INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (007, #A4 ON DIVISIBILITY OF SOME POWER SUMS Tamás Lengyel Department of Mathematics, Occidental College, 600 Campus Road, Los Angeles, USA lengyel@oxy.edu Received: 5/4/07, Accepted: 9/7/07, Published: 9/9/07 Abstract We determine the exact power of a prime p which divides the power sum n + n + +(b m n provided that m and b are positive integers, p divides b, and m is large enough.. Introduction Let n and k be positive integers, p be a prime, and let d (k and ρ p (k denote the number of ones in the binary 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 rational n/k we set ρ p (n/k = ρ p (n ρ p (k. Let b and m be positive integers. In this paper we determine the p-adic order of n + n + + (b m n for any positive integer n in the exponent, provided that p divides b. Our original motivation was to find the -adic order of the power sum O n ( m = n +3 n + +( m n in order to prove the congruence S(c n, m 3 m mod m+ for Stirling numbers of the second kind, with integer c so that c n > m and m. Thus we first consider the case with p =. We observe that ρ (O n ( m m by an easy induction proof on m. In fact, more can be said. For n even, the same proof yields ρ (O n ( m = m, too. Clearly, O ( m = (m but in general, the odd case seems more difficult. We set S n (x = and determine the exact -adic order of S n ( m by using Bernoulli polynomials in Theorem in Section 3. x k n
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY (007, #A4 We generalize Theorem and its proof in Theorem 3 in Section 4 for any prime p. We also obtain Theorem 4 in order to get a lower bound on the p-adic order of S n (b m and Theorem 5 to determine the exact order for any large enough m.. An Odd Divisibility Property There is a general divisibility property that we can apply here to prove that S (b m S n (b m for n odd. Of course, this already implies that ρ p (S n (b m m. So, in general, we write S n = S n (c where c is an arbitrary odd positive integer. We can easily prove that S n is divisible by S. Note that S = ( c+. Then, by two different grouping of the terms in S n we get (( n ( n c c + c ( n + (c n + ( n + (c n + + + + c n, and (( n ( n ( n c + c c + 3 c + ( n + c n + ( n + (c n + + + +, and the proof is complete since c and c + are relatively prime. We note that Faulhaber had already known in 63 (cf. [] that S n (c can be expressed as a polynomial in S (c and S (c, although with fractional coefficients. In fact, S n (c can be written as a polynomial in c(c + or ( c(c +, if n is even or odd, respectively. This gives rise to the appearance of factors such as b m and b m in S n (b m, depending on whether n is even or odd. 3. The Exact -adic Order Now we discuss the case with p =. Theorem For m and n, we have that { ρ (S n ( m m, if n is even or n =, = (m, if n 3 odd. We note that clearly, S ( m = m ( m. For m =, we have O n ( = S n ( =, and in general, for n, the -adic order of O n ( m and S n ( m are the same, as it easily follows from O n ( m = S n ( m n S n ( m ; thus ρ (O n ( m = ρ (S n ( m.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY (007, #A4 3 Proof of Theorem. The statement is true for n = or m = so we assume that n and m from now on. The Bernoulli polynomials [3] are defined by It is well known [] that B m (x = x m i + ( i ( k (x + k m. ( k k n = B n+(x + B n+ (0. ( n + The usual Bernoulli numbers can be defined as B n = B n (0, and the initial values are B 0 =, B = /, B = /6, B 3 = 0, B 4 = /30, B 5 = 0, etc. Note that B n =0 for every odd integer n 3. We form the difference in the numerator of ( and then, for B n+ (x +, we use the binomial expansion of (x + + k n+ and focus on terms with (x + with small exponents. We have ( ( i ( n + B n+ (x + B n+ (0 = ( (x k + k n+ k n+ i + k =0 ( ( i = ( k (n + (x + k n i + k ( n + ( n + + (x + k n + (x + k n+, =3 so that B n+ (x + B n+ (0 n + = i + ( ( i ( k (x + k n + n k (x + k n ( n + (x + k. n+ =3 Now we rewrite this with x = m and get that S n ( m = i + ( ( i ( k m k n + n k m k n + =3 ( n m k. n+ (3 If n is even then we only need the first term in the last parenthetical expression, otherwise we need the first two terms. Let n be even, then the term with = contributes m i + ( i ( k k n = m B n (0 = m k n i + ( i i!s(n, i (4
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY (007, #A4 4 by identity (, a standard formula for the Stirling numbers of the second kind, and S(n, n + = 0. The other terms are all divisible by m. Clearly, ρ ( i! i+ 0 if i 4. Indeed, in this case i d (i ρ (i + 0 since i log (i +. Therefore, we need only the -adic order of 3 ( i i + i!s(n, i = S(n, + 3 S(n, 3 S(n, 3, 3 n+ +3 = m, by the ( which yields ρ (S n ( m = ρ m (3S(n, 3 + 7 = ρ ( m identity S(n, 3 = (3n n +, n. Note that ρ (S n ( m = ρ ( m B n (0 = m also follows by simply noting the well-known fact about the Bernoulli numbers that ρ (B n (0 = for even n by a theorem by von Staudt [5]. Theorem (von Staudt, [5] For n = and n even, we have B n p prime p n mod. p Proof. Clearly, for n even, the denominator of B n is the product of the primes p with (p n, and thus, it must be square-free. It follows that ρ p (B n for all primes p, and it is nonnegative unless (p n. Now assume that n 3 is odd. The first two terms in the parenthesis of (3 contribute m n+ i i+ ( k( i k k n + m n n+ i i+ ( k( i k k n = m B n + m n n+ ( i i!s(n, i = i+ m B n + m nb n. The -adic order is (m since B n = 0 and ρ (B n = since n 3 is odd. The other terms of (3 with 3 are all divisible by m since m ρ ( > m for m, as in this case. log > m Remark. The above proof can be generalized to the case in which m is replaced by (c m with any odd integer c. 4. The General Case: The Exact p-adic Order We note that S n ( m = n+ ( = m B n n+ by (3 with an observation similar to (4, and in general, for any positive integer b, ( n S n (b m = b m B n+. (5 = We now prove the generalized version of Theorem.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY (007, #A4 5 Theorem 3 For m, n, and b positive integers with p b, p prime, and m = m ρ p (b, we have that m + ρ p (B n, if n =, ρ p (S n (b m = m + ρ p (B n, if n is even and ρ p (B n = 0 or, m + ρ p (B n + ρ p (n/, if n 3 odd and ρ p (B n = 0 or. (6 Proof. We have already proved the statement for p = in Theorem and Remark. If p 3 then the case with n = is easy to check. Thus, we can also assume that n. We now prove the theorem with ρ p (b =, i.e., if m = m. The general case with ρ p (b easily follows by replacing m by m in the proof below. First, if n is even then all terms with 5 on the right hand side of (5 are divisible by p m+ since m ρ p ( m + as for m. If = 3 and p = 3 then +log p m 3m + ρ 3 (B n ρ 3 (3 m + ρ 3 (B n + for m and n even. If = 3 and p 5 then clearly 3m ρ p (3 m +. The term with = works since B n = 0 except for n = when m ρ p ( m +. The term with = 4 also works for n 4 since B n 3 = 0 except for n = 4 when 4m ρ p (4 m +. Next, if n is odd then we have two cases. Case. If n = 3 then for = 3 and 4 we have either p = 3 and thus, m + ρ 3 (B 4 m + ρ 3 (B +, i.e., m m for m ; or p 5 and thus, m + ρ p (B 4 m + ρ p (B +, i.e., m m + again. Case. If n 5 odd then we rewrite ( ( n as n n for. All terms with 5 on the right hand side of (5 are divisible by p m+ρp(n/+ since m + ρ p (n ρ p (( m + ρ p (n/ + as +ρ p(( for m. If p = 3 then for the term with = 4, we m get that 4m + ρ 3 (B n 3 + ρ 3 (n ρ 3 (4 3 m + ρ 3 (B n + ρ 3 (n/ + since 4m m for m and ρ 3 (B k = for k even. If p 5 then for the term with = 4, we get that 4m + ρ p (n m + ρ p (n/ + since 4m m + for m. The term with = 3 makes no contribution to (5 as B n = 0. We obtain a lower bound and the exact p-adic order of S n (b m in the next two theorems. Theorem 4 For m, n, and b positive integers with p b, p prime, and m = m ρ p (b, we have that { ρ p (S n (b m m, if n is even or n =, m + ρ p (n/, if n 3 odd.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY (007, #A4 6 Theorem 5 For m, n, and b positive integers so that m is sufficiently large and p b, p prime, and m = m ρ p (b, we have that { ρ p (S n (b m m + ρ p (B n, if n is even or n =, = m + ρ p (B n + ρ p (n/, if n 3 odd. The proof of Theorem 3 shows how to extend it to those of Theorems 4 and 5. A result by Andrews [6] implies that ρ p (B n can be arbitrary large. For example, if (p n and ρ p (n = l > 0 then ρ p (B n l, and this suggests that it might be difficult to get the exact order of ρ p (S n (b m with a formula, similar to (6, which is uniformly valid in all m. Remark. We intended to find the p-adic order of S n (x for special integers of the form x = b m with p b, however, the above theorems remain true for x = c b m with p b and p c, by adusting identity (5. In this case, S n (c b m + mod p also follows if m is sufficiently large. We note that if p divides b for some prime p, and we calculate B n+ ( n / in (5 p- adically, such as by using a theorem discovered independently by Anton, Stickelberger, and Hensel [4] on binomial coefficients modulo powers of p, then we can find further and more refined congruential properties of S n (b m. Acknowledgment The author wishes to thank Gregory Tollisen for making helpful suggestions and comments that improved the presentation of this paper. References [] L. Comtet, Advanced Combinatorics, D. Reidel Publishing Co., Dordrecht, 974. [] A. W. F. Edwards, A quick route to sums of powers, Amer. Math. Monthly, 93(6:45 455, 986. [3] R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics, Addison-Wesley Publishing Company, Reading, MA, second edition, 994. [4] A. Granville, Arithmetic properties of binomial coefficients I, Binomial coefficients modulo prime powers, In Organic Mathematics (Burnaby, BC, 995, volume 0 of CMS Conf. Proc., pages 53 76. Amer. Math. Soc., Providence, RI, 997. Electronic version (a dynamic survey: http://www.dms.umontreal.ca/ andrew/binomial/. [5] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, The Clarendon Press Oxford University Press, New York, fifth edition, 979. [6] R. Thangadurai, Adams theorem on Bernoulli numbers revisited, J. Number Theory, 06(:69 77, 004.