Tewodros Amdeberhan, Dante Manna and Victor H. Moll Department of Mathematics, Tulane University New Orleans, LA 70118

Transcription

1 The -adic valuation of Stirling numbers Tewodros Amdeberhan, Dante Manna and Victor H. Moll Department of Mathematics, Tulane University New Orleans, LA 7011 Abstract We analyze properties of the -adic valuations of the Stirling numbers of the second kind. 1 Introduction Divisibility properties of integer sequences have always been objects of interest for number theorists. Nowadays these are expressed in terms p-adic valuations. Given a prime p and a positive integer m, there exist unique integers a, n, with a not divisible by p and n 0, such that m = ap n. The number n is called the p-adic valuation of m, denoted by n = ν p (m). Thus, ν p (m) is the highest power of p that divides m. The graph in figure 1 shows the function ν (m) Figure 1: The -adic valuation of m 1

2 One of the most celebrated examples is the expression for the p-adic valuation of factorials. This is due to Legendre [], who established Here s p (m) is the sum of the base p-digits of m. In particular, ν p (m!) = m s p(m). (1.1) p 1 ν (m!) = m s (m). (1.) The reader will find in [7] details about this identity. Figure shows the graph of ν (m!) exhibiting its linear growth: ν (m!) m. This can be verified using the binary expansion of m: m = a 0 + a 1 + a a r r, with a j {0, 1}, a r 0, so that r m < r+1. Therefore s (m) = O(log (m)) and we have ν (m!) lim = 1. (1.) m m The error term s (m) = m ν (m!), shown in Figure, shows a regular pattern Figure : The -adic valuation of m!. Legendre s result (1.) provides an elementary proof of Kummer s identity (( )) m ν = s (k) + s (m k) s (m). (1.) k Not many explicit identities of this type are known. The function ν p is extended to Q by defining ν p ( a b ) = νp (a) ν p (b). The p-adic metric is then defined by r p := p νp(r), for r Q. (1.5) It satisfies the ultrametric inequality r 1 + r 1 p Max { r 1 p, r p }. (1.) The completion of Q under this metric, denoted by Q p, is the field of p-adic numbers. The set Z p := {x Q p : x p 1} is the ring of p-adic integers. It plays the role of Z inside Q p.

3 Our interest in -adic valuations started with the sequence m ( )( m k m + k b l,m := k m k m k=l )( k l for m N and 0 l m, that appeared in the evaluation of the definite integral N 0, (a; m) = In [], it was shown that the polynomial defined by satisfies 0 ), (1.7) dx (x + ax. (1.) + 1) m+1 m P m (a) := m b l,m a l (1.9) l=0 P m (a) = m+/ (a + 1) m+1/ N 0, (a; m)/π. (1.) The reader will find in [] more details on this integral. The results on the -adic valuations of b l,m are expressed in terms of The coefficients A l,m can be written as A l,m := l! m! m l b l,m. (1.11) m m A l,m = α l (m) (k 1) β l (m) (k + 1), (1.1) k=1 for some polynomials α l, β l with integer coefficients and of degree l and l 1, respectively. The next remarkable property was conjectured in [] and established by J. Little in []. k=1 Theorem 1.1. All the zeros of α l (m) and β l (m) lie on the vertical line Re m = 1. The next theorem, presented in [1], gives -adic properties of A l,m. Theorem 1.. The -adic valuation of A l,m satisfies ν (A l,m ) = ν ((m + 1 l) l ) + l, (1.1) where (a) k = a(a + 1)(a + ) (a + k 1) is the Pochhammer symbol. The identity (a) k = (a + k 1)! (a 1)! and Legendre s identity (1.) yields the next expression for ν (A l,m ). (1.1)

4 Corollary 1.. The -adic valuation of A l,m is given by ν (A l,m ) = l s (m + l) + s (m l). (1.15) Among the other examples of -adic valuations, we mention the results of H. Cohen [] on the partial sums of the polylogarithmic series Li k (x) := Cohen proves that the sum 1 Lk(n) := satisfies and j=1 n j=1 x j j k. (1.1) j j k (1.17) ν (L 1 ( m )) = m + m, for m, (1.1) ν (L ( m )) = m + m 1, for m. (1.19) The graph in figure shows the linear growth of ν (L 1 (m)) and the error term ν (L 1 (m)) m Figure : The -adic valuation of L 1 (m) In this paper we analyze the -adic valuation of the Stirling numbers of the second kind S(n, k), defined for n N and 0 k n as the number of ways to partition a set of n elements into exactly k nonempty subsets. Figure shows the function ν (S(n, k)) for k = 75 and k = 1. These graphs indicate the complexity of this problem. Section gives a larger selection of these type of pictures. Main conjecture. We describe an algorithm that leads to a description of the function ν (S(n, k)). Figure shows the graphs of this function for k = 75 and k = 1. The conjecture is stated here and the special case k = 5 is established in section. 1 Cohen uses the notation s k (n), employed here in a different context.

5 Figure : The data for k = 75 and k = 1. Definition 1.. Let k N be fixed and m N. Then for 0 j < m define C m,j := { m i + j : for i N such that m i + j k}. (1.0) For example, for k = 5, m = and j = we have C, = { i + : i 0}. (1.1) We use the notation ν (C m,j ) = {ν (S( m i + j, k)) : i N and m + i k}. (1.) The classes C m,j form a partition of {n N : n k} into classes modulo m. For example, for m = and k = 5, we have the four classes C,0 = { i : i N, i }, C, = { i + : i N}, C,1 = { i + 1 : i N}, C, = { i + : i N}. The class C m,j is called constant if ν (C m,j ) consists of a single value. This single value is called the constant of the class C m,j. For example, Corollary. shows that ν (S(i + 1, 5)) = 0, independently of i. Therefore, the class C,1 is constant. Similarly C, has constant valuation 0. We now introduce inductively the concept of m-level. For m = 1, the 1-level consists of the two classes C 1,0 = {i : i N, i k} and C 1,1 = {i + 1 : i N, i + 1 k}, (1.) that is, the even and odd integers greater or equal than k. Assume that the (m 1) level has been defined and it consists of the s classes C m 1,i1, C m 1,i,, C m 1,is. (1.) Each class C m 1,ij splits into two classes modulo m, namely, C m,ij and C m,ij+m 1. The m-level is formed by the non-constant classes modulo m. Example. We describe the case of Stirling numbers S(n, ). Start with the fact that the -level consists of the classes C,7, C,, C,9 and C,1. These split into the eight classes C 5,7, C 5,, C 5,, C 5,, C 5,9, C 5,5, C 5,1, and C 5,0, 5

6 modulo. Then one checks that C 5,, C 5,, C 5,5 and C 5,0 are all constant (with constant value for each of them). The other four classes form the 5-level: We are now ready to state our main conjecture. {C 5,7, C 5,, C 5,9, C 5,1 }. (1.5) Conjecture 1.5. Let k N be fixed. Define m 0 = m 0 (k) N by m0 1 < k m0. Then the -adic valuation of the Stirling numbers of the second kind S(n, k) satisfies: 1) The first index for which there is a constant class is m 0 1. That is, every class is part of the j-level for 1 j m 0. ) For any m m 0, the m-level consists of m0 classes. Each one of them produces a single non-constant class for the (m + 1)-level, keeping the number of classes at each level constant. Example. We illustrate this conjecture for S(n, 11). Here m 0 = in view of < 11. The conjecture predicts that the -level is the first with constant classes and that each level after that has exactly four classes. First of all, the four classes C,0, C,1, C,, C, have non-constant -adic valuation. Thus, they form the -level. To compute the -level, we observe that ν (C, ) = ν (C,5 ) = {0} and ν (C, ) = ν (C, ) = {1}, so there are four constant classes. The remaining four classes C,0, C,1, C, and C,7 form the -level. Observe that each of the four classes from the -level splits into a constant class and a class that forms part of the -level. This process continues. At the next step, the classes of the -level split in two giving a total of classes modulo. For example, C, splits into C, and C,. The conjecture states that exactly one of these classes has constant -adic valuation. Indeed, the class C, satisfies ν (C, ). Figure 5 illustrates this process. Elementary formulas. Throughout the paper we will use several elementary properties of S(n, k), listed below: Relation to Pochhammer n x n = S(n, k)(x k + 1) k (1.) k=0 An explicit formula S(n, k) = 1 k 1 ( ) k ( 1) i (k i) n (1.7) k! i i=0 The generating function 1 (1 x)(1 x)(1 x) (1 kx) = S(n, k)x n (1.) n=1

7 Figure 5: The splitting for k = 11 The recurrence S(n, k) = S(n 1, k 1) + ks(n 1, k) (1.9) Lengyel [9] conjectured, and De Wannemacker [1] proved, a special case of the -adic valuation of S(n, k): ν (S( n, k)) = s (k) 1, (1.0) independently of n. Here s (k) is the sum of the binary digits of k. Legendre s result (1.) then shows that ν (S( n, k)) = ν ( k 1 /k!). (1.1) From here we obtain as a companion of (1.0). Indeed, the recurrence (1.9) yields and using (1.1) we have ν (S( n + 1, k + 1) = s (k) 1 (1.) S( n + 1, k + 1) = S( n, k) + (k + 1)S( n, k + 1), (1.) ν ((k + 1)S( n, k + 1)) = ν ( k /k!) > ν ( k 1 /k!) = ν (S( n, k)), as claimed. Similar arguments can be used to obtain the value of ν (S(m, k)) for values of m near a power of. For instance, ν (S( n +, k + )) = s (k) 1 if ν (k) 0 and ν (S( n +, k + )) = s (k + 1) 1 if ν (k + 1) > 1. 7

8 In the general case, De Wannemacker [1] established the inequality ν (S(n, k)) s (k) s (n), 0 k n. (1.) Figure shows the difference ν (S(n, k)) s (k) + s (n) in the cases k = 1 and k = Figure : De Wannemacker difference for k = 1 and k = 19 The elementary cases This section presents, for sake of completeness, the -adic valuation of S(n, k) for 1 k. The formulas for S(n, k) come from (1.7). The arguments are all elementary. Lemma.1. The Stirling numbers of order 1 are given by S(n, 1) = 1, for all n N. Therefore ν (S(n, 1)) = 0. (.1) Lemma.. The Stirling numbers of order are given by S(n, ) = n 1, for all n N. Therefore ν (S(n, )) = 0. (.) Lemma.. The Stirling numbers of order are given by S(n, ) = 1 (n 1 n + 1). (.) Moreover, ν (S(n, )) = { 0 if n is odd, 1 if n is even. (.)

9 Proof. Iterate the recurrence (1.9) to obtain and with N = n 1 we have N 1 n 1 = S(n, ) k ( n k 1) N S(n N, ), (.5) k=1 If n is odd, then S(n, ) is odd and ν (S(n, )) = 0. For n even, the recurrence (1.9) yields n S(n, ) = n 1 k ( n k 1). (.) k=1 S(n, ) = n 1 + n + S(n, ). (.7) As an inductive step, assume that S(n, ) = T n, with T n odd. Then (.7) yields 1 S(n, ) = n + n + T n, (.) and we conclude that S(n, )/ is an odd integer. Therefore ν (S(n, )) = 1 as claimed. We now present a second proof of this result using elementary properties of the valuation ν. In particular, we use the ultrametric inequality ν (x 1 + x ) Min {ν (x 1 ), ν (x ) }. (.9) The inequality is strict unless ν(x 1 ) = ν (x ). This inequality is equivalent to (1.). The powers of modulo satisfy m ( 1) m+1 mod, because k 1 mod. Therefore, m + 1 = t + + ( 1) m+1 for some t Z. Now ν (t) = + ν (t) > ν ( + ( 1) m+1 ), (.) and the ultrametric inequality (.9) yields ν ( m + 1) = ν ( + ( 1) m+1 ) = { if m is odd, 1 if m is even. (.11) Using S(n, ) = n n and ν ( n ) = n > ν ( n 1 + 1), we conclude that ν (S(n, )) = ν ( n n ) 1 = ν ( n 1 + 1) 1. (.1) Lemma. now follows from (.11). We now discuss the Stirling number of order. Lemma.. The Stirling numbers of order are given by S(n, ) = 1 (n 1 n n+1 1). (.1) Moreover, ν (S(n, )) = { 1 if n is odd, 0 if n is even. (.1) 9

10 Proof. The expression for S(n, ) comes from (1.7). To establish the formula for ν (S(n, )), we use the recurrence (1.9) in the case k = : S(n, ) = S(n 1, ) + S(n 1, ). (.15) For n even, the value S(n 1, ) is odd, so that S(n, ) is odd and ν (S(n, )) = 0. For n odd, S(n, ) is even, since S(n 1, ) is even. Then (.15), written as 1 S(n, ) = 1 S(n 1, ) + S(n 1, ), (.1) and the value ν (S(n 1, )) = 1, show that the right hand side of (.1) is odd, yielding ν (S(n, )) = 1. The Stirling numbers of order 5 The elementary cases discussed in the previous section are the only ones for which the -adic valuation ν (S(n, k)) is easy to compute. The graph in figure 7 shows ν (S(n, 5)) The explicit formula (1.7) yields Figure 7: The -adic valuation of S(n, 5) S(n, 5) = 1 (5n 1 n + n n+1 + 1). (.1) We now discuss the valuation ν (S(n, 5)) in terms of the m-levels introduced in Section 1. The 1-level consists of two classes: {C 1,0, C 1,1 }. None of these classes are constant, so we split them into {C,0, C,1, C,, C, }. The parity of S(n, 5) determines two of them. Lemma.1. The Stirling numbers S(n, 5) satisfy { 1 mod if n 1, or mod, S(n, 5) 0 mod if n, or 0 mod, (.) for all n N.

11 Proof. The recurrence S(n, 5) = S(n 1, ) + 5S(n 1, 5) and the parity { 1 mod if n 0 mod, S(n, ) 0 mod if n 1 mod, (.) give the result by induction. Corollary.. The -adic valuations of the Stirling numbers S(n, 5) satisfy ν (S(n + 1, 5)) = ν (S(n +, 5)) = 0 for all n N. (.) The corollary states that the classes C,1 and C, are constant, so the level is level : {C,0, C, }. (.5) This confirms part of the main conjecture: here m 0 = in view of < 5 and the first level where we find constant classes is m 0 1 =. Remark. Corollary. reduces the discussion of ν (S(n, 5)) to the indices n 0 or mod. These two branches can be treated in parallel. Introduce the notation and consider the table of values This starts as and after a while it continues as We observe that q i = q i+ for most indices. q n := ν (S(n, 5)), (.) X := {q i, q i+ : i }. (.7) X = {1, 1,,, 1, 1,,, 1, 1,, 7, 1, 1,...}, (.) X = {...,1, 1,,, 1, 1, 11,, 1, 1,,,...}. (.9) Definition.. The index i is called exceptional if q i q i+. The first exceptional index is i = 7 where q = 11 q 1 =. The list of exceptional indices continues as {7, 9, 71,,...}. Conjecture.. The set of exceptional indices is {j + 7 : j 1}. 11

12 We now consider the class C,0 := {q i = ν (S(i), 5) : i }, (.) where we have omitted the first term S(, 5) = 0. The class C,0 starts as C,0 = {1,, 1,, 1,, 1,, 1,, 1,, 1,, 1,, 1,, 1,,...}, (.11) and it splits according to the parity of the index i into C, = {q i+ : i 1} and C,0 = {q i : i 1}. (.1) It is easy to check that C,0 is constant. Proposition.5. The Stirling numbers of order 5 satisfy ν (S(i, 5)) = 1 for all i 1. (.1) Proof. We analyze the identity S(i, 5) = 5 i 1 i + i i (.1) modulo. Using 5 1 and 5 7 1, we obtain 5 i 1 1. Also, i i+1 mod 0. Finally, i 1 i 17 i 1. Therefore 5 i 1 i + i i mod. (.15) We obtain that S(i, 5) = t+1 for some t N, and this yields S(i, 5) = (t+1). Therefore ν (S(i, 5)) = 1. We now consider the class C,. Proposition.. The Stirling numbers of order 5 satisfy ν (S(i +, 5)) for all i 1. (.1) Proof. We analyze the identity S(i +, 5) = 5 i+ i+ + i+ i (.17) modulo. Using 5 1, 5 9, 1, 17 and 1 modulo, we obtain S(i +, 5) 0 mod. (.1) Therefore S(i +, 5) = t for some t N, and this yields ν (S(i +, 5). 1

13 Note. Lengyel [9] established that ν (k!s(n, k)) = k 1, (.19) for n = a q, a odd and q k. In the special case k = 5, this yields ν (S(n, 5)) = 1 for n = a q and q. These values of n have the form n = a q, so this is included in Proposition.5. Remark. A similar argument yields We conclude that ν (S(i +, 5)) = 1 and ν (S(i + 7, 5)). (.0) level : {C,, C,7 }. (.1) This illustrates the main conjecture: each of the classes of the -level produces a constant class and a second one in the -level. We now consider the class C, and its splitting as C, and C,1. The data for C, starts as C, = {,,,,,,,,,, 5,,,,,,,, 11,,,,...}. (.) This suggests that the values with even index are all. This is verified below. Proposition.7. The Stirling numbers of order 5 satisfy ν (S(1i +, 5)) = for all 1. (.) Proof. We analyze the identity S(1i +, 5) = 5 1i+ 1i+ + 1i+ 1i (.) modulo. Using 5 1 1, 5 1, 1 1 and 17, we obtain 5 1i+ 1i+ + 1i+ 1i mod. (.5) Therefore S(1i +, 5) = t + for some t N. This gives S(1i +, 5) = (t + 1), and it follows that ν (S(1i +, 5)) =. Note. A similar argument shows that ν (S(1i + 1, 5)), ν (S(1i + 7, 5)) = and ν (S(1i + 15, 5)). Therefore the -level is {C,1, C,15 }. This splitting process of the classes can be continued and, according to our main conjecture, the number of elements in the m-level is always constant. To prove the statement similar to Propositions.5 and.7, we must analyze the congruence S( m i + j, 5) 5 m i+j 1 m i+j + m i+j m i+j mod m+. (.) We present a proof of this conjecture, for the special case k = 5, in the next section. Lundell [11] studied the Stirling-like numbers T p (n, k) = k ( ) k ( 1) k j j n, (.7) j j=0 1

14 where the prime p is fixed, and the index j is omitted in the sum if it is divisible by p. Clarke [5] conjectured that ν p (k! S(n, k)) = ν p (T(n, k)). (.) From this conjecture he derives an expression for ν (S(n, 5)) in terms of the zeros of the form f 0,5 (x) = 5 + x + 5 x in the ring of -adic integers Z. Theorem.. Let u 0 and u 1 be the -adic zeros of the function f 0,5. Then, under the assumption that conjecture (.) holds, we have { 1 + ν (n n 0 ) if n is even, ν (S(n, 5)) = (.9) 1 + ν (n n 1 ) if n is odd. Here u 0 is the unique zero of f 0,5 that satsifies u 0 Z and u 1 is the other zero of f 0,5 and satisfies u Z. Clarke also obtained in [5] similar expressions for ν (S(n, )) and ν (S(n, 7)) in terms of zeros of the functions f 0, = 0 x 5 x and f 0,7 = x x + 7 x. Proof of the main conjecture for k = 5 The goal of this section is to prove the main conjecture in the case k = 5. The parameter m 0 is in view of < 5. In the previous section we have verified that m 0 1 = is the first level for constant classes. We now prove this splitting of classes. Theorem.1. Assume m m 0. Then the m-level consists of exactly two split classes: C m,j and C m,j+ m 1. They satisfy ν (C m,j ) > m and ν (C m,j+ m 1) > m. Then exactly one, call it C 1, satisfies ν (C 1 ) = {m } and the other one, call it C, satisfies ν (C ) > m. The proof of this theorem requires several elementary results of -adic valuations. Lemma.. For m N, ν ( 5 m 1 ) = m +. Proof. Start at m = 1 with ν () =. The inductive step uses 5 m+1 1 = (5 m 1) (5 m + 1). Now 5 k + 1 mod so that 5 m + 1 = α 1 with α 1 odd. Thus ν (5 m+1 1) = ν (5 m 1) + ν (5 m + 1) = (m + ) + 1 = m +. 1

15 The same type of argument produces the next lemma. Lemma.. For m N, ν ( m 1) = m +. Lemma.. For m N, ν (5 m m ) = m +. Proof. The inductive step uses 5 m+1 m+1 = (5 m m ) ( ) (5 m 1) + ( m + 1). Therefore ν (5 m 1) = m + and m 1 mod, thus ν ( m + 1) = 1. We conclude that We obtain and this concludes the inductive step. ν ((5 m 1) + ( m + 1)) = Min{m +, 1} = 1. ν (5 m+1 m+1 ) = m +, (.1) The recurrence (1.9) for the Stirling numbers S(n, 5) is S(n, 5) = 5S(n 1, 5) + S(n 1, ). Iterating yields the next lemma. Lemma.5. Let t N. Then t 1 S(n, 5) 5 t S(n t, 5) = 5 j S(n j 1, ). (.) j=0 Proof of theorem.1. We have already checked the conjecture for the -level. The inductive hypothesis states that there is an (m 1)-level survivor of the form C m,k = {ν (S( m n + k, 5)) : n 1}, (.) where ν (S( m n + k, 5)) > m. At the next level, C m,k splits into the two classes C m+1,k = {ν (S( m+1 n + k, 5)) : n 1} and C m+1,k+ m = {ν (S( m+1 n + k + m, 5)) : n 1}, and every element of each of these two classes is greater or equal to m 1. We now prove that one of these classes reduces to the singleton {m 1} and that every element in the other class is strictly greater than m 1. 15

16 The first step is to use Lemma.5 to compare the values of S( m+1 n + k, 5) and S( m+1 n + k + m, 5). Define M = m 1 and N = m+1 n + k, (.) and use (1.9) to write S( m+1 n + k + m, 5) 5 m S( m+1 n + k, 5) = The next proposition establishes the -adic valuation of the right hand side. M 5 M j S(N + j, ). (.5) j=0 Proposition.. With the notation as above, M 5 M j S(N + j, ) = m 1. (.) ν j=0 Proof. The explicit formula (1.7) yields S(n, ) = n 1 + n 1 n 1. Thus M 5 M j S(N + j, ) = N 1 (5 M+1 M+1 ) + N 1 (5 M+1 M+1 ) j=0 The results in Lemmas.,. and. yield N 1 (5M+1 M+1 ) 1 (5M+1 1). M 5 M j S(N + j, ) = N 1 α 1 + N 1 α N m+ α m α, (.7) j=0 with α j odd integers. Write this as M 5 M j S(N + j, ) = N 1 ( N 1 ) α 1 + α m ( α N + 1 ) T 1 + T. j=0 Then ν (T 1 ) = N 1 > m = ν (T ), and we obtain M 5 M j S(N + j, ) = m 1. (.) We conclude that with α 5 odd. Define ν j=0 S( m+1 n + k + m, 5) 5 m S( m+1 n + k, 5) = m 1 α 5, (.9) X := m+1 S( m+1 n + k + m, 5) and Y := m+1 S( m+1 n + k, 5). (.) Then X and Y are integers and X Y 1 mod, so that they have opposite parity. If X is even and Y is odd, we obtain ν ( S( m+1 n + k + m, 5) ) > m 1 and ν ( S( m+1 n + k, 5) ) = m 1. (.11) The case X odd and Y even is similar. This completes the proof. 1

17 There are four classes at the first level corresponding to the residues modulo, two of which are constant. The complete determination of the valuation ν (S(n, 5)) is now determined by the choice of class when we move from level m to m+1. We consider only the branch starting at indices congruent to 0 modulo ; the case of modulo is similar. Now there is single class per level that we write as C m,j = {q m i+j : i N}, (.1) where j = j(m) is the index that corresponds to the non-constant class at the m-level. The first few examples are listed below. C, = {q i+ : i N} C, = {q i+ : i N} C,1 = {q 1i : i N} C 5, = {q i : i N} C, = {q i : i N} C 7,15 = {q 1i 0 : i N} C,15 = {q 5i 0 : i N} C 9,15 = {q 51i 5 : i N} C,15 = {q i : i N} We have observed a connection between the indices j(m) and the set of exceptional indices I 1 in (5.). Conjecture.7. Construct a list of numbers {c i : i N} according to the following rule: let c 1 = (the first index in the class C, ), and then define c j as the first value on C m,j that is strictly bigger than c j 1. The set C begins as Then, starting at 15, the number c i I 1. C = {, 1,, 0, 9, 15, 1,, 110,...}. (.1) 5 Some approximations In this section we present some approximations to the function ν (S(n, 5)). These approximations were derived empirically, and they support our belief that -adic valuations of Stirling numbers can be well approximated by simple integer combinations of -adic valuations of integers. For each prime p, define λ p (m) = 1 ( 1 ( 1) m mod p ). (5.1) First approximation. Define f 1 (m) := m λ (m) + 50λ (m + 1). (5.) Then ν (S(m, 5)) and ν (f 1 (m)) agree for most values. The first time they differ is at m = 15 where ν (S(15, 5)) ν (f 1 (15)) =. The first few indices for which ν (S(m, 5)) ν (f 1 (m)) are {15, 7, 1,, 799,...}. 17

18 Conjecture 5.1. Define and Then ν (S(m, 5)) = ν (f 1 (m)) unless m I 1. x 1 (m) = m + m + 1 (5.) I 1 = {x 1 (m) : m 0}. (5.) The parity of the exceptions in I 1 is easy to establish: every third element is odd and the even indices of I 1 are on the arithmetic progression 5m Second approximation. We now consider the error Observe that Err 1 (m, 5) = 0 for m outside I 1. and Define Err 1 (m, 5) := ν (S(m, 5)) ν (f 1 (m)). (5.5) m (m) := (m + ) mod, α m := λ (m + )(1 + λ (m)) + λ (m + 1)λ (m), ( ) m f (m) = m + + 0λ (m + 1) + 7λ (m)λ (m). (5.) m The next conjecture improves the prediction of Conjecture 5.1. Conjecture 5.. Consider the set I = {x (m) : m 0}, where Then unless m I. x (m) = m m + 1. (5.7) Err 1 (x 1 (m), 5) = ( 1) αm ν (f (m)), (5.) We now present one final improvement. Define Define β(m) = α m + ( 1) m+1 λ (m) and Err (m, 5) := Err 1 (x 1 (m), 5) ( 1) αm ν f (m)). (5.9) f (m) = 1 λ(m) m + + λ (m)(5λ (m) + λ (m + 1) + λ (m + 1)). (5.) Conjecture 5.. Err (x (m), 5) agrees with ( 1) β(x(m)) ν (f (x (m))) for most values of m N. 1

19 A sample of pictures In this section we present data that illustrate the wide variety of behavior for the -adic valuation of Stirling numbers S(n, k). Several features are common to all. For instance we observe the appearance of an empty region from below the graph. The graph of ν (S(n, 1)) shows the basic function ν (n) in the interior of the graph. The dark objects in ν (S(n, 195)) and ν (S(n, 0)) correspond to an oscillation between two consecutive values. We are lacking an explanation of these features Figure : The data for S(n, 0) and S(n, 1) Figure 9: The data for S(n, 1) and S(n, 195). 7 Conclusions We have presented a conjecture that describes the -adic valuation of the Stirling numbers S(n, k). This conjecture is established for k = 5. Acknowledgements. The last author acknowledges the partial support of NSF-DMS The second author was partially supported as a graduate student by the same grant. The work of the first author was done while visiting Tulane University in the Spring of 00. The authors wish to thank Valerio de Angelis for the diagrams in the paper. 19

20 Figure : The data for S(n, 5) and S(n, 0) Figure 11: The data for S(n, 79) and S(n, ). References [1] T. Amdeberham, D. Manna, and V. Moll. The -adic valuation of a sequence arising from a rational integral. Preprint, 00. [] G. Boros and V. Moll. An integral hidden in Gradshteyn and Ryzhik. Jour. Comp. Applied Math., :1, [] G. Boros and V. Moll. Irresistible Integrals. Cambridge University Press, New York, 1st edition, 00. [] G. Boros, V. Moll, and J. Shallit. The -adic valuation of the coefficients of a polynomial. Scientia, 7:7 50, 001. [5] F. Clarke. Hensel s lemma and the divisibility by primes of Stirling-like numbers. J. Number Theory, 5:9, [] H. Cohen. On the -adic valuation of the truncated polylogarithmic series. Fib. Quart., 7:117 11, [7] R. Graham, D. Knuth, and O. Patashnik. Concrete Mathematics. Addison Wesley, Boston, nd edition,

21 Figure 1: The data for S(n, 5) and S(n, 5). [] A. M. Legendre. Theorie des Nombres. Firmin Didot Freres, Paris, 10. [9] T. Lengyel. On the divisiblity by of the Stirling numbers of the second kind. Fib. Quart., :19 01, 199. [] J. Little. On the zeroes of two families of polynomials arising from certain rational integrals. Rocky Mountain Journal, 5:5 11, 005. [11] A. Lundell. A divisiblity property for Stirling numbers. J. Number Theory, :5 5, 197. [1] S. De Wannemacker. On the -adic orders of Stirling numbers of the second kind. INTEGERS, 5(1):A 1, 005. [1] S. De Wannemacker. Annihilating polynomials for quadratic forms and Stirling numbers of the second kind. Math. Nachrichten, 00. 1