THE DENOMINATORS OF POWER SUMS OF ARITHMETIC PROGRESSIONS. Bernd C. Kellner Göppert Weg 5, Göttingen, Germany

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1 #A95 INTEGERS 18 (2018) THE DENOMINATORS OF POWER SUMS OF ARITHMETIC PROGRESSIONS Bernd C. Kellner Göppert Weg 5, Göttingen, Germany Jonathan Sondow 209 West 97th Street, New Yor, NY 10025, USA Received: 5/15/17, Revised: 9/10/18, Accepted: 11/17/18, Published: 11/23/18 Abstract In a recent paper the authors studied the denominators of polynomials that represent power sums by Bernoulli s formula. Here we extend our results to power sums of arithmetic progressions. In particular, we obtain a simple explicit criterion for integrality of the coe cients of these polynomials. As applications, we obtain new results on the sequence of denominators of the Bernoulli polynomials. A consequence is that certain quotients of successive denominators are infinitely often integers, which we characterize. 1. Introduction For positive integers n and x, define the power sum xx 1 S n (x) := n = 0 n + 1 n + + (x 1) n, and for integers m 1 and r 0 define the more general power sum of an arithmetic progression xx 1 Sm,r(x) n := (m + r) n = r n + (m + r) n + + ((x 1)m + r) n. In particular, we have S n 1,0(x) = S n (x) and, more generally, S n m,0(x) = m n S n (x). (1)

2 INTEGERS: 18 (2018) 2 Bazsó et al. [2, 3] considered the generalized Bernoulli formula Sm,r(x) n = B mn n+1 x + r r B n+1, (2) n + 1 m m where the nth Bernoulli polynomial B n (x) is defined by the series te xt e t and is given by the formula 1 = 1 X n=0 B n (x) = B n (x) tn n! ( t < 2 ) n B x n, (3) B = B (0) 2 Q being the th Bernoulli number. Thus, Sm,r(x) n is a polynomial in x of degree n + 1 with rational coe cients. Remar. Bazsó et al. required r and m to be coprime. However, since the forward di erence B n (x) := B n (x + 1) B n (x) equals nx n 1 (cf. [7, Eq. (5), p. 18]), the telescoping sum of these di erences with x = + r m implies (2) at once for any r/m 2 Q. For a polynomial f(x) 2 Q[x], define its denominator, denoted by denom f(x), to be the smallest d 2 N such that d f(x) 2 Z[x]. This includes the usual definition of denom(q) for q 2 Q. In the classical case of Bernoulli s formula S n (x) = 1 n + 1 B n+1(x) B n+1, the authors [6, Thms. 1 and 2] determined the denominator of the polynomial S n (x). From now on, let p denote a prime. Theorem 1 (Kellner and Sondow [6]). For n Then we have the relation and the remarable formula = Y p apple M n s p(n) p 1, denote := denom B n (x) B n. (4) denom S n (x) = (n + 1) +1 8 >< p with M n := >: n + 1, if n is odd, 2 n + 1, if n is even, 3 where s p (n) denotes the sum of the base-p digits of n, as defined in Section 4. Moreover, is odd if and only if n = 2 ( 0). (6) (5)

3 INTEGERS: 18 (2018) 3 The first few values of are (see [11, Seq. A195441]) = 1, 1, 2, 1, 6, 2, 6, 3, 10, 2, 6, 2, 210, 30, 6, 3, 30, 10, 210, 42, 330,.... The sequence ( ) n 1 and its properties will play a central role in this paper. The denominators are involved in formulas for related denominators in an essential way. As implied by the product formula (5), it turns out that the values of obey certain divisibility properties. This culminates in the fact that certain quotients of successive denominators are infinitely often integers, as we will see. Here we extend Theorem 1 to the denominator of Sm,r(x), n as follows. Theorem 2. We have denom S n m,r(x) = n + 1 gcd(n + 1, m n ) +1 gcd(+1, m). (7) In particular, denom S n m,r(x) divides denom S n (x) and is independent of r. Moreover, for any integers r 1, r 2 0, S n m,r 1 (x) S n m,r 2 (x) 2 Z[x]. The next theorem shows exactly when S n m,r(x) itself lies in Z[x]. Theorem 3. For n 1, denote Then we have the equivalence := denom B n (x), := denom(b n ). S n m,r(x) 2 Z[x] if and only if m as well as the equalities and where rad() := Q p p. = lcm(, ) (8) = lcm +1, rad(n + 1), (9) The first few values of and are (see [11, Seqs. A and A027642]) = 2, 6, 2, 30, 6, 42, 6, 30, 10, 66, 6, 2730, 210, 30, 6, 510, 30, 3990,..., = 2, 6, 1, 30, 1, 42, 1, 30, 1, 66, 1, 2730, 1, 6, 1, 510, 1, 798, 1, 330,.... Example 1. Set m = = 2, 6, 2, 30, 6 for n = 1, 2, 3, 4, 5, respectively. Then certainly m, so Z[x] contains the polynomials Sm,r(x) n with r = 0 (which

4 INTEGERS: 18 (2018) 4 satisfy (1)): S 1 2,0(x) = x 2 x = (x2 x) = 2 S 1 (x), S6,0(x) 2 = 6 (2x 3 3x 2 + x) = (2x3 3x 2 + x) = 6 2 S 2 (x), S2,0(x) 3 = 2 (x 4 2x 3 + x 2 ) = (x4 2x 3 + x 2 ) = 2 3 S 3 (x), S30,0(x) 4 = (6x 5 15x x 3 x) = (6x5 15x x 3 x) = 30 4 S 4 (x), S6,0(x) 5 = 648 (2x 6 6x 5 + 5x 4 x 2 ) = (2x6 6x 5 + 5x 4 x 2 ) = 6 5 S 5 (x) as well as those with r = 1: S 1 2,1(x) = x 2, S 2 6,1(x) = 12x 3 12x 2 + x, S 3 2,1(x) = 2x 4 x 2, S 4 30,1(x) = x x x x x, S 5 6,1(x) = 1296x x x x 3 273x 2 170x. Remar. Bazsó and Mező [2, Eqs. (7), (8) and Thm. 2, pp ] defined a very complicated formula F (n) in order to give a somewhat tautological characterization of when S n m,r(x) 2 Z[x]. With their formula they computed a few values of F (n) that apparently equal, but without recognizing this relation. They were not aware of advanced results lie those in our Theorems 2 and 3. As an immediate by- product of our theorems, we obtain a new product formula for from (9) by applying Theorem 1. (Other explicit product formulas for this denominator, based on (8), were already given in [6, Thm. 4].) Corollary 1. For n 1, the denominator of the nth Bernoulli polynomial equals = Y Y p p. p n+1 p - n+1 p apple M n+1 s p(n+1) p Remar. The first author [5] has shown that the condition s p (n) p is su cient in (5) to define as a product over all primes: Y = p. (10) s p(n) p

5 INTEGERS: 18 (2018) 5 (So one can remove the restrictions p apple M n in (5) and p apple M n+1 in Corollary 1.) Moreover, if n + 1 is composite, then (see [5, Thm. 1]) Finally, we obtain new properties of and. rad(n + 1). (11) Corollary 2. The sequences ( ) n 1 and ( ) n 1 satisfy the following conditions: (i) We have the relations (ii) We have the divisibilities = lcm +1, rad(n + 1), if n 3 is odd, = lcm +1, rad(n + 1), if n 2 is even. +1, if n 1 is odd, +1, if n 2 is even. Theorem 4. For odd n 1, the quotients (see [11, Seq. A286516]) +1 = 1, 2, 3, 2, 5, 3, 7, 2, 3, 5, 11, 1, 13, 7, 15, 2, 17, 3, 19, 5, 7,... are odd, except that +1 = 2 if and only if n = 2 1 ( 2). Moreover, if p is an odd prime and n = 2`p 1, then +1 2 {1, p} (, ` 1), and more precisely, +1 = p ( 1, 1 apple ` < log 2 p), while +1 = 1 ( 1, ` L p ), where L p > log 2 p is a constant depending on p.

6 INTEGERS: 18 (2018) 6 Theorem 5. For even n (see [11, Seq. A286517]) 2, all terms are odd in the sequence +1 = 3, 5, 7, 3, 11, 13, 5, 17, 19, 7, 23, 5, 3, 29, 31, 11, 35, 37,.... In particular, if p is an odd prime and n = p 1, then +1 = p ( 1). More generally, if p 6= q are odd primes and n = p q` 1, then +1 2 {1, p, q, pq} (, ` 1) with the following cases: +1 = p ( L 0 p,q, 1 apple ` < log q p), = q (1 apple < log D p q, ` L 00 n+1 = 1 ( L 0 D p,q, ` L 00 p,q), n+1 p,q), where L 0 p,q > log p q and L 00 p,q > log q p are constants depending on p and q. Theorems 4 and 5 immediately imply the following result. Corollary 3. Statements (i), (ii) (respectively, (iii), (iv)) below hold for infinitely many odd (respectively, even) values of n: (i) /+1 = p for a given prime p 2. (ii) = +1. (iii) /+1 = p for a given prime p 3. (iv) = Preliminaries Let Z p be the ring of p-adic integers, Q p be the field of p-adic numbers, and v p (s) be the p-adic valuation of s 2 Q p (see [9, Chap. 1.5, pp ]). If s 2 Z, then p e s means that p e s but p e+1 - s, or equivalently, e = v p (s).

7 INTEGERS: 18 (2018) 7 The Bernoulli numbers satisfy the following properties (cf. [4, Chap. 9.5, pp ]). The first few nonzero values are B 0 = 1, B 1 = 1 2, B 2 = 1 6, B 4 = 1 30, B 6 = 1 42, (12) while B n = 0 for odd n 3. For even n 2 the von Staudt Clausen theorem states that the denominator of B n equals = Y p (n 2 2N). (13) p 1 n Thus, all nonzero Bernoulli numbers have a squarefree denominator. Moreover, for even n 2 the p-adic valuation of the divided Bernoulli number B n /n is Bn (vp (n) + 1), if p 1 n, v p = (14) n 0, else. Now let m, n, and r be positive integers. The Bernoulli polynomials satisfy as Appell polynomials the general relation B n (x + y) = of which (3) is a special case, as well as the reflection formula n B (y) x n, (15) B n (1 x) = ( 1) n B n (x) (16) (see [8, Chap. 3.5, pp ]). Further, denote by B n m,r the number B n m,r := m n B n r m 1 n B n = B m r n. (17) Almvist and Meurman [1, Thm. 2, p. 104] showed that B n m,r 2 Z. (18) Actually, (18) holds for all r 2 Z (cf. [4, Thm , pp ]). We also point out an analog to (15) for r 1, r 2 2 Z, namely, B n m,r 1+r 2 = n Bm,r 1 r2 n + Bm,r n 2. The integers B n m,r satisfy a useful divisibility property, which we need later on. The following lemma is part of [4, Thm , pp ], but we give here a clearer and simpler proof.

8 INTEGERS: 18 (2018) 8 Lemma 1. If m, n 1, r 2 Z, a prime p - m, and 0 apple e apple v p (n), then B n m,r 0 (mod p e ). (19) Proof. It su ces to prove the case e = v p (n). If e = 0, then we are trivially done. So let p e n with n > e = v p (n) 1. We split the proof into two cases as follows. Case p r: From (12) and (17) we deduce that 1 n Bm,r n = B m r n 1 n 1 = r n + n B n m n r 1. =1 Since p r, we have v p (r n ) n and v p (r /) 1 for all 1. If B n 6= 0, then v p (B n ) 1, since the denominator is squarefree. In this case we obtain v p n B n r Considering all summands, we finally infer that (19) holds. Case p - r: Since n 2, we have by (3) and (16) that e. 1 n B n (1) B n = B = 0, which we use in the second step below. Set u := m/r 2 Z p. As in the first case above, we derive that 1 n r n Bm,r n = B u 1 n = B (u 1) = n n 1 2 (u 1) + X n 1 n =2 2 1 B (u 1). In both cases p = 2 and p 3, we have n 2 (u 1) 0 (mod pe ).

9 INTEGERS: 18 (2018) 9 Since (14) implies B / 2 Z p if 2 is even and p 1 -, we get r n B n m,r 1 =2 2 p 1 n 1 n 1 B (u 1) (mod p e ). Now fix one of the above sum. We then have the decomposition = a (p 1) p t = a '(p t+1 ), where p - a, t = v p (), and '( ) is Euler s totient function. By assumption u is a unit in Z p and so is û := u a 2 Z p. Euler Fermat s theorem shows that u û '(pt+1) 1 (mod p t+1 ). Thus, v p (u 1) t + 1. Since v p (B /) = (t + 1) by (14), we achieve finally that v p n B (u 1) e (t + 1) + (t + 1) = e, implying that and showing the result. r n B n m,r 0 (mod p e ) 3. Proof of Theorem 2 Before giving the proof of Theorem 2, we need several lemmas with some complementary results. The next lemma easily shows a related partial result toward Theorem 2, while the full proof of this theorem requires much more e ort. Lemma 2. We have denom (n + 1)S n m,r(x) = denom m n (B n+1 (x) B n+1 ) = +1 gcd(+1, m). Proof. By rewriting Sm,r(x) n as given in (2), and using (3) and (15), we easily derive that X n n + 1 r (n + 1)Sm,r(x) n = m n B x n+1 m X n n + 1 r = m n B B + B x n+1 m

10 INTEGERS: 18 (2018) 10 = n + 1 r m n m B B x n+1 {z m } (20) Bm,r 2 Z by (18) + m n B n+1 (x) B n+1. (21) By applying the simple observation that if f(x) 2 Z[x] and g(x) 2 Q[x], then we infer that denom f(x) + g(x) = denom g(x), (22) denom (n + 1)S n m,r(x) = denom m n (B n+1 (x) B n+1 ). Finally, from (4) and (5) we deduce that denom m n (B n+1 (x) B n+1 ) = +1 gcd(+1, m n ) = +1 gcd(+1, m), the latter equation holding because +1 is squarefree. This completes the proof. Lemma 3. For positive integers apple n, define the rational number c n, := 1 n. 1 Then we have the following properties: (i) Symmetry: c n, = c n,n+1. (ii) Denominator: (iii) Integrality: denom(c n, ) gcd(n + 1, ), denom(c n, ) apple n If = 1 or = n or n + 1 is prime, then c n, 2 Z. Proof. We first observe that c n, = 1 n 1 = 1 n + 1 n + 1, (23)

11 INTEGERS: 18 (2018) 11 which shows the symmetry in (i). From (23) it also follows that denom(c n, ) must divide both of the integers n + 1 and. Thus, denom(c n, ) gcd(n + 1, ). Since < n + 1, we then infer that denom(c n, ) apple (n + 1)/2. This shows (ii). If = 1 or = n, then c n, = 1. If n + 1 is prime, then gcd(n + 1, ) = 1 as apple n, so denom(c n, ) = 1. This proves (iii). Lemma 4. If m, n, r 1 and 0 apple apple n, then m n n + 1 r B n + 1 m B 2 Z. (24) Proof. If = 0, then the quantity in (24) vanishes by B 0 (x) B 0 = 0. For 1 apple apple n, we can rewrite the quantity in (24) by (17) and (23) as c n, m n B m,r, (25) n 1 where Bm,r 2 Z by (18) and c n, = 1 2 Q. We have to show that (25) lies in Z. If = 1 or = n or n + 1 is prime, then c n, 2 Z by Lemma 3. We can now assume that n 3, 1 < < n, and d := denom(c n, ) > 1. For each prime power divisor p e d we consider two cases, which together imply the integrality of (25). Case p - m: Since d, we have p e B m,r by Lemma 1. Case p m: We show that p e m n, or equivalently, n + 1 > e +. (26) As p e, by symmetry in Lemma 3 we also have p e n + 1, so e < n + 1 and (26) holds. This completes the proof. Proof of Theorem 2. To prove the last statement, it su ces to show that for r 0 By (20) and (21) we have S n m,r(x) = S n m,r(x) S n m,0(x) 2 Z[x]. (27) mn n + 1 B n+1(x) B n+1 + h(x), (28) where (n + 1)h(x) = f(x) 2 Z[x] as given by (20). By Lemma 4 it turns out that the coe cients of h(x) are already integral, and thus h(x) 2 Z[x]. Since by (2) relation (27) follows. S n m,0(x) = mn n + 1 B n+1(x) B n+1,

12 INTEGERS: 18 (2018) 12 Applying the rule (22) to (28) and using (4) along with the fact that B n+1 (x) is monic, we then infer that denom Sm,r(x) n m n = denom. (29) (n + 1)+1 We have to show that (29) implies (7). In the following trivial cases we are done: case m = 1; cases n = 1, 3, since D 2 = D 4 = 1; and case n = 2, since n + 1 = 3 and D 3 = 2. So let m 2 and n 4. If a prime power p e n + 1, then e < n. Consequently, we deduce that gcd(n + 1, m n ) = gcd(n + 1, m n 1 ). (30) Then by splitting m n into m n 1 m in (29) and applying (30) and the fact that +1 is squarefree, we infer that (7) holds. Since denom S n (x) = (n+1) +1 by Theorem 1, we see at once that (7) implies denom S n m,r(x) denom S n (x). As a result of (29), the denominator of S n m,r(x) is independent of r. This completes the proof of Theorem Proofs of Theorem 3 and Corollary 2 Before we give the proofs, we need some definitions and lemmas. Recall that, given a prime p, any positive integer n can be written in base p as a unique finite p-adic expansion n = p + + t p t (0 apple j apple p 1). This expansion defines the sum-of-digits function which satisfies the congruence s p (n) := t, s p (n) n (mod p 1). (31) Actually, these properties hold for any integer base b 2 in place of a prime p. The following lemma (see [9, Chap. 5.3, p. 241]) shows the relation between s p (n) and s p (n + 1). Lemma 5. If n 1 and p is a prime, then s p (n + 1) = s p (n) + 1 (p 1) v p (n + 1).

13 INTEGERS: 18 (2018) 13 In particular, while Lemma 6. If n s p (n + 1) apple s p (n) if and only if p n + 1, s p (n + 1) = s p (n) + 1 if and only if p - n + 1. (32) 1, then lcm(, ) lcm +1, rad(n + 1). Proof. Set L n := lcm +1, rad(n + 1). Since and are squarefree by (5) and (13), we show that p lcm(, ) implies p L n. Moreover, since rad(n + 1) L n, we may assume that p - n + 1. If p, then by (5) we have s p (n) p. Applying (32) followed by (10), we obtain p +1, and finally p L n. Since D 1 = 2 by (12) and = 1 for odd n 3, we have L n for odd n 1. So tae n 2 even. If p, then p 1 n by (13), so also p 1 s p (n) by (31). Thus s p (n) p 1. As p - n + 1 by assumption, (32) implies s p (n + 1) p, so p +1 by (10). Finally p L n. This proves the lemma. Lemma 7. If n 1, then lcm +1, rad(n + 1) lcm(, ). Proof. As D 1 = D 2 = 1, and D 1 = 2 by (12), the case n = 1 holds. So assume n 2 and set L n := lcm(, ). If n + 1 is not prime, then (11) implies rad(n + 1) L n. Otherwise, p = n + 1 = rad(n + 1) is an odd prime and so n is even. By (13) we have p, so rad(n + 1) L n. It remains to show that +1 L n. As +1 is squarefree by (5), it su ces to show for any prime p +1 that p L n. By (5) again we have s p (n + 1) p, and as rad(n + 1) L n we may assume that p - n + 1. Then by (32) we obtain s p (n) = s p (n + 1) 1 p 1. If s p (n) p, then p by (10), so p L n. Otherwise, s p (n) = p 1 and so p 1 n by (31). Moreover, n must be even, as n odd would imply p = 2, contradicting p - n + 1. Hence p by (13), and finally p L n. This completes the proof. Proof of Theorem 3. To show the equivalence, we have to prove that denom S n m,r(x) = 1 if and only if m. By (7), we have denom S n m,r(x) = 1 if and only if n + 1 gcd(n + 1, m n ) = +1 gcd(+1, m) = 1,

14 INTEGERS: 18 (2018) 14 which in turn is true if and only if n + 1 m n and +1 m. Moreover, n + 1 m n if and only if rad(n + 1) m. Indeed, p n + 1 m n implies p m, proving the ) direction. Conversely, if p rad(n + 1) m, then p n m n. But p e n + 1 with e apple n, so finally n + 1 m n. It follows that denom S n m,r(x) = 1 if and only if lcm +1, rad(n + 1) m. By Lemmas 6 and 7, together with the proof of [6, Thm. 4], we have This proves the theorem. lcm +1, rad(n + 1) = lcm(, ) =. Proof of Corollary 2. (i), (ii) If n 3 is odd, then = 1. Hence, (8) and (9) yield = lcm +1, rad(n + 1). Together with D 1 = D 2 = 1, this implies that +1 for all odd n 1, as desired. Similarly, for even n 2, we have +1 = +1 by (8). Then (9) gives = lcm +1, rad(n + 1), so +1, as claimed. 5. Proofs of Theorems 4 and 5 Let a, b 2 be integers that are multiplicatively independent, that is, a e 6= b f for all integers e, f 1. Senge and Straus [10, Thm. 3] showed that for a given constant A the number of integers n satisfying s a (n) + s b (n) < A is finite. Steward [12, Thm. 1, p. 64] proved the e ective lower bound s a (n) + s b (n) > log log n log log log n + C 1 (33) for n > 25, where C > 0 is an e ectively computable constant depending on a and b. This bound leads to the following lemma. Lemma 8. If p 6= q are primes, then lim s p q = 1.!1 In particular, there exists a positive integer L p,q > log q p such that s p q p ( L p,q ).

15 INTEGERS: 18 (2018) 15 Proof. By taing a = p, b = q, 5, and n = q > 25, we derive from (33) that s p (q ) > log( log q) log log( log q) + C 2 =: f() with some constant C > 0 depending on p and q. Since f() is strictly increasing for all su ciently large, we infer that lim!1 s p q = 1. Therefore, there exists a positive integer L p,q such that s p q p for L p,q. On the other hand, since s p (m) = m for 0 apple m < p, we have that s p (q ) < p for 1 apple < log q p, implying that L p,q > log q p, as claimed. Proof of Theorem 4. If n = 1, then D 1 /D 2 = 1. By (6), if n 3 is odd and n + 1 is not a power of 2, then and +1 are both even. Since by (5) they are squarefree, /+1 must be odd. Liewise, if n = 2 1 for some 2, then /+1 must be twice an odd number. If an odd prime p divides, then s p (n) p by (5). Since p - 2 = n + 1, we infer by (32) that s p (n + 1) > p. Hence by (10) the prime p also divides +1, so indeed /+1 = 2. Now, let n = 2`p 1 with p an odd prime and, ` 1. Then we have rad(n+1) = 2p, and by (6) that and +1 are both even. Thus, /+1 2 {1, p} by Corollary 2 part (i). We consider two cases. Case 1 apple ` < log 2 p: Since s p (n + 1) = s p (2`) < p, we infer that p - +1 by (10) implying /+1 = p. Case ` > log 2 p: Lemma 8 implies a constant L p := L p,2 > log 2 p such that s p (n + 1) = s p (2`) p for all ` L p. Hence p +1 by (10) and /+1 = 1. This proves the theorem. Proof of Theorem 5. It is shown in [6, Thm. 4] that is even and squarefree for all n 1. (This also follows from (8) for even n 2, since 2, and from (9) for odd n 1, since 2 rad(n + 1), all terms in question being squarefree.) Hence if n 2 is even, so that +1, then the quotient must be odd. Let p be an odd prime. If n = p 1 for some 1, then we have rad(n+1) = p and s p (n + 1) = s p (p ) = 1 < p. Thus p - +1 by (10). Since n is even, we have +1 = +1 by (8) and so p By Corollary 2 part (i) we finally obtain /+1 = p. Now, let p 6= q be odd primes and n = p q` 1 with, ` 1. We then have rad(n + 1) = pq and by Corollary 2 part (i) that /+1 2 {1, p, q, pq}. Note that s p (n + 1) = s p (q`) and s q (n + 1) = s q (p ). By Lemma 8 we define L 0 p,q := L q,p > log p q and L 00 p,q := L p,q > log q p. We consider the following statements by using (10): If 1 apple ` < log q p, then s p (q`) < p and p Otherwise, if ` L 00 p,q, then s p (q`) p and p +1.

16 INTEGERS: 18 (2018) 16 If 1 apple < log p q, then s q (p ) < q and q Otherwise, if L 0 p,q, then s q (p ) q and q +1. All three cases of the theorem follow from the arguments given above, since +1 = +1. This completes the proof of the theorem. 6. Conclusion The numbers B n m,r = m n B n r m B n, shown by Almvist and Meurman to be integers, play here a ey role in proofs. By their result, the polynomial B n (x) B n, with an extra factor, taes integer values at rational arguments x = r/m. In the present paper, the numbers Bm,r n reveal their natural connection with the power sums of arithmetic progressions Sm,r(x). n Moreover, the divisibility properties of Bm,r n are important in attaining our results in Theorems 2 and 3. Acnowledgment. We than the anonymous referee for several suggestions. References [1] G. Almvist and A. Meurman, Values of Bernoulli polynomials and Hurwitz s zeta function at rational points, C. R. Math. Acad. Sci. Soc. R. Can. 13 no. 2 3 (1991), [2] A. Bazsó and I. Mező, On the coe cients of power sums of arithmetic progressions, J. Number Theory 153 (2015), [3] A. Bazsó, Á. Pintér, and H. M. Srivastava, A refinement of Faulhaber s theorem concerning sums of powers of natural numbers, Appl. Math. Lett. 25 no. 3 (2012), [4] H. Cohen, Number Theory, Volume II: Analytic and Modern Tools, GTM 240, Springer Verlag, New Yor, [5] B. C. Kellner, On a product of certain primes, J. Number Theory 179 (2017), [6] B. C. Kellner and J. Sondow, Power-sum denominators, Amer. Math. Monthly 124 (2017), [7] N. E. Nørlund, Vorlesungen über Di erenzenrechnung, J. Springer, Berlin, [8] V. V. Prasolov, Polynomials, D. Leites, transl., 2nd edition, ACM 11, Springer Verlag, Berlin, [9] A. M. Robert, A Course in p-adic Analysis, GTM 198, Springer Verlag, New Yor, 2000.

17 INTEGERS: 18 (2018) 17 [10] H. G. Senge and E. G. Straus, PV-numbers and sets of multiplicity, Period. Math. Hungar. 3 (1973), [11] N. J. A. Sloane, ed., The On-Line Encyclopedia of Integer Sequences, [12] C. L. Stewart, On the representation of an integer in two di erent bases, J. Reine Angew. Math. 319 (1980),