 0 for all even integers l with 2 ≤ l ≤ p −
3, then p is called a regular prime .Note that 3 is a regular prime. If p is not regular it is called irregular . The first irregularprimes are 37 and 59. We should emphasize that the cardinality and density of regularprimes are largely unknown. Although one can argue heuristically that asymptoticallymore than half of primes should be regular, it is not even known that there are infinitelymany regular primes, while infinitely many primes are known to be irregular. We referthe interested readers to [13] and [31] for the history and basic facts on the regular andirregular primes. We can now state the first main result of this paper as follows. Theorem 1.1.
Let p ≥ be a prime and let a and k be integers such that ≤ a ≤ p − and ≤ k ≤ ap − . Then each of the following is true: (i). If k ≡ ǫ k (mod p − , then v p ( s ( ap, ap − k )) = ( v p ( k ) + 1) ǫ k . (ii). If ≤ k ≤ a ( p − − and k ǫ k (mod p − , then v p ( s ( ap, ap − k )) ≥ ( v p ( k ) + 1) ǫ k + 1 , where the equality holds if and only if v p (cid:0) B (cid:4) h k i (cid:5)(cid:1) = 0 . In particular, if p is regular,then v p ( s ( ap, ap − k )) = ( v p ( k ) + 1) ǫ k + 1 . (iii). If a ≥ and a ( p −
1) + 2 ≤ k ≤ ap − , then v p ( s ( ap, ap − k )) ≥ a + k − ap. Let a and k be integers such that 1 ≤ a ≤ p − { , a − } ≤ k ≤ ap . FromTheorem 1.1, we can see that if p is regular, then v p ( s ( ap, k )) ≤ v p ( k ) ≤
3. Since s ( ap, k ) = ( ap − H ( ap − , k − Corollary 1.2.
Let p ≥ be a regular prime and let a and k be integers such that ≤ a ≤ p − and max { , a − } ≤ k ≤ ap − . Then v p ( H ( ap − , k )) ≤ − a. So for any regular prime p ≥ a and k with 5 ≤ a ≤ p − a − ≤ k ≤ ap −
1, one has v p ( H ( ap − , k )) < − c log ( ap −
1) with c =
12 log p . Thisgives a partial support to Conjecture (1.1) due to Leonetti and Sanna.Consequently, we state the second main result of this paper as follows. S.F. HONG AND M. QIU
Theorem 1.3.
Let p ≥ be a prime and let a and n be positive integers such that ( a, p ) = 1 . Let k be an odd integer with ≤ k ≤ ap n − . Then v p ( s ( ap n , ap n − k )) (cid:26) = v p ( s ( ap n , ap n − k + 1)) + v p ( ap n − k ) + n, if v p ( s ( ap n , ap n − k + 1)) ≤ n − ≥ v p ( ap n − k ) + 3 n, if v p ( s ( ap n , ap n − k + 1)) ≥ n. On the other hand, Lengyel [19] proved the following interesting result.
Theorem 1.4. [19]
For any prime p , any integer a ≥ with ( a, p ) = 1 , and any even k ≥ with the condition ∃ n ∈ Z + : n > p k + log p a such that v p ( s ( ap n , ap n − k )) < n or k = 1 with n = 1 , then for n ≥ n one has v p ( s ( ap n +1 , ap n +1 − k )) = v p ( s ( ap n , ap n − k )) + 1 . Meanwhile, for any odd k ≥
3, Lengyel conjectured in [19] that for any integer n ≥ n ( p, k ) with some sufficiently large n ( p, k ), one has v p ( s ( ap n +1 , ap n +1 − k )) = v p ( s ( ap n , ap n − k )) + 2 . (1.2)Now by Theorems 1.3 and 1.4, one yields the following analogous result which is thethird main result and proves partially Lengyel’s Conjecture (1.2). Theorem 1.5.
Let p ≥ be a prime and let a be a positive integer such that ( a, p ) = 1 .Let k ≥ be an odd integer with the condition ∃ n ∈ Z + : n > p ( k −
1) + log p a such that v p ( s ( ap n , ap n − ( k − < n . Then for any positive integer n with n ≥ n one has v p ( s ( ap n +1 , ap n +1 − k )) = v p ( s ( ap n , ap n − k )) + 2 . Furthermore, we have v p ( s ( ap n , ap n − k )) = v p ( s ( ap n , ap n − k )) + 2( n − n ) . The paper is organized as follows. We reveal some useful properties of Stirling numbersof the first kind and generalized harmonic numbers in the next section. Then we proveTheorems 1.3 and 1.5 and Theorem 1.1 in Sections 3 and 4, respectively. Finally, for anypositive integer n and k such that 2 ≤ k ≤ ap n −
2, we propose three conjectures on the p -adic valuation of s ( ap n , k ).2. Preparatory lemmas on Stirling numbers of the first kind andgeneralized harmonic numbers
Let n and k be positive integers. Some basic identities involving Stirling numbers ofthe first kind can be listed as follows (see [5]): s ( n + 1 , k ) = ns ( n, k ) + s ( n, k − , s ( n,
0) = 0 , s ( n, k ) = 0 if k ≥ n + 1 , n X k =1 s ( n, k ) = n ! , s ( n,
1) = ( n − , s ( n, n ) = 1 , s ( n, n −
1) = (cid:18) n (cid:19) ,s ( n,
2) = ( n − n − X k =1 k and s ( n, n −
2) = 14 (3 n − (cid:18) n (cid:19) if n ≥ . The following two results are known.
N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 5 Lemma 2.1. [1]
Let n and k be positive integers. If n + k is odd, then s ( n, k ) = 12 n X i = k +1 ( − n − i n i − k (cid:18) i − i − k (cid:19) s ( n, i ) . Lemma 2.2. [3]
Let p be a prime with p ≥ . Let n and k be positive integers with k ≤ p − . Then H np ≡ p H n (mod p ) and H np + k ≡ H np + H k (mod p ) . As introduced in [27], for any given nonnegative integer m , we define the m -th Stirlingnumbers of the first kind , denoted by s m ( n, k ) with n and k being nonnegative integers,as follows: ( x + m )( x + m + 1) · · · ( x + m + n −
1) := n X k =0 s m ( n, k ) x k . One notices that the m -th Stirling number of the first kind is actually a special case ofthe generalized Stirling number of Hsu and Shiue [12]. The following two basic resultson s m ( n, k ) will be needed in the proof of Theorem 1.1. Lemma 2.3. [27]
Let m, k and n be nonnegative integers. Then s ( m + n, k ) = k X i =0 s ( m, i ) s m ( n, k − i ) . Lemma 2.4. [27]
Let m, k and n be nonnegative integers. Then s m ( n, k ) = n X i = k s ( n, i ) (cid:18) ii − k (cid:19) m i − k . Let n be a positive integer and let r be a nonnegative integer. Then the generalizedharmonic number , denoted by H ( r ) n , is defined by H ( r ) n := n X i =1 i r . Washington proved the following congruences which were later extended by Hong [9].
Lemma 2.5. [32]
Let p be an odd prime and let r be an integer. (i). If r ≥ is odd and p ≥ r + 4 , then H ( r ) p − ≡ − r ( r + 1)2( r + 2) B p − r − p (mod p ) . (ii). If r ≥ is even and p ≥ r + 3 , then H ( r ) p − ≡ rr + 1 B p − r − p (mod p ) . From Lemma 2.5 and noticing the fact that v p (cid:0) B l (cid:1) ≥ p − ∤ l , one can derivethe following corollary immediately. S.F. HONG AND M. QIU
Corollary 2.6.
Let p ≥ be a prime. Let r be an integer with ≤ r ≤ p − . Then v p (cid:0) H ( r ) p − (cid:1) ≥ ǫ r (2.1) with the equality holding if and only if v p (cid:0) B p − − ⌈ r ⌉ (cid:1) = 0 . Let n and k be integers with n ≥ k ≥
0. Let S k = S k ( x , ..., x n ) and σ k = σ k ( x , ..., x n ) denote the k th Newton sum and the k th elementary symmetric function ofthe n variables x , ..., x n , respectively. That is, S k = S k ( x , ..., x n ) := n X i =1 x ki and σ k = σ k ( x , ..., x n ) := , if k = 0 , P ≤ i < ··· n. Then the well known Newton-Girard formula can be stated as follows.
Lemma 2.7. [29]
Let n be a positive integer and let k be a nonnegative integer. (i). If ≤ k ≤ n , then k X i =1 ( − k − i S i σ k − i + ( − k kσ k = 0 . (ii). If k > n , then n X i =0 ( − i S k − i σ i = 0 . For any integer r with 0 ≤ r ≤ p −
1, it is easy to see that S r (cid:16) , , · · · , p − (cid:17) = p − X i =1 i r = H ( r ) p − and σ r (cid:16) , , · · · , p − (cid:17) = H ( p − , r ) , where H ( p − ,
0) = 1. Therefore by using Lemma 2.7 and Corollary 2.6, we can deducethe following result.
Lemma 2.8.
Let p be an odd prime and let r be an integer. (i). If r ≥ is odd and p ≥ r + 4 , then H ( p − , r ) ≡ r H ( r ) p − (mod p ) . (2.2)(ii). If r ≥ is even and p ≥ r + 3 , then H ( p − , r ) ≡ − r H ( r ) p − (mod p ) . (2.3) N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 7 Proof . Since H ( p − ,
1) = H (1) p − , Lemma 2.8 is obviously true when r = 1. Now let2 ≤ r ≤ p −
3. Using Lemma 2.7 and also noting that H ( p − ,
0) = 1, one derives that H ( p − , r ) = 1 r r X i =1 ( − i − H ( p − , r − i ) H ( i ) p − = 1 r r − X i =1 ( − i − H ( p − , r − i ) H ( i ) p − − ( − r r H ( r ) p − . (2.4)If r = 2, then it follows from (2.4) that H ( p − , r ) = H ( p − ,
2) = 12 H ( p − , H (1) p − − H (2) p − = 12 (cid:0) H (1) p − (cid:1) − H (2) p − . (2.5)From (2.1), one yields that v p (cid:0) H (2) p − (cid:1) ≥ v p (cid:16)(cid:0) H (1) p − (cid:1) (cid:17) = 2 v p (cid:0) H (1) p − (cid:1) ≥ . (2.6)Using (2.5) and (2.6), we then deduce that H ( p − , r ) = H ( p − , ≡ − H (2) p − = − r H ( r ) p − (mod p ) . Hence Lemma 2.9 is true when r = 2.Now let 3 ≤ r ≤ p −
3. Suppose that Lemma 2.8 holds for any integer e with2 ≤ e ≤ r −
1. In the following, we show that Lemma 2.8 is still true for the r case.For any integer i with 1 ≤ i ≤ r −
1, by the inductive hypothesis and Corollary 2.6,one obtains that v p ( H ( p − , r − i )) ≥ v p (cid:0) H ( i ) p − (cid:1) ≥ r is even, then one can conclude that v p (cid:0) H ( p − , r − i ) H ( i ) p − (cid:1) ≥
2. Since 2 ≤ r ≤ p − r is a p -adic unit and so v p ( r ) = 0. Hence v p (cid:16) r r − X i =1 ( − i − H ( p − , r − i ) H ( i ) p − (cid:17) ≥ . (2.7)Thus (2.4) and (2.7) imply the truth of (2.3). So (2.3) is proved.If r is odd, then one of i and r − i must be odd. So it follows from Corollary 2.6 andthe inductive hypothesis that either v p (cid:0) H ( i ) p − (cid:1) ≥ v p ( H ( p − , r − i )) ≥
2. Hence v p (cid:0) H ( p − , r − i ) H ( i ) p − (cid:1) ≥
3, which infers that v p (cid:16) r r − X i =1 ( − i − H ( p − , r − i ) H ( i ) p − (cid:17) ≥ . (2.8)Therefore (2.2) follows from (2.4) and (2.8).This completes the proof of Lemma 2.8. (cid:3) Combining Lemma 2.5 with Lemma 2.8, one can easily obtain the following congru-ences.
Lemma 2.9.
Let p be an odd prime and let r be an integer. (i). If r ≥ is odd and p ≥ r + 4 , then H ( p − , r ) ≡ − r + 12( r + 2) B p − r − p (mod p ) . S.F. HONG AND M. QIU (ii). If r ≥ is even and p ≥ r + 3 , then H ( p − , r ) ≡ − r + 1 B p − r − p (mod p ) . Evidently, one has s ( p,
1) = ( p − ≡ − p ) by Wilson theorem, and s ( p, p ) = 1.Also it is known that s ( p, k ) ≡ p ) if 2 ≤ k ≤ p − p -adic valuation of s ( p, k ) for any integer k with 2 ≤ k ≤ p − Corollary 2.10.
Let p ≥ be a prime and let k be an integer with ≤ k ≤ p . Then v p ( s ( p, k )) = (cid:26) , if k = 1 or k = p, , if k = p − , (2.9) and if ≤ k ≤ p − , then v p ( s ( p, k )) ≥ ǫ k − (2.10) with the equality holding if and only if v p (cid:0) B p − − ⌊ k ⌋ (cid:1) = 0 . Proof . Since v p ( s ( p, v p (( p − v p ( s ( p, p − v p (cid:0)(cid:0) p (cid:1)(cid:1) = 1 and v p ( s ( p, p )) = v p (1) = 0, the equality (2.9) is clearly true.Now let 2 ≤ k ≤ p −
2. Note that v p (cid:0) B p − − ⌊ k ⌋ (cid:1) ≥
0. So replacing r by k − v p ( H ( p − , k − ≥ ǫ k − (2.11)with the equality holding if and only if v p (cid:0) B p − − ⌊ k ⌋ (cid:1) = 0. Since v p ( s ( p, k )) = v p (( p − H ( p − , k − v p ( H ( p − , k − , (2.10) follows immediately from (2.11).This finishes the proof of Corollary 2.10. (cid:3) For any positive integer m , it follows from Lemma 2.4 that s m ( n, k ) ≡ s ( n, k ) (mod m ).Now together with Theorem 1.3, we can obtain the following result. Lemma 2.11.
Let p ≥ be a prime. Let a and n be positive integers such that ( a, p ) = 1 .For any integer k with ≤ k ≤ ap n , if ∤ k , then v p ( s p n ( ap n , ap n − k )) ≥ n, (2.12) and if | k , then s p n ( ap n , ap n − k ) ≡ s ( ap n , ap n − k ) (mod p n ) . (2.13) Proof . Since s p n ( ap n , ap n ) = s ( ap n , ap n ) = 1, (2.13) is obviously true when k = 0. If k = 1, then by Lemma 2.4, one obtains that s p n ( ap n , ap n −
1) = s ( ap n , ap n −
1) + ap n . Since v p ( s ( ap n , ap n − n , it follows that v p ( s p n ( ap n , ap n − n . Thus (2.12) isproved when k = 1. For the case that k = ap n , we have v p ( s p n ( ap n , v p (cid:16) (( a + 1) p n − p n − (cid:17) = ap n − d p ( a ) p − ≥ a ( p n − p − a n − X i =0 p i ≥ an. Hence (2.12) is clearly true when k = ap n is odd. If k = ap n is even, then a is evenand so a ≥
2, which infers that v p ( s p n ( ap n , ≥ n . Thus together with the fact that s ( ap n ,
0) = 0, we can derive that (2.13) holds when k is even. N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 9 Now let 2 ≤ k ≤ ap n −
1. From Lemma 2.4, one deduces that s p n ( ap n , ap n − k ) = s ( ap n , ap n − k ) + s ( ap n , ap n − k + 1)( ap n − k + 1) p n + ap n X i = ap n − k +2 s ( ap n , i ) (cid:18) ii − ap n + k (cid:19) p n ( i − ap n + k ) ≡ s ( ap n , ap n − k ) + s ( ap n , ap n − k + 1)( ap n − k + 1) p n (mod p n ) . (2.14)If k is odd, then by Theorem 1.3, one gets that v p ( s ( ap n , ap n − k )) ≥ n . Therefore itfollows from (2.14) that v p ( s p n ( ap n , ap n − k )) ≥ n as desired. If k is even, then k − v p ( s ( ap n , ap n − k + 1)) = v p ( s ( ap n , ap n − ( k − ≥ n . Thus (2.14) impliesthe truth of (2.13) as required.The proof of Lemma 2.11 is complete. (cid:3) Proofs of Theorems 1.3 and 1.5
In this section, we present the proofs of Theorems 1.3 and 1.5. We begin with theproof of Theorem 1.3.
Proof of Theorem 1.3.
Let p ≥ a and n be positive integers suchthat ( a, p ) = 1. Let k be an odd integer with 1 ≤ k ≤ ap n −
1. Since k is odd, replacingin Lemma 2.1 n and k by ap n and ap n − k , respectively, gives us that s ( ap n , ap n − k ) = 12 ap n X i = ap n − k +1 ( − ap n − i ( ap n ) i − ap n + k (cid:18) i − i − ap n + k (cid:19) s ( ap n , i ) . (3.1)It follows immediately from (3.1) that v p ( s ( ap n , ap n − k )) ≥ n .Note that p ≥ a, p ) = 1. If k = 1, then v p ( s ( ap n , ap n − k + 1)) = v p ( s ( ap n , ap n )) = v p (1) = 0and v p ( s ( ap n , ap n − k )) = v p ( s ( ap n , ap n − v p (cid:16)(cid:18) ap n (cid:19)(cid:17) = v p (cid:16) ap n ( ap n − (cid:17) = n. So Theorem 1.3 is clearly true when k = 1.Now let 3 ≤ k ≤ ap n −
1. By (3.1), one deduces that s ( ap n , ap n − k ) = 12 ( ap n ( ap n − k ) s ( ap n , ap n − k + 1) + L ) , (3.2)where L = ap n − k +3 X i = ap n − k +2 ( − ap n − i ( ap n ) i − ap n + k (cid:18) i − i − ap n + k (cid:19) s ( ap n , i ) + ∆ (3.3)with ∆ being an integer such that v p (∆) ≥ n . Since p ≥ a, p ) = 1 and 3 ≤ k ≤ ap n −
1, it is easy to check that for i = ap n − k + 2 and i = ap n − k + 3, one has v p (cid:16)(cid:18) i − i − ap n + k (cid:19)(cid:17) ≥ v p ( ap n − k ) . (3.4)Notice that k − v p ( s ( ap n , ap n − k + 2)) = v p ( s ( ap n , ap n − ( k − ≥ n. (3.5) We claim that v p (∆) ≥ v p ( ap n − k ) + 3 n. (3.6)Then it follows from (3.3) to (3.5) and claim (3.6) that v p ( L ) ≥ min n v p (cid:16) ap n − k +3 X i = ap n − k +2 ( − ap n − i ( ap n ) i − ap n + k (cid:18) i − i − ap n + k (cid:19) s ( ap n , i ) (cid:17) , v p (∆) o ≥ v p ( ap n − k ) + 3 n. (3.7)If v p ( s ( ap n , ap n − k + 1)) ≥ n , then one can easily get that v p ( ap n ( ap n − k ) s ( ap n , ap n − k + 1)) ≥ v p ( ap n − k ) + 3 n. (3.8)Hence by (3.2), (3.7) and (3.8), we obtain that v p ( s ( ap n , ap n − k )) ≥ min { v p ( ap n ( ap n − k ) s ( ap n , ap n − k + 1)) , v p ( L ) }≥ v p ( ap n − k ) + 3 n as desired.If v p ( s ( ap n , ap n − k + 1)) ≤ n −
1, then from (3.7) one derives that v p ( ap n ( ap n − k ) s ( ap n , ap n − k + 1)) ≤ v p ( ap n − k ) + 3 n − < v p ( L ) . (3.9)Since p ≥
5, by (3.2) and (3.9) together with the isosceles triangle principle (see, forexample, [15]), we arrive at v p ( s ( ap n , ap n − k )) = v p ( ap n ( ap n − k ) s ( ap n , ap n − k + 1) + L )= v p ( ap n ( ap n − k ) s ( ap n , ap n − k + 1))= v p ( s ( ap n , ap n − k + 1)) + v p ( ap n − k ) + n as expected. So to finish the proof of Theorem 1.3, it remains to show the truth of claim(3.6) that will be done in what follows.If k = 3, then ∆ is empty sum and so ∆ = 0. Since v p (0) = + ∞ , claim (3.6) is trueif k = 3. In the following one lets k ≥
5. One has∆ = ap n X i = ap n − k +4 ( − ap n − i ∆ i (3.10)where ∆ i := ( ap n ) i − ap n + k (cid:18) i − i − ap n + k (cid:19) s ( ap n , i ) . Let i be any integer with ap n − k + 4 ≤ i ≤ ap n . Then one can write i := ap n − k + j forsome integer j with 4 ≤ j ≤ k . So v p (∆ i ) ≥ nj + v p (cid:16)(cid:18) ap n − k + j − j (cid:19)(cid:17) . (3.11)In what follows, we show that the following p -adic estimate holds: v p (∆ i ) ≥ v p ( ap n − k ) + 3 n. (3.12)The proof of (3.12) is divided into the following two cases. Case 1. v p ( ap n − k ) ≤ n . It follows from the hypothesis j ≥ v p (∆ i ) ≥ jn ≥ n ≥ v p ( ap n − k ) + 3 n . Thus (3.12) is true in this case. N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 11 Case 2. v p ( ap n − k ) ≥ n + 1. We have (cid:18) ap n − k + j − j (cid:19) = ( ap n − k + j − ap n − k + j − · · · ( ap n − k ) j ! . Subcase 2.1. v p ( j ′ ) < v p ( ap n − k ) for any integer j ′ with 1 ≤ j ′ ≤ j −
1. Then v p ( ap n − k + j ′ ) = v p ( j ′ ). It implies that v p (cid:16)(cid:18) ap n − k + j − j (cid:19)(cid:17) = v p ( ap n − k ) − v p ( j ) . (3.13)Since j ≥ p ≥
5, one can deduce that j ≥ v p ( j ) + 4. In fact, if v p ( j ) = 0, then j ≥ v p ( j ) + 4 and if v p ( j ) ≥
1, then j ≥ p v p ( j ) ≥ v p ( j ) + 4 as expected. Now by (3.11)and (3.13), we have v p (∆ i ) ≥ nj + v p ( ap n − k ) − v p ( j ) ≥ v p ( ap n − k ) + 4 n + v p ( j )( n − > v p ( ap n − k ) + 3 n as required. So (3.12) holds in this case. Subcase 2.2. v p ( j ′ ) ≥ v p ( ap n − k ) ≥ n + 1 for some integer j ′ with 1 ≤ j ′ ≤ j − p v p ( ap n − k ) ≥ v p ( ap n − k ) + 2 and by (3.11), one has v p (∆ i ) ≥ nj ≥ nj ′ + n ≥ np v p ( j ′ ) + n ≥ np v p ( ap n − k ) + n ≥ v p ( ap n − k ) + 3 n as desired. Therefore (3.12) is true in this case. So (3.12) is proved.Finally, from (3.10) and (3.12) one can deduce immediately that v p (∆) ≥ min ap n − k +4 ≤ i ≤ ap n { v p (∆ i ) } ≥ v p ( ap n − k ) + 3 n as (3.6) claimed. This completes the proof of claim (3.6) and that of Theorem 1.3. (cid:3) Let i be an integer such that 1 ≤ i ≤ ap n −
1. We remark that if ap n + i is odd, thenusing Theorem 1.3 we can deduce that v p ( s ( ap n , i )) ≥ n since one may write i = ap n − i ′ for some integer i ′ with 1 ≤ i ′ ≤ ap n −
1, where i ′ is odd if and only if ap n + i is odd.We can now use Theorem 1.3 to show Theorem 1.5. Proof of Theorem 1.5.
Let p be a prime with p ≥
5. Let a and k be positive integerssuch that ( a, p ) = 1 and k ≥ ∃ n ∈ Z + : n > p ( k −
1) + log p a such that v p ( s ( ap n , ap n − ( k − < n . (3.14)From condition (3.14), one can easily get that k < p n , which infers that v p ( k ) < n .Let n be an integer with n ≥ n . Since k − ≥ k −
1, it follows from Theorem 1.4 that v p ( s ( ap n , ap n − ( k − v p ( s ( ap n , ap n − ( k − n − n < n and v p ( s ( ap n +1 , ap n +1 − ( k − v p ( s ( ap n , ap n − ( k − v p ( s ( ap n , ap n − ( k − n + 1 − n < n + 1 . Note that k is odd and v p ( k ) < n ≤ n . Since v p ( s ( ap n +1 , ap n +1 − ( k − < n + 1 and v p ( s ( ap n , ap n − ( k − < n , Theorem 1.3 together with (3.15) give us that v p ( s ( ap n +1 , ap n +1 − k )) = v p ( s ( ap n +1 , ap n +1 − k + 1)) + v p ( ap n +1 − k ) + n + 1= v p ( s ( ap n , ap n − k + 1)) + v p ( k ) + n + 2= v p ( s ( ap n , ap n − k + 1)) + v p ( ap n − k ) + n + 2= v p ( s ( ap n , ap n − k )) + 2as Theorem 1.5 expected. Hence v p ( s ( ap n , ap n − k )) = v p ( s ( ap n , ap n )) + 2( n − n ) . This concludes the proof of Theorem 1.5. (cid:3) Proof of Theorem 1.1
In this section, we use the lemmas given in section 2 and Theorem 1.3 to supply theproof of Theorem 1.1.
Proof of Theorem 1.1.
We prove Theorem 1.1 by induction on the positive integer a with a ≤ p − a = 1. Then 2 ≤ k ≤ p −
2. It infers that k ǫ k (mod p − h k i = k and v p ( k ) = 0. Hence only part (ii) happens. So we need only to show part (ii), i.e., toshow that v p ( s ( p, p − k )) ≥ ( v p ( k ) + 1) ǫ k + 1 = ǫ k + 1with the equality holding if and only if v p (cid:0) B ⌊ k ⌋ (cid:1) = 0. But replacing k by p − k inCorollary 2.10 tells that for any integer k with 2 ≤ k ≤ p −
2, one has v p ( s ( p, p − k )) ≥ ǫ p − k − + 1 = ǫ k + 1with the equality being true if and only if v p (cid:0) B p − − ⌊ p − k ⌋ (cid:1) = 0. Since p − − j p − k k = p − − j p − − k − k = p − − (cid:16) p −
12 + j − k − k(cid:17) = − j − k − k = 2 j k k , it then follows that v p (cid:0) B p − − ⌊ p − k ⌋ (cid:1) = 0 holds if and only if v p (cid:0) B ⌊ k ⌋ (cid:1) = 0. ThereforeTheorem 1.1 is true when a = 1. Now let 2 ≤ a ≤ p −
1. Assume that Theorem 1.1 holdsfor the a − a case.We begin with the proof of part (iii). (iii). Let a and k be integers with a ≥ a ( p −
1) + 2 ≤ k ≤ ap −
2. By setting t = ap − k , one finds that showing the truth of part (iii) is equivalent to showing that v p ( s ( ap, t )) ≥ a − t (4.1)holds for any integer t such that 2 ≤ t ≤ a −
2. We prove (4.1) by induction on theinteger a ≥
4. First, let a = 4. Then t = 2, and so s ( ap, t ) = s (4 p,
2) = (4 p − p − X i =1 i = (4 p − H p − . (4.2) N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 13 Since p ≥
5, using Lemma 2.2, one can easily deduce that v p ( H p − ) ≥ −
1. Then by v p ((4 p − v p ( s (4 p, ≥ a = 4.In what follows, we let 5 ≤ a ≤ p −
1. Then p >
5. Assume that (4.1) holds for the a − a case. From Lemma 2.3, we get that s ( ap, t ) = t X i =1 s ( p, i ) s p (( a − p, t − i ) . (4.3)For any integer i with 1 ≤ i ≤ t , it follows from Lemma 2.4 that s p (( a − p, t − i ) = ( a − p X j = t − i s (( a − p, j ) (cid:18) jj − t + i (cid:19) p j − t + i . (4.4)Let j be any integer such that 0 ≤ j ≤ ( a − p . If we can show that v p ( s (( a − p, j )) ≥ a − − j, (4.5)then from (4.4) and (4.5), one can derive that v p ( s p (( a − p, t − i )) ≥ a − t since i ≥ j = 0, then (4.5) is obviously true since s (( a − p,
0) = 0. If j = 1, then one has v p ( s (( a − p, j )) = v p ( s (( a − p, v p ((( a − p − a − a − − j. So (4.5) holds when j = 1. If 2 ≤ j ≤ a −
3, then it follows from the induction assumptionof (4.1) that (4.5) is true. If j = a −
2, then ( a − p + j is odd and so we can deduce fromTheorem 1.3 that v p ( s (( a − p, j )) ≥ a − − j . If j ≥ a −
1, then (4.5) is clearlytrue since v p ( s (( a − p, j )) ≥
0. So (4.5) holds for all integers j with 0 ≤ j ≤ ( a − p .Hence part (iii) is proved.Now we turn our attention to the proofs of parts (i) and (ii). Let k be an integer suchthat 2 ≤ k ≤ a ( p −
1) + 1. Assume that parts (i) and (ii) hold for all even integers k with 2 ≤ k ≤ a ( p − k with 3 ≤ k ≤ a ( p −
1) + 1. To do so, let k be an odd integer. Then k p −
1) and k − ≤ k − ≤ a ( p − k ≡ p − k − ≡ p − v p ( s ( ap, ap − ( k − v p ( s ( ap, ap − k )) = v p ( s ( ap, ap − k + 1)) + v p ( k ) + 1 = v p ( k ) + 1as expected. So part (i) is true for any odd integer k with k ≡ p − k p − k − p − v p ( s ( ap, ap − ( k − ≥ v p ( B h k − i ) = 0. Since k is odd and one may write k = h k i + l ( p −
1) for somenonnegative integer l , we can deduce that h k − i = D j k kE = D j h k i + l ( p − kE = D j h k i k + l ( p − E = D j h k i kE = 2 j h k i k . Thus v p ( B h k − i ) = 0 holds if and only if v p (cid:0) B (cid:4) h k i (cid:5)(cid:1) = 0. From this together withTheorem 1.3, one then deduces that v p ( s ( ap, ap − k )) ≥ v p ( k ) + 2 with the equality being true if and only if v p (cid:0) B (cid:4) h k i (cid:5)(cid:1) = 0. Hence part (ii) holds for anyodd integer k with k p − k with 2 ≤ k ≤ a ( p − k be even and 2 ≤ k ≤ a ( p − s ( p, i ) = 0 if i ≥ p + 1 and s p (( a − p, ap − k − i ) = 0 if i ≤ p − k −
1, replacing m by p , n by ( a − p and k by ap − k in Lemma 2.3 gives us that s ( ap, ap − k ) = ap − k X i =1 s ( p, i ) s p (( a − p, ap − k − i )= min { p,ap − k } X i =max { ,p − k } s ( p, i ) s p (( a − p, ap − k − i ) . (4.6)Now let i be an integer such that max { , p − k } ≤ i ≤ min { p, ap − k } . Then 0 ≤ ap − k − i ≤ ( a − p . By Corollary 2.10, we know that v p ( s ( p, i )) ≥ ≤ i ≤ p − ap − k − i = ( a − p − ( k + i − p ), where 0 ≤ k + i − p ≤ min { k, ( a − p } .From the inductive hypothesis of parts (i) and (ii) together with the truth of part (iii)and Lemma 2.11, one can deduce that v p ( s p (( a − p, ap − k − i )) = v p ( s p (( a − p, ( a − p − ( k + i − p ))) ≥ k + i − p ≡ p − V k := { i ∈ Z : 2 ≤ i ≤ p − k + i − p p − } . Then v p ( s ( p, i )) ≥ v p ( s p (( a − p, ap − k − i )) ≥ i ∈ V k . This infers that forany i ∈ V k , we have s ( p, i ) s p (( a − p, ap − k − i ) ≡ p ) . (4.7)Now we begin to prove part (i) for the case of even number k . (i). Let k be an even integer such that 2 ≤ k ≤ a ( p −
1) and k ≡ p − k = l ( p −
1) for an integer l with 1 ≤ l ≤ a . We claim that if k = l ( p − l with 1 ≤ l ≤ a , then s ( ap, ap − k ) ≡ (cid:18) al (cid:19) s ( p, l (mod p ) . (4.8)Since s ( p,
1) = ( p − v p ( s ( p, v p (( p − ≤ l ≤ a ≤ p − v p ( s ( ap, ap − k )) = 0 when 2 ≤ k ≤ a ( p −
1) and k ≡ p − a with 2 ≤ a ≤ p − a = 2. Then l = 1 or l = 2. If l = 1, then k = p −
1. So p − k = 1 and2 p − k = p + 1. By (4.6) and s ( p, p ) = 1, we obtain that s (2 p, p − ( p − p X i =1 s ( p, i ) s p ( p, p + 1 − i )= s ( p,
1) + s p ( p,
1) + p − X i =2 s ( p, i ) s p ( p, p + 1 − i ) . (4.9) N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 15 For any integer i with 2 ≤ i ≤ p −
1, one has k + i − p = i − p − i ∈ V p − . Then by (4.7), we derive that p − X i =2 s ( p, i ) s p ( p, p + 1 − i ) ≡ p ) . (4.10)Using Lemma 2.11, we get that s p ( p, ≡ s ( p,
1) (mod p ). Hence it follows from (4.9)and (4.10) that s (2 p, p − ( p − ≡ s ( p,
1) = (cid:18) (cid:19) s ( p,
1) (mod p )as (4.8) claimed. If l = 2, then k = 2( p −
1) and 2 p − k = 2. Likewise, since s p ( p, ≡ s ( p,
1) (mod p ) and v p ( s ( p, ≥ v p ( s p ( p, s (2 p, p − p − s (2 p,
2) = s ( p, s p ( p,
1) + s ( p, s p ( p, ≡ s ( p, = (cid:18) (cid:19) s ( p, (mod p ) . Thus claim (4.8) is true when a = 2.Now let 3 ≤ a ≤ p −
1. Assume that claim (4.8) holds for the a − a case. We divide this into the followingthree cases. Case 1. l = 1. Then k = p −
1. So p − k = 1 and ap − k − a − p . From (4.6)and s p (( a − p, ( a − p ) = s ( p, p ) = 1, one derives that s ( ap, ap − k ) = p X i =1 s ( p, i ) s p (( a − p, ap − k − i )= s ( p,
1) + s p (( a − p, ( a − p − k ) + p − X i =2 s ( p, i ) s p (( a − p, ap − k − i ) . (4.11)Since k = p −
1, by Lemma 2.11 and the inductive hypothesis of claim (4.8), we get that s p (( a − p, ( a − p − k ) = s p (( a − p, ( a − p − ( p − ≡ s (( a − p, ( a − p − ( p − ≡ (cid:18) a − (cid:19) s ( p,
1) (mod p ) . (4.12)For any integer i with 2 ≤ i ≤ p −
1, one has 1 ≤ k + i − p = i − ≤ p − k + i − p p − i ∈ V p − . It then follows from (4.7) that p − X i =2 s ( p, i ) s p (( a − p, ap − k − i ) ≡ p ) . (4.13)Therefore by (4.11) to (4.13), we arrive at s ( ap, ap − k ) ≡ s ( p,
1) + (cid:18) a − (cid:19) s ( p,
1) = (cid:18) a (cid:19) s ( p,
1) (mod p )as (4.8) asserted. Thus claim (4.8) is proved when l = 1. Case 2. ≤ l ≤ a −
1. Then 2( p − ≤ k = l ( p − ≤ ( a − p − p − k < p + 1 ≤ ap − k ≤ ( a − p −
1. From (4.6) and s ( p, p ) = 1, one derives that s ( ap, ap − k ) = p X i =1 s ( p, i ) s p (( a − p, ap − k − i )= s ( p, s p (( a − p, ap − k −
1) + s p (( a − p, ( a − p − k )+ p − X i =2 s ( p, i ) s p (( a − p, ap − k − i ) . (4.14)Since k = l ( p −
1) is even and 2 ≤ l ≤ a −
1, by using Lemma 2.11 together with theinductive hypothesis of claim (4.8), we deduce that s p (( a − p, ap − k −
1) = s p (( a − p, ( a − p − ( k − ( p − ≡ s (( a − p, ( a − p − ( k − ( p − ≡ s (( a − p, ( a − p − ( l − p − ≡ (cid:18) a − l − (cid:19) s ( p, l − (mod p ) (4.15)and s p (( a − p, ( a − p − k ) ≡ s (( a − p, ( a − p − k ) ≡ s (( a − p, ( a − p − l ( p − ≡ (cid:18) a − l (cid:19) s ( p, l (mod p ) . (4.16)For any integer i with 2 ≤ i ≤ p −
1, one has ( l − p −
1) + 1 ≤ k + i − p ≤ l ( p − − k + i − p p − i ∈ V k . So from (4.7), one obtains that p − X i =2 s ( p, i ) s p (( a − p, ap − k − i ) ≡ p ) . (4.17)Thus by (4.14) to (4.17), we derive that s ( ap, ap − k ) ≡ s ( p, (cid:18) a − l − (cid:19) s ( p, l − + (cid:18) a − l (cid:19) s ( p, l ≡ (cid:18) al (cid:19) s ( p, l (mod p )as (4.8) asserted. Thus claim (4.8) is proved when 2 ≤ l ≤ a − Case 3. l = a . Then k = a ( p − p − k < ≤ ap − k = a ≤ p −
1. So by (4.6),one deduces that s ( ap, ap − k ) = s ( ap, a ) = a X i =1 s ( p, i ) s p (( a − p, a − i )= s ( p, s p (( a − p, a −
1) + a X i =2 s ( p, i ) s p (( a − p, a − i ) . (4.18)Using Lemma 2.11 and the inductive hypothesis of claim (4.8), we obtain that s p (( a − p, a −
1) = s p (( a − p, ( a − p − ( a − p − ≡ s (( a − p, ( a − p − ( a − p − ≡ s ( p, a − (mod p ) . (4.19) N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 17 For any integer i with 2 ≤ i ≤ a ≤ p −
1, one has k + i − p = a ( p − − ( p − i − ≡ i − p − i ∈ V k . Hence by (4.7), one gets that a X i =2 s ( p, i ) s p (( a − p, ap − k − i ) ≡ p ) . (4.20)It then follows from (4.18) to (4.20) that s ( ap, ap − k ) = s ( ap, a ) ≡ s ( p, s p (( a − p, a − ≡ s ( p, a (mod p ) . Namely, claim (4.8) is true when l = a . This completes the proof of part (i).Finally, we prove part (ii) for the case of even number k . (ii). Let k be an even integer such that 2 ≤ k ≤ a ( p −
1) and k p − l with 1 ≤ l ≤ a such that ( l − p − ≤ k ≤ l ( p − − h k i = k − ( l − p − l − p −
1) + 2 ≤ k ≤ l ( p − − l with 1 ≤ l ≤ a , then s ( ap, ap − k ) ≡ a (cid:18) a − l − (cid:19) s ( p, l − s ( p, p − h k i ) (mod p ) . (4.21)Since 3 ≤ p −h k i ≤ p − h k i is even, by Corollary 2.10, one gets that v p ( s ( p, p −h k i )) ≥ v p ( B h k i ) = 0. Thus for any even integer k with2 ≤ k ≤ a ( p − v p ( s ( ap, ap − k )) ≥ k p −
1) with the equality being true if and only if v p ( B h k i ) = 0. So to finish theproof of part (ii), it remains to show the truth of claim (4.21). We proceed this withinduction on the integer a with 2 ≤ a ≤ p − a = 2, one has l = 1 or l = 2. If l = 1, then 2 ≤ k ≤ p −
3. So h k i = k ,3 ≤ p − k ≤ p − p + 3 ≤ p − k ≤ p −
2. Since s p ( p, p ) = s ( p, p ) = 1, it followsfrom (4.6) that s (2 p, p − k ) = p X i = p − k s ( p, i ) s p ( p, p − k − i )= s ( p, p − k ) + s p ( p, p − k ) + p − X i = p − k +1 s ( p, i ) s p ( p, p − k − i ) . (4.22)For any integer i with 4 ≤ p − k + 1 ≤ i ≤ p −
1, we have 1 ≤ k + i − p ≤ p −
4. It impliesthat i ∈ V k . Hence (4.7) tells us that s ( p, i ) s p ( p, p − k − i ) ≡ p ). So it followsthat p − X i = p − k +1 s ( p, i ) s p ( p, p − k − i ) ≡ p ) . (4.23)Since k is even, by Lemma 2.11, we know that s p ( p, p − k ) ≡ s ( p, p − k ) (mod p ). Thus(4.22) and (4.23) give us that s (2 p, p − k ) ≡ s ( p, p − k ) ≡ (cid:18) (cid:19) s ( p, p − h k i ) (mod p )as (4.21) claimed. Hence claim (4.21) is true when a = 2 and l = 1. If l = 2, then p + 1 ≤ k ≤ p − −
2. So one has h k i = k − ( p − p − k < ≤ p − k ≤ p −
1. Thus by (4.6), we get that s (2 p, p − k ) = p − k X i =1 s ( p, i ) s p ( p, p − k − i )= s ( p, s p ( p, p − k −
1) + p − k − X i =2 s ( p, i ) s p ( p, p − k − i )+ s ( p, p − k − s p ( p,
1) + s ( p, p − k ) s p ( p, . (4.24)Note that 2 p − k − p − ( k − ( p − p − h k i and h k i is even. So by Lemma 2.11,one deduces that s p ( p, ≡ s ( p,
1) (mod p ) and s p ( p, p − k −
1) = s p ( p, p − h k i ) ≡ s ( p, p − h k i ) (mod p ) . (4.25)If 2 ≤ i ≤ p − k − ≤ p −
3, then 3 ≤ k + i − p ≤ p −
2. It infers that i ∈ V k . So from(4.7), one obtains that p − k − X i =2 s ( p, i ) s p ( p, p − k − i ) ≡ p ) . (4.26)Since 4 ≤ p − k ≤ p −
1, one has v p ( s ( p, p − k )) ≥
1. It then follows from the fact v p ( s p ( p, s ( p, p − k ) s p ( p, ≡ p ). Thus by (4.24) to (4.26), wearrive at s (2 p, p − k ) ≡ s ( p, s ( p, p − ( k − ( p − ≡ (cid:18) (cid:19) s ( p, s ( p, p − h k i ) (mod p ) . This completes the proof of claim (4.21) when a = 2.In what follows, we let 3 ≤ a ≤ p −
1. Assume that claim (4.21) holds for the a − a case, we consider the following threecases. Case 1. l = 1. Then 2 ≤ k ≤ p −
3. So h k i = k , p − k ≥ a − p + 3 ≤ ap − k ≤ ap −
2. Since s p (( a − p, ( a − p ) = s ( p, p ) = 1, by (4.6) we get that s ( ap, ap − k )= p X i = p − k s ( p, i ) s p (( a − p, ap − k − i )= s ( p, p − k ) + s p (( a − p, ( a − p − k ) + p − X i = p − k +1 s ( p, i ) s p (( a − p, ap − k − i ) . (4.27)Note that k is even and 2 ≤ k ≤ p − ≤ ( a − p −
1. So by using Lemma 2.11 and theinductive hypothesis of claim (4.21), one derives that s p (( a − p, ( a − p − k ) ≡ s (( a − p, ( a − p − k ) ≡ ( a − s ( p, p − k ) (mod p ) . (4.28) N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 19 For any integer i with 4 ≤ p − k + 1 ≤ i ≤ p −
1, we have 1 ≤ k + i − p ≤ p −
4, and so i ∈ V k . By (4.7), one gets that p − X i = p − k +1 s ( p, i ) s p ( p, p − k − i ) ≡ p ) . (4.29)Hence it follows from (4.27) to (4.29) that s ( ap, ap − k ) ≡ s ( p, p − k ) + ( a − s ( p, p − k ) ≡ as ( p, p − k ) ≡ a (cid:18) a − (cid:19) s ( p, p − h k i ) (mod p )as (4.21) claimed. Hence claim (4.21) is proved when l = 1. Case 2. ≤ l ≤ a −
1. Then ( l − p −
1) + 2 ≤ k ≤ l ( p − −
2. Hence one has h k i = k − ( l − p − p − k < p + 4 ≤ ap − k ≤ ap −
3. From (4.6) and s ( p, p ) = 1we know that s ( ap, ap − k ) = p X i =1 s ( p, i ) s p (( a − p, ap − k − i )= s ( p, s p (( a − p, ap − k −
1) + s p (( a − p, ( a − p − k )+ p − X i =2 s ( p, i ) s p (( a − p, ap − k − i ) . (4.30)Notice that ap − k − a − p − ( k − ( p − l − p −
1) + 2 ≤ k ≤ l ( p − − ≤ l ≤ a −
1. So 2 ≤ ( l − p −
1) + 2 ≤ k − ( p − ≤ ( l − p − − h k − ( p − i = h k i . Thus using Lemma 2.11 and the induction assumption of claim(4.21), we obtain that s p (( a − p, ap − k −
1) = s p (( a − p, ( a − p − ( k − ( p − ≡ s (( a − p, ( a − p − ( k − ( p − ≡ ( a − (cid:18) a − l − (cid:19) s ( p, l − s ( p, p − h k i ) (mod p ) (4.31)and s p (( a − p, ( a − p − k ) ≡ s (( a − p, ( a − p − k ) ≡ ( a − (cid:18) a − l − (cid:19) s ( p, l − s ( p, p − h k i ) (mod p ) . (4.32)For any integer i with 2 ≤ i ≤ p −
1, one has ( l − p − ≤ k + i − p ≤ l ( p − −
3. If k + i − p ≡ p − k + i − p = ( l − p − i = p − ( k − ( l − p − p − h k i and ap − k − i = ( a − p − ( l − p − k + i − p p − i = p − h k i and i ∈ V k . It implies that s ( p, i ) s p (( a − p, ap − k − i ) ≡ p ) when2 ≤ i ≤ p − i = p − h k i . Hence one gets that p − X i =2 s ( p, i ) s p (( a − p, ap − k − i ) ≡ s ( p, p − h k i ) s p (( a − p, ( a − p − ( l − p − p ) . (4.33) Since 2 ≤ l ≤ a − l − p −
1) is even, it follows from the truth of Lemma 2.11and claim (4.8) that s p (( a − p, ( a − p − ( l − p − ≡ s (( a − p, ( a − p − ( l − p − ≡ (cid:18) a − l − (cid:19) s ( p, l − (mod p ) . (4.34)Therefore by (4.30) to (4.34), we conclude that s ( ap, ap − k ) ≡ (cid:16) ( a − (cid:18) a − l − (cid:19) + ( a − (cid:18) a − l − (cid:19) + (cid:18) a − l − (cid:19)(cid:17) s ( p, l − s ( p, p − h k i ) ≡ a (cid:18) a − l − (cid:19) s ( p, l − s ( p, p − h k i ) (mod p )as (4.21) asserted. Thus claim (4.21) is proved when 2 ≤ l ≤ a − Case 3. l = a . Then ( a − p −
1) + 2 ≤ k ≤ a ( p − −
2. So h k i = k − ( a − p − p − k < ≤ a + 2 ≤ ap − k ≤ p + a − ≤ p −
4. From (4.6), one gets that s ( ap, ap − k )= min { p,ap − k } X i =1 s ( p, i ) s p (( a − p, ap − k − i )= s ( p, s p (( a − p, ap − k −
1) + X i ∈ W k s ( p, i ) s p (( a − p, ap − k − i ) , (4.35)where W k := { i ∈ Z : 2 ≤ i ≤ min { p, ap − k }} .Since ap − k − a − p − ( k − ( p − a − p −
1) + 2 ≤ k − ( p − ≤ ( a − p − −
2, it follows from Lemma 2.11 and the inductive hypothesis of claim(4.21) that s p (( a − p, ap − k −
1) = s p (( a − p, ( a − p − ( k − ( p − ≡ s (( a − p, ( a − p − ( k − ( p − ≡ ( a − s ( p, a − s ( p, p − h k i ) (mod p ) . (4.36)Since 2 ≤ h k i = k − ( a − p − ≤ p − a ≥
3, one has 2 ≤ p − h k i ≤ p − p − h k i ≤ ap − k −
2. It implies that p − h k i ∈ W k . For the integer i with i = p − h k i , wehave ap − k − i = ap − k − ( p − h k i ) = ( a − p − ( a − p − s p (( a − p, ap − k − ( p − h k i )) = s p (( a − p, ( a − p − ( a − p − ≡ s (( a − p, ( a − p − ( a − p − ≡ s ( p, a − (mod p ) . (4.37)If a + 2 ≤ ap − k ≤ p −
1, then W k = { i ∈ Z : 2 ≤ i ≤ ap − k } . For any integer i with i ∈ W k , one has ( a − p −
1) + 3 ≤ k + i − p ≤ a ( p − −
3. So if k + i − p ≡ p − k + i − p = ( a − p − i = p − ( k − ( a − p − p − h k i . If k + i − p p − i = p − h k i and i ∈ V k , which infers that N THE p -ADIC PROPERTIES OF STIRLING NUMBERS OF THE FIRST KIND 21 s ( p, i ) s p (( a − p, ap − k − i ) ≡ p ). Hence one obtains that X i ∈ W k s ( p, i ) s p (( a − p, ap − k − i ) ≡ s ( p, p − h k i ) s p (( a − p, ap − k − ( p − h k i )) ≡ s ( p, a − s ( p, p − h k i ) (mod p ) . (4.38)If p ≤ ap − k ≤ p + a −
3, then W k = { i ∈ Z : 2 ≤ i ≤ p } and ( a − p −
1) + 2 ≤ k ≤ ( a − p . Since k is even, one has ( a − p −
1) + 2 ≤ k ≤ ( a − p − k = ( a − p .So by the truth of part (iii) and s (( a − p,
0) = 0, we know that v p ( s (( a − p, ( a − p − k )) ≥ a − k − ( a − p ≥ . It then follows from Lemma 2.11 that s p (( a − p, ( a − p − k ) ≡ s (( a − p, ( a − p − k ) ≡ p ) . (4.39)Likewise, for any integer i with i ∈ W k , one has k + i − p ≡ p −
1) if and only if i = p − h k i . Let U k := { i ∈ W k : i = p − h k i and i = p } . Then ∅ 6 = U k ⊆ V k . Hence by(4.7) together with (4.37) and (4.39), one gets that X i ∈ W k s ( p, i ) s p (( a − p, ap − k − i )= X i ∈ U k s ( p, i ) s p (( a − p, ap − k − i ) + s ( p, p ) s p (( a − p, ( a − p − k )+ s ( p, p − h k i ) s p (( a − p, ap − k − ( p − h k i )) ≡ s ( p, p − h k i ) s p (( a − p, ap − k − ( p − h k i )) ≡ s ( p, a − s ( p, p − h k i ) (mod p ) . (4.40)Therefore (4.35) and (4.36) together with (4.38) and (4.40) give us that s ( ap, ap − k ) ≡ ( a − s ( p, a − s ( p, p − h k i ) + s ( p, p − h k i ) s ( p, a − ≡ as ( p, a − s ( p, p − h k i ) ≡ a (cid:18) a − a − (cid:19) s ( p, a − s ( p, p − h k i ) (mod p )as desired. The proof of claim (4.21) is complete. So part (ii) is proved.This finishes the proof of Theorem 1.1. (cid:3) Concluding remarks
In [27], we gave a formula for v ( s (2 n , k )) with k being an integer such that 1 ≤ k ≤ n .In [26], Qiu, Feng and Hong presented a formula for v ( s ( a n , k )) with k being an integersuch that 1 ≤ k ≤ a n , where a ∈ { , } . In this paper, we arrive at an exact expressionor a lower bound of v p ( s ( ap, k )) with a and k being integers such that 1 ≤ a ≤ p − ≤ k ≤ ap . It is natural to consider the p -adic valuation of the Stirling number s ( ap n , k ),where a, n and k being integers such that 1 ≤ a ≤ p − , n ≥ ≤ k ≤ ap n . Forany odd prime p and any positive integer k , recall that ǫ k is defined by ǫ k := 0 if k iseven and ǫ k := 1 if k is odd, and h k i denotes the integer such that 0 ≤ h k i ≤ p − k ≡ h k i (mod p − Conjecture 5.1.
Let p be an odd prime. Let a, n, m, k be positive integers such that ≤ a ≤ p − , ≤ m ≤ n and ≤ k ≤ ap n − . Then each of the following is true: (i). If ≤ k ≤ a ( p − p m − + 1 < ap m , then v p ( s ( ap n , ap m − k )) = ap − p n − p m ) − ( n − m )( ap m − k ) + m + ( m + v p ( k )) ǫ k + T k , where T k := n − − v p ( ⌊ k ⌋ ) , if k ≡ ǫ k (mod p − v p ( B ⌊ h k i ⌋ ) , if k ǫ k (mod p − . (ii). If a ≥ and a ( p −
1) + 2 ≤ k ≤ ap − , then v p ( s ( ap n , ap − k )) ≥ ap − p n − p ) − ( n − ap − k ) + a + k − ap. From Theorem 1.1, we can see that for all primes p ≥ n = 1 and part (i) of Conjecture 5.1 also holds for n = 1 and k ≡ ǫ k (mod p − p = 3.Letting m = n , Conjecture 5.1 becomes the following conjecture. Conjecture 5.2.
Let p be an odd prime. Let a, n, k be positive integers such that ≤ a ≤ p − and ≤ k ≤ a ( p − p n − + 1 . Then v p ( s ( ap n , ap n − k )) = n n + ( n + v p ( k )) ǫ k − − v p ( ⌊ k ⌋ ) , if k ≡ ǫ k (mod p − n + ( n + v p ( k )) ǫ k + v p ( B ⌊ h k i ⌋ ) , if k ǫ k (mod p − . On the other hand, Corollary 4 in [16] gives us that v p ( s ( ap n , ap m )) = ap − p n − p m ) − a ( n − m ) p n . So we suggest the following conjecture as the conclusion of this paper.
Conjecture 5.3.
Let p be a prime. Let a, n, m and k be positive integers such that ≤ a ≤ p − , ≤ m ≤ n and ≤ k ≤ a ( p − p m − + 1 < ap m . Then v p ( s ( ap n , ap m − k )) = v p ( s ( ap n , ap m )) + v p ( s ( ap n , ap n − k )) + (cid:0) j k k − (cid:1) ( n − m ) . Acknowledgement
The authors would like to thank the anonymous referee for careful reading of the manu-script and helpful comments and suggestions.
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Mathematical College, Sichuan University, Chengdu 610064, P.R. China
E-mail address : [email protected]; [email protected]; [email protected] Mathematical College, Sichuan University, Chengdu 610064, P.R. ChinaSchool of Science, Xihua University, Chengdu 610039, P.R. China
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