Distribution modulo one and denominators of the Bernoulli polynomials
aa r X i v : . [ m a t h . N T ] A ug DISTRIBUTION MODULO ONEAND DENOMINATORS OFTHE BERNOULLI POLYNOMIALS
BERND C. KELLNER
Abstract.
Let {·} denote the fractional part and n ≥ p the one-to-onecorrespondence X ν ≥ (cid:26) np ν (cid:27) > ⇐⇒ p | denom( B n ( x ) − B n ) , where B n ( x ) − B n is the n th Bernoulli polynomial without constantterm and denom( · ) is its denominator, which is squarefree. Introduction
Recently, the properties of the denominators of the Bernoulli poly-nomials B n ( x ) and B n ( x ) − B n have sparked interest of several authors,see [1, 4, 5, 6]. Various related sequences of these denominators werestudied in [5, 6].The Bernoulli polynomials are defined by te xt e t − X n ≥ B n ( x ) t n n ! ( | t | < π ) , where B n ( x ) = n X k =0 (cid:18) nk (cid:19) B k x n − k ( n ≥
0) (1)and B k = B k (0) is the k th Bernoulli number (cf. [7, Chap. 3.5, pp. 112]).Let p denote always a prime. The denominators of the Bernoullipolynomials can be explicitly characterized as follows, which involvesthe function s p ( n ) giving the sum of base- p digits of n . Mathematics Subject Classification.
Key words and phrases.
Distribution modulo one, Bernoulli polynomials, de-nominator, sum of base- p digits. Theorem 1 (Kellner and Sondow [5]) . For n ≥ , we have denom (cid:0) B n ( x ) − B n (cid:1) = Y p ≤ n +1 λn s p ( n ) ≥ p p with λ n := ( , if n is odd, , if n is even.The bound n +1 λ n is sharp for odd n = 2 p − and even n = 3 p − , when p is an odd prime, respectively. While the sharp bounds n +1 λ n in Theorem 1 give a stronger result, theauthor [4] has subsequently shown that they can be omitted yielding amore suitable formula:denom (cid:0) B n ( x ) − B n (cid:1) = Y s p ( n ) ≥ p p ( n ≥ . (2)In this short note, we want to focus on the following theorem relating(2) to sums of fractional parts; the latter denoted by { · } . Theorem 2. If n ≥ , then for any prime p we have X ν ≥ (cid:26) np ν (cid:27) > ⇐⇒ p | denom (cid:0) B n ( x ) − B n (cid:1) . (3) This is a one-to-one correspondence, since the denominator of theBernoulli polynomial B n ( x ) − B n is squarefree. At first glance, this remarkable connection between the distributionmodulo one and the denominators of the Bernoulli polynomials seemsto be “mysterious”. It remains an open question, whether there existsa more general law that is hidden behind this relation.Clearly, the sequence ( { np − ν } ) ν ≥ eventually converges to zero forany n ≥ p . We have a partition into a geometric seriesand a finite sum of fractional parts, which turns into X ν ≥ (cid:26) np ν (cid:27) = np ℓ ( p −
1) + ℓ X ν =1 (cid:26) np ν (cid:27) , (4)where ℓ ≥ p ℓ ≤ n < p ℓ +1 . For p > n the right-hand side of(4) reduces to n/ ( p − ≤
1, implying that (3) only holds for finitelymany primes p . Similarly, we can derive from (4) the same bounds asin Theorem 1. ISTRIBUTION MODULO ONE AND DENOMINATORS 3
Lemma 1. If n ≥ and p is a prime, then p > n + 1 λ n = ⇒ X ν ≥ (cid:26) np ν (cid:27) ≤ , where λ n is defined as in Theorem 1.Proof. As pointed out above, for p > n the claim already holds. Wehave to distinguish between two cases as follows.Case p > n +12 : We obtain the bounds 2 p − ≥ n ≥ p . Thus n = p + r with 0 ≤ r ≤ p −
2. The right-hand side of (4) yields with a = 1 that np ( p −
1) + (cid:26) np (cid:27) = ap + rp ( p −
1) + rp = r + ap − ≤ , (5)showing the result for the first case.Case p > n +13 and n even: With the first case there remain thebounds 3 p − ≥ n ≥ p . If p = 2, then only n = p can hold. Bydefinition, the left-hand side of (4) evaluates to 1 / ( p −
1) = 1. For odd p ≥ n = 2 p + r with 0 ≤ r ≤ p −
3. The right-hand sideof (4) then becomes (5) with a = 2. This completes the second caseand shows the result. (cid:3) On the other side, the next theorem shows that the values of thesum of fractions can be arbitrarily large for powers of n . Theorem 3.
Let n > and p be a prime. If n is not a power of p ,then X ν ≥ (cid:26) n k p ν (cid:27) → ∞ as k → ∞ . In view of Theorem 2, a weaker version of Theorem 3, that the abovesum is greater than 1 for all sufficiently large values of k , already impliesthe following corollary. Corollary 1.
Let P be a finite set of primes and Π := Y p ∈P p. If n > is not a power of any p ∈ P , then there exists a constant M depending on n and P such that Π | denom (cid:0) B n k ( x ) − B n k (cid:1) ( k ≥ M ) . The aim of this paper is to give a somewhat elementary and directproof of Theorem 2 using properties of fractional parts. This results innew variants of proofs given in the next sections.
BERND C. KELLNER Preliminaries
Let Z p be the ring of p -adic integers, and Q p be the field of p -adicnumbers. Define ord p s as the p -adic valuation of s ∈ Q p . Let [ · ]denote the integer part. For n ≥ (cid:26)(cid:26) np (cid:27)(cid:27) := X ν ≥ (cid:26) np ν (cid:27) . By Legendre’s formula (see [8, Chap. 5.3, p. 241]) we haveord p n ! = X ν ≥ (cid:20) np ν (cid:21) . (6)Writing n/ ( p −
1) as a geometric series and using { x } = x − [ x ] yield (cid:26)(cid:26) np (cid:27)(cid:27) = np − − ord p n ! . (7)Consequently, we have (cid:26)(cid:26) np (cid:27)(cid:27) ∈ N ⇐⇒ p − | n. (8) Lemma 2.
Let p be a prime. If a ≥ and ≤ r < p , then (cid:26)(cid:26) ap + rp (cid:27)(cid:27) = (cid:26)(cid:26) ap (cid:27)(cid:27) + (cid:26)(cid:26) rp (cid:27)(cid:27) . Proof.
Since 0 ≤ r < p , we observe by (6) thatord p ( ap + r )! = X ν ≥ (cid:20) ap + rp ν (cid:21) = a + ord p a ! . Hence, the result follows easily by (7). (cid:3)
Lemma 3. If n ≥ and p is a prime, then (cid:26)(cid:26) np (cid:27)(cid:27) = ℓ X j =0 (cid:26)(cid:26) n j p (cid:27)(cid:27) , where n = n + n p + · · · + n ℓ p ℓ is the p -adic expansion of n .Proof. This follows by applying Lemma 2 iteratively. (cid:3)
Lemma 4. If n > and p ≤ n is a prime, then (cid:26)(cid:26) np (cid:27)(cid:27) ≤ ⇐⇒ ord p (cid:18) nk (cid:19) ≥ for all < k < n , where p − | k . ISTRIBUTION MODULO ONE AND DENOMINATORS 5
Proof.
Set K := { k ∈ N : 0 < k < n, p − | k } , where K 6 = ∅ by as-sumption. We consider the following two cases.Case nn np oo ≤
1: Applying (7) and (8) we infer thatord p (cid:18) nk (cid:19) = (cid:26)(cid:26) n − kp (cid:27)(cid:27)| {z } > + (cid:26)(cid:26) kp (cid:27)(cid:27)| {z } ≥ − (cid:26)(cid:26) np (cid:27)(cid:27)| {z } ≤ > k ∈ K ) . (9)Case nn np oo >
1: We will show that an integer k ∈ K exists withord p (cid:0) nk (cid:1) = 0. Generally, there exist integers k with 1 ≤ k ≤ n suchthat the p -adic expansions yield n = n + n p + · · · + n ℓ p ℓ ,k = k + k p + · · · + k ℓ p ℓ (10)with 0 ≤ k j ≤ n j for j = 0 , . . . , ℓ . Thus, we have the p -adic expansion n − k = ( n − k ) + ( n − k ) p + · · · + ( n ℓ − k ℓ ) p ℓ . (11)With that we achieve by Lemma 3 and using (7) that (cid:26)(cid:26) np (cid:27)(cid:27) = (cid:26)(cid:26) n − kp (cid:27)(cid:27) + (cid:26)(cid:26) kp (cid:27)(cid:27) , (12)implying by (9) that ord p (cid:18) nk (cid:19) = 0 . (13)Since nn np oo >
1, we obtain by Lemma 3 and (7) that n + n + · · · + n ℓ > p − . Hence, we can choose k j with 0 ≤ k j ≤ n j for j = 0 , . . . , ℓ such that k + k + · · · + k ℓ = p − , satisfying (10) – (13) but with 0 < k < n . By Lemma 3 and (7) it thenfollows that nn kp oo = 1. Consequently, p − | k by (8) and therefore k ∈ K . This completes the second case.Finally, both cases imply the claimed equivalence. (cid:3) Remark.
Actually, Eqs. (9), (12), and (13) reflect Kummer’s theoremthat ord p (cid:0) nk (cid:1) equals the number of carries when adding k to n − k inbase p . Carlitz [3] gave a more general result of Lemma 4 in context ofthe function s p ( n ), whose proof depends on Lucas’s theorem, namely (cid:18) nk (cid:19) ≡ (cid:18) n k (cid:19)(cid:18) n k (cid:19) · · · (cid:18) n ℓ k ℓ (cid:19) (mod p ) . BERND C. KELLNER Proof of Theorem 2
While the Bernoulli numbers B n = 0 for odd n ≥
3, the theorem ofvon Staudt–Clausen asserts for even n ≥ B n + X p − | n p ∈ Z implying that denom( B n ) = Y p − | n p. (14)The p -adic valuation of a nonzero polynomial f ( x ) = r X k =0 a k x k ∈ Q [ x ] \{ } of degree r is given by ord p f ( x ) = min ≤ k ≤ r ord p a k . Define the polynomials e B n ( x ) := B n ( x ) − B n , e B n,p ( x ) := n − X k =22 | kp − | k (cid:18) nk (cid:19) B k x n − k . Lemma 5. If n ≥ and p is a prime, then ord p e B n,p ( x ) = ( − , if nn np oo > , ≥ , else.In particular, if p = 2 and n ≥ is odd, then ord p e B n,p ( x ) = − .Proof. Set K := { k ∈ N : 0 < k < n, p − | k } . Note that K = ∅ ⇔ p > n . In this case, we have e B n,p ( x ) = 0 and thus ord p e B n,p ( x ) = ∞ ,as well as nn np oo ≤ K = ∅ and p ≤ n . For all coefficients of e B n,p ( x ),we obtain by (14) thatord p (cid:18) nk (cid:19) B k = − p (cid:18) nk (cid:19) ≥ − k ∈ K ) . (15)For odd primes p the claim follows at once by Lemma 4. For p = 2there remain two cases as follows.Case p = 2 and n ≥ n − ∈ K and n = (cid:0) nn − (cid:1) is odd,it follows by (15) that ord p e B n,p ( x ) = −
1. On the other side, we canwrite n = n ′ p + 1 with some n ′ ≥
1. By Lemma 2 and (7) we obtain nn np oo = nn n ′ p oo + 1 >
1; showing the claim for this case.
ISTRIBUTION MODULO ONE AND DENOMINATORS 7
Case p = 2 and n ≥ n is even, we have for odd ℓ with0 < ℓ < n that (cid:18) nℓ (cid:19) ≡ nℓ (cid:18) n − ℓ − (cid:19) ≡ . Therefore, with K = { , , . . . , n − } ,ord p (cid:18) nk (cid:19) ≥ k ∈ K ) ⇐⇒ ord p (cid:18) nk (cid:19) ≥ < k < n ) . With that we can apply Lemma 4 to show the claim for that case. (cid:3)
Proof of Theorem 2.
Note that B = 1 and B = − .Cases n = 1 ,
2: As e B ( x ) = x and e B ( x ) = x − x by (1), we have forall primes p that ord p e B n ( x ) = 0, while nn np oo ≤ n ≥
3: Since B k = 0 for odd k ≥
3, we deduce from (1) that e B n ( x ) = x n − n x n − + n − X k =22 | k (cid:18) nk (cid:19) B k x n − k . (16)For even k ≥ B k ∈ Z p if p − ∤ k . Thus weinfer from (16) thatord p e B n ( x ) = min (cid:16) , ord p n , ord p e B n,p ( x ) (cid:17) . (17)Since ord p e B n,p ( x ) ≥ − p n ≥ −
1, we getord p e B n ( x ) ∈ {− , } . (18)If ord p n = −
1, then p = 2 and n ≥ p e B n,p ( x ) = − (cid:26)(cid:26) np (cid:27)(cid:27) > ⇐⇒ ord p e B n ( x ) = − ⇐⇒ p | denom (cid:0) e B n ( x ) (cid:1) . This shows (3) for the case n ≥ e B n ( x ) issquarefree. This completes the proof. (cid:3) BERND C. KELLNER Proof of Theorem 3 and Corollary 1
All previous results have been deduced without involving the func-tion s p ( n ) directly. However, using an alternative form of Legendre’sformula of ord p n !, we have besides thatord p n ! = n − s p ( n ) p − (cid:26)(cid:26) np (cid:27)(cid:27) = s p ( n ) p − . (19)For two multiplicatively independent integers a, b ≥
2, a positiveinteger n cannot have few nonzero digits in both bases a and b simul-taneously. Steward [10, Thm. 1, p. 64] gave a lower bound such that s a ( n ) + s b ( n ) > log log n log log log n + c − n >
25) (20)with an effectively computable constant c depending on a and b . Seealso Bugeaud [2, Thm. 6.9, p. 134] for a related result. However, theweaker result of Senge and Straus [9, Thm. 3], that for any constant C the number of integers n satisfying s a ( n ) + s b ( n ) < C (21)is finite, would already suffice for our purpose. Proof of Theorem 3.
We can write n = ˜ n p r with some r ≥ n > p ∤ ˜ n by assumption. Taking a = p and b = n in view of (20)and (21), we have s p ( n k ) = s p (˜ n k ) > s n ( n k ) = 1 for all k ≥ s p ( n k ) → ∞ as k → ∞ . By (19) the result follows. (cid:3)
Proof of Corollary 1.
From Theorem 3 we infer that for each prime p ∈ P there exists a constant m p such that (cid:26)(cid:26) n k p (cid:27)(cid:27) > k ≥ m p ) . Set M := max p ∈P m p . Then by Theorem 2 the result follows. (cid:3) References [1] O. Bordell`es, F. Luca, P. Moree, and I. E. Shparlinski,
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