The elementary symmetric functions of reciprocals of the elements of arithmetic progressions
aa r X i v : . [ m a t h . N T ] M a r THE ELEMENTARY SYMMETRIC FUNCTIONS OFRECIPROCAL ARITHMETIC PROGRESSIONS
CHUNLIN WANG AND SHAOFANG HONG ∗ Abstract.
Let a and b be positive integers. In 1946, Erd˝os and Niven provedthat there are only finitely many positive integers n for which one or more ofthe elementary symmetric functions of 1 /b, / ( a + b ) , ..., / ( an − a + b ) areintegers. In this paper, we show that for any integer k with 1 ≤ k ≤ n , the k -th elementary symmetric function of 1 /b, / ( a + b ) , ..., / ( an − a + b ) is notan integer except that either b = n = k = 1 and a ≥
1, or a = b = 1 , n = 3and k = 2. This strengthens the Erd˝os-Niven theorem and answers an openproblem raised by Chen and Tang in 2012. Introduction
A well-known result in number theory states that for any integer n > P ni =1 1 n is not an integer. Let a and b be positive integers.In 1946, Erd˝os and Niven [4] proved that there are only finitely many posi-tive integers n for which one or more of the elementary symmetric functions of1 /b, / ( a + b ) , ..., / ( an − a + b ) are integers. Chen and Tang [1] proved that noneof the elementary symmetric functions of 1 , / , ..., /n is an integer if n ≥ , / , ..., / (2 n −
1) is an integer if n ≥ { b + ai } n − i =0 such that one or more elementary symmetric functionsof 1 /b, / ( a + b ) , ..., / ( an − a + b ) are integers (see Problem 1 of [1]). For anyinteger k with 1 ≤ k ≤ n , let S a,b ( n, k ) denote the k -th elementary symmetricfunction of 1 /b, / ( a + b ) , ..., / ( an − a + b ). That is, S a,b ( n, k ) := X ≤ i <...
Let a, b, n and k be positive integers with ≤ k ≤ n . Then S a,b ( n, k ) is not an integer except that either b = n = k = 1 , or a = b = 1 , n = 3 and k = 2 , in which case S a,b ( n, k ) is an integer. ∗ Hong is the corresponding author and was supported partially by National Science Founda-tion of China Grant ∗ Clearly, Theorem 1.1 strengthens the Erd˝os-Niven theorem and answers com-pletely Problem 1 of [1]. The key tool of the current paper is to use an effectiveresult of Dusart [3] on the distribution of primes, see Lemma 2.3 below.The paper is organized as follows. First, in Section 2, we show several lemmaswhich are needed for the proof of Theorem 1.1. Finally, in Section 3, we give thedetails of the proof of Theorem 1.1.As usual, we denote by ⌊ x ⌋ and ⌈ x ⌉ the biggest integer no more than x and thesmallest integer no less than x , respectively. Let v p denote the p -adic valuationon the field Q of rational numbers, i.e., v p ( a ) = b if p b divides a and p b +1 doesnot divide a . 2. Preliminary lemmas
In this section, we show some preliminary lemmas, which are needed for theproof of Theorem 1.1. We begin with the following well-known result.
Lemma 2.1. [4] [5]
Let a, b, n and k be positive integers with n ≥ . Then S a,b ( n, is not an integer. Lemma 2.2.
Let a, b, n and k be positive integers such that ≤ k ≤ n . If either n ≤ ba (cid:0) e a ( √ b +1 − /b − (cid:1) or k ≥ ea log an + bb + eb , then < S a,b ( n, k ) < .Proof. Evidently, S a,b ( n, k ) >
0. It remains to show that S a,b ( n, k ) <
1. If k = n ,it is easy to see that S a,b ( n, k ) <
1. In the following, we assume that 2 ≤ k ≤ n − n ≤ ba e a ( √ b +1 − /b − ba . By the multinomial expansion theorem,we deduce that S a,b ( n, k ) = X ≤ i <...
But n ≤ ba e a ( √ b +1 − /b − ba gives us that a log an + bb ≤ √ b +1 − b . It thenfollows from (2.3) and k ≥ S a,b ( n, k ) < b ( k − (cid:16) √ b + 1 − b (cid:17) k − + 1 k ! (cid:16) √ b + 1 − b (cid:17) k = √ b + 1 − b (cid:16) √ b +1 − b (cid:17) k − ( k − √ b + 1 − b (cid:16) √ b +1 − b (cid:17) k − k ! . (2.4)Obviously 0 < √ b +1 − b < √ b . Hence for all k ≥
2, wehave 0 < (cid:16) √ b +1 − b (cid:17) k − k ! ≤ (cid:16) √ b +1 − b (cid:17) k − ( k − ≤ . (2 . S a,b ( n, k ) < √ b + 1 − b + ( √ b + 1 − b = 1as desired. So Lemma 2.2 is true if n ≤ ba e a ( √ b +1 − /b − ba .Consequently, let k ≥ ea log an + bb + eb . The multinomial expansion theoremtogether with (2.2) tells us that S a,b ( n, k ) ≤ k ! (cid:16) n − X i =0 b + ai (cid:17) k < k ! (cid:18) b + 1 a log an + bb (cid:19) k . (2 . k ≥ ea log an + bb + eb , one haslog k ! = k X i =2 log i > Z k log x d x > k (log k − ≥ k log (cid:18) b + 1 a log an + bb (cid:19) = log (cid:18) b + 1 a log an + bb (cid:19) k , which implies that the right-hand side of (2.6) is strictly less than 1. So (2.6)concludes the desired result S a,b ( n, k ) < . Thus Lemma 2.2 is proved in thiscase.The proof of Lemma 2.2 is complete. (cid:3)
Lemma 2.3. [2] [3]
For any real number x ≥ , there is a prime number p satisfying that x < p ≤ x (1 +
12 log x ) . Lemma 2.4.
Let a and b be positive integers. Let n be an integer satisfying that n > if a ≤ and b ≤ a ( √ e +1) e a − + 1 , and n > ba (cid:0) e a ( √ b +1 − /b − (cid:1) otherwise. Then for any integer k with ≤ k < ea log an + bb + eb , there is a prime p such that nk +1 < p ≤ nk and p > ak + 2 a + 6 . CHUNLIN WANG AND SHAOFANG HONG ∗ Proof.
To show that there is a prime p such that nk +1 < p ≤ nk , it suffices to showthat the following two inequalities hold: nk + 1 ≥ . nk + 1 ≥ k. (2 . p such that nk + 1 < p ≤ nk + 1 (cid:0) ( n/ ( k + 1)) (cid:1) ≤ nk + 1 (cid:0) k (cid:1) = nk . So we need only to show that (2.7) and (2.8) hold, which will be done in whatfollows.First we prove that (2.7) is true. To do so, let f ( x ) = x − ea log ax + bb − eb + 1) . Then f ′ ( x ) = 1 − e/ ( ax + b ). Further, f ′ ( x ) > ax + b > e . To prove(2.7), it is sufficient to show that f ( n ) >
0. Actually, let ∆ := ea log an + bb + eb . Since∆ > k , we have f ( n ) − n = − < − k + 1). So − n < − k + 1)and (2.7) follows immediately. It remains to show that f ( n ) >
0. We divide theproof into the following three cases:
Case 1. a ≤ b ≤ a ( √ e +1) e a − + 1 and n > f (120000) > f ( n ) > Case 2. a ≤ b > a ( √ e +1) e a − + 1 and n > ba (cid:0) e a ( √ b +1 − /b − (cid:1) . Since a ≥ √ b + 1 − /b ≥ b >
1, then for any real number x > ba (cid:0) e a ( √ b +1 − /b − (cid:1) , we have ax + b > be a ( √ b +1 − /b ≥ (cid:16) a ( √ e + 1) e a − (cid:17) e a > √ e +1) ae a e a − > e, which implies that f ′ ( x ) >
0. Since b ≥ a ( √ e +1) e a − + 1 and √ b +1 − b < √
2, onededuces that f ( n ) > f (cid:16) ba (cid:0) e a ( √ b +1 − /b − (cid:1)(cid:17) = ba (cid:0) e a ( √ b +1 − /b − (cid:1) − e ( √ b + 1 − b − eb + 1) > √ e + 1) + e a − a − √ e − e ( e a − a ( √ e + 1) − > Case 3. a >
18 and n > ba (cid:0) e a ( √ b +1 − /b − (cid:1) . Since b ≥ √ b + 1 − /b ≥ √ − b , it follows that for any real number YMMETRIC FUNCTIONS OF RECIPROCAL ARITHMETIC PROGRESSIONS 5 x > ba (cid:0) e a ( √ b +1 − /b − (cid:1) , we have ax + b > be a ( √ b +1 − /b ≥ e √ − > e and so f ′ ( x ) >
0. Hence f ( n ) > f (cid:16) ba (cid:0) e a ( √ b +1 − /b − (cid:1)(cid:17) > e √ − − − √ e − e + 1) > . So (2.7) is proved.Second, we show that (2.8) is true. Since k < ea log an + bb + eb < ea (log an +1)+ eb ,so to show that (2.8) holds, we need only to show2 (cid:0) log n − log( k + 1) (cid:1) > ea log an + ea + eb . (2 . n − k + 1) > e a + e n (cid:16) log aa + 1 a + 1 b (cid:17) , (2 . g ( x ) = x − (cid:16) exa + e log aa + ea + eb + 1 (cid:17) − . Then g ′ ( x ) = 1 − / ( x + log a + a/b + a/e + 1). Since log xx ≤ e for any real number x ≥
1, we have g (9) ≥ − (cid:0) ea +1+ ea + eb +1 (cid:1) − ≥ − e + e + e +2) > g ′ ( x ) > x ≥
9. Therefore g ( x ) > x >
9. Underthe assumption, one can conclude that n > e , that is log n >
9. It then followsthat e a + e n (cid:16) log aa + 1 a + 1 b (cid:17) < . k < ea (log an + 1) + eb , one haslog n − k + 1) − > log n − (cid:16) e log ana + ea + eb + 1 (cid:17) − g (log n ) > , which means (2.10) is true. Thereby (2.8) is proved. This concludes that thereis a prime p such that nk +1 < p ≤ nk . Finally, we show that for any prime p with p > nk +1 , one has p > ak + 2 a + 6.To do so, we need only to show that nk +1 > ak + 2 a + 6. Let h ( x ) = x − (cid:16) ea log ax + bb + eb + 1 (cid:17)(cid:16) e log ax + bb + aeb + 2 a + 6 (cid:17) . Then h ′ ( x ) = 1 − e ax + b log ax + bb − ae + 3 abe + 6 beb ( ax + b ) . Since k < ea log an + bb + eb , we have n − ( k + 1)( ak + 2 a + 6) > h ( n ). So, to provethat nk +1 > ak + 2 a + 6, we only need to show that h ( n ) >
0, which we will do inthe following.
CHUNLIN WANG AND SHAOFANG HONG ∗ If a ≤
18 and b ≤ a ( √ e +1) e a − + 1, then for any real number x > h ′ ( x ) ≥ − e log( ax + b ) ax + b − e a + 3 ae + 6 eax + b (2.12) > − e log 120000120000 − e ea + 3 a + 6120000 > − × − ×
18 + 6)120000 > . This implies that h ( n ) > h (120000) > n > a ≤
18 and b > a ( √ e +1) e a − + 1, since 1 ≤ ( √ b + 1 − /b < √ b ≥
2, it then follows that for any real number x > ba (cid:0) e a ( √ b +1 − /b − (cid:1) , we have h ′ ( x ) > − e b log ax + bbax + bb − e (2 ea + 3 a + 6) ax + b> − e b ae a − e (2 ea + 3 a + 6) be a > − e + 3 e + eae a e a − (cid:0) √ e + 1) + e a − a (cid:1) > − e + 9 e √ e + 1) > . Hence for any integer n with n > ba (cid:0) e a ( √ b +1 − /b − (cid:1) , one has h ( n ) > h (cid:16) ba (cid:0) e a ( √ b +1 − /b − (cid:1)(cid:17) > √ e + 1) + ( e a − /a − (cid:0) √ e + e + 1 (cid:1)(cid:0) √ ae + ae + 2 a + 6 (cid:1) > . If a >
18, then by (2.12) and noting that √ b +1 − b ≥ √ − b , it follows that for any real number x > ba (cid:0) e a ( √ b +1 − /b − (cid:1) , one has h ′ ( x ) > − e ( √ − ae ( √ − a − e (2 ea + 3 a + 6) e ( √ − a = 1 − ae ( √ − a (cid:0) √ e + 3 e + 6 ea (cid:1) > − e √ − (cid:0) √ e + 3 e + 6 e (cid:1) > . It then follows from the hypothesis n > ba (cid:0) e a ( √ b +1 − /b − (cid:1) that h ( n ) > h (cid:16) ba (cid:0) e a ( √ b +1 − /b − (cid:1)(cid:17) > (cid:0) e a ( √ − − (cid:1) /a − (cid:0) √ e + e + 1 (cid:1)(cid:0) √ ae + ae + 2 a + 6 (cid:1) > . By the above discussion, we can now conclude that h ( n ) >
0. Hence one getsthat p > nk +1 > ak + 2 a + 6 as desired. YMMETRIC FUNCTIONS OF RECIPROCAL ARITHMETIC PROGRESSIONS 7
This completes the proof of Lemma 2.4. (cid:3)
Now we consider the p -adic valuation of S a,b ( n, k ) for the prime p appeared inLemma 2.4. Lemma 2.5.
Let a, b, n and k be positive integers such that ≤ k ≤ n . If thereis a prime p satisfying that nk +1 < p ≤ nk and p > ak + 2 a + bp , then one has v p ( S a,b ( n, k )) = − k .Proof. Since p > ak +2 a + bp , a and p are relatively prime. Hence there is a uniqueinteger r ∈ { , , , ..., p − } such that p | ( ar + b ). Let a = ar + bp . Then a < a + bp .Evidently, one can split the sum S a,b ( n, k ) into two parts: S a,b ( n, k ) = S + S , where S = X ≤ i < ···
1. Since ar + b ≡ p ), it followsthat i ′ ≡ r (mod p ), i.e. i ′ = r + pi ′′ with 0 ≤ i ′′ ≤ (cid:4) n − r − p (cid:5) . But nk +1 < p ≤ nk implies that k − r + 1 p ≤ n − r − p < k + 1 − r + 1 p . Then one deduces immediately that (cid:4) n − r − p (cid:5) = k + t , where t = (cid:26) − , if n − − rp < k , otherwise = (cid:26) − , if p ( ak + a ) > a ( n −
1) + b , otherwise . Thus the set of all the terms divided by p in { ai + b } n − i =0 is given as follows: { b + ar, b + ar + ap, ..., b + ar + ap ( k + t ) } = { pa , pa + pa, ..., pa + pa ( k + t ) } . Therefore one can rewrite the sum S as follows: S = X ≤ l < ···
1, then by (2.13) v p ( S ) = − k + v p (cid:18) k − Y i =0 ai + a (cid:19) = − k. If t = 0, then (2.13) gives us that CHUNLIN WANG AND SHAOFANG HONG ∗ v p ( S ) = v p (cid:18) X ≤ i < ··· ak + 2 a yields that v p (cid:16) k X i =0 ( a + ai ) (cid:17) = v p ( k + 1) + v p ( ak + 2 a ) − v p (2) = 0 . This infers that v p (cid:18) P ki =0 ( a + ai ) Q ki =0 ( a + ai ) (cid:19) = 0 . Therefore v p ( S ) = − k as claimed. The claim is proved.Let’s now consider v p ( S ). Since p > p (cid:0) ak + 2 a + 2 bp (cid:1) > nk + 1 · a ( k + 2) + p · bp > a ( n −
1) + b, it follows that v p ( ai + b ) ≤ ≤ i ≤ n −
1. Then v p ( S ) = v p (cid:16) X ≤ i < ···
11 + ai j (cid:19) ≥ − k. (2.14)Finally, by the above claim and (2.14), we can derive immediately that v p ( S a,b ( n, k )) = v p ( S + S ) = − k as required. This ends the proof of Lemma 2.5. (cid:3) YMMETRIC FUNCTIONS OF RECIPROCAL ARITHMETIC PROGRESSIONS 9 Proof of Theorem 1.1
This section is devoted to the proof of Theorem 1.1.
Proof of Theorem 1.1 . Clearly S a,b ( n, k ) = 1 if b = n = k = 1 and S a,b ( n, k ) isnot an integer if n = k = 1 and b ≥
2. By Lemma 2.1 we know that S a,b ( n, k ) isnot an integer if k = 1 and n ≥
2. So we let 2 ≤ k ≤ n in what follows.First let a >
18 or b > a ( √ e +1) e a − + 1. If either 2 ≤ n ≤ ba e a ( √ b +1 − /b − ba or k ≥ ea log an + bb + eb , then by Lemma 2.2, one has that 0 < S a,b ( n, k ) < S a,b ( n, k ) is not an integer. If n > ba e a ( √ b +1 − /b − ba and k < ea log an + bb + eb , then by Lemma 2.4 there is a prime p satisfying nk +1 < p ≤ nk and p > ak + 2 a + 6. Hence bp < b ( k +1) n . But from n > ba e a ( √ b +1 − /b − ba and k < ea log an + bb + eb one derives that b ( k + 1) n < b ( ea log an + bb + eb + 1) n = e log(1 + anb ) anb + b + en < . So p > ak + 2 a + 6 > ak + 2 a + 2 b/p . It then follows from Lemma 2.5 that v p (cid:0) S a,b ( n, k ) (cid:1) = − k <
0. Thus S a,b ( n, k ) is not an integer if n > ba e a ( √ b +1 − /b − ba and k < ea log an + bb + eb . This concludes that S a,b ( n, k ) is not an integer if a > b > a ( √ e +1) e a − + 1.Consequently, let a ≤ , b ≤ a ( √ e +1) e a − + 1 and n > k ≥ ea log an + bb + eb , then by Lemma 2.2, one has 0 < S a,b ( n, k ) <
1. If k < ea log an + bb + eb , then by Lemma 2.4 there is a prime p satisfying nk +1 < p ≤ nk and p >ak +2 a +6. Hence bp < b ( k +1) n <
3, which gives that p > ak +2 a +6 > ak +2 a +2 b/p .Then by Lemma 2.5 we obtain that v p (cid:0) S a,b ( n, k ) (cid:1) = − k <
0. So S a,b ( n, k ) is notan integer if a ≤ , b ≤ a ( √ e +1) e a − + 1 and n > S a,b ( n, k ) is not an integer if a ≤ b ≤ a ( √ e +1) e a − + 1, 2 ≤ k < ea log an + bb + eb and ba (cid:0) e a ( √ b +1 − /b − (cid:1) < n ≤ nk +1 , nk ] where 2 ≤ k < e log 120001 + e <
35 and n ≤ p i denote the i -th prime. For 2 ≤ k ≤
34, define i k to be the integer satisfying that kp i k ≥ ( k + 1) p i k − and kp i +1 < ( k + 1) p i for all integers i with i k ≤ i ≤ p = 119993, the biggest prime less than 120000. We list all the valuesof i k and p i k in the following Table 1. Evidently, Table 1 gives us the observationthat ( k + 1) /p i k < / ≤ k ≤
34. We claim that for any integer k with2 ≤ k ≤
34, if kp i k ≤ n ≤ p such that p ≥ p i k and nk +1 < p ≤ nk . Actually, if p i k > nk +1 , then we have done. If p i k ≤ nk +1 , thenthe fact nk +1 < i with i k ≤ i < p i ≤ nk +1 < p i +1 . Since kp i +1 < ( k + 1) p i ≤ n , we have p i +1 < nk . So letting p := p i +1 givers us the desired result nk +1 < p < nk and the claim is proved. ∗ Table 1. k i k p i k
11 11 29 29 37 37 53 127 127 127 127 k
13 14 15 16 17 18 19 20 21 22 23 i k
31 35 35 35 47 48 48 48 63 63 67 p i k
127 149 149 149 211 223 223 223 307 307 331 k
24 25 26 27 28 29 30 31 32 33 34 i k
67 67 67 67 67 67 100 100 100 100 100 p i k
331 331 331 331 331 331 541 541 541 541 541Let’s continue the proof of Theorem 1.1. Let 13 ≤ a ≤
18. Then one caneasily check that b ≤ a ( √ e +1) e a − + 1 <
2. So b = 1. Furthermore we have k < ea log an + bb + eb ≤ ea log 120000 a + e <
6. It then follows that n > ba e a ( √ b +1 − /b − ba = e ( √ − a − a > kp i k . So the above claim infers that there is a prime p such that nk +1 < p ≤ nk . Besides, since k <
6, one has p > n/ ( k + 1) ≥ e ( √ − a − a > a + 2 >ak + 2 a + 2 b/p for all integers a with 13 ≤ a ≤
18. Then applying Lemma 2.5yields v p (cid:0) S a,b ( n, k ) (cid:1) = − k <
0. Hence S a,b ( n, k ) is not an integer in this case.Let 2 ≤ a ≤
12 and b ≤ min { , a ( √ e +1) e a − + 1 } . Then k < ea log an + bb + eb ≤ ea log(120000 a + 1) + e ≤ e log 240001 + e <
20. Define k a := (cid:4) ea log(120000 a +1) + e (cid:5) and n a := max (cid:8) k a p i ka , a ( k a + 1)( k a + 2) + (cid:6) b ( k a +1) p ika (cid:7)(cid:9) . Then the value of n a for 2 ≤ a ≤
12 can be listed as follows: a n a a with 2 ≤ a ≤
12. If n a ≤ n ≤ ≤ k ≤ k a ,then n ≥ k a p i ka ≥ kp i k . It follows that p i ka ≤ nk a ≤ nk . If p i ka ≤ nk +1 , then bythe above claim we know that there is a prime p ′ satisfying that nk +1 < p ′ ≤ nk .Clearly, p ′ > p i ka . If p i ka > nk +1 , then nk +1 < p i ka ≤ nk . This concludes that wecan always choose a prime p ≥ p i ka such that nk +1 < p ≤ nk . For such prime p , wehave p > nk +1 ≥ n a k a +1 ≥ ak + 2 a + 2 b/p . Hence by Lemma 2.5, S a,b ( n, k ) is notan integer. If n ≤ n a − k ≤ k a , then by direct computations using Maple12 (see Program 1 in Appendix) and the following recursive formulas: S a,b (1 ,
1) = 1 b , S a,b ( n,
1) = S a,b ( n − ,
1) + 1 b + ( n − a , (3 . S a,b ( n, k ) = S a,b ( n − , k ) + 1 b + ( n − a S a,b ( n − , k −
1) for 2 ≤ k ≤ n − . S a,b ( n, n ) = 1 b + ( n − a S a,b ( n − , n − , (3 . S a,b ( n, k ) is not an integer in this case. YMMETRIC FUNCTIONS OF RECIPROCAL ARITHMETIC PROGRESSIONS 11
Since a ( √ e +1) e a − + 1 <
28 if 9 ≤ a ≤
12, the above proof implies that S a,b ( n, k ) is not an integer if 9 ≤ a ≤
12. Hence to complete the proof forthe case a ≥
2, one may let 2 ≤ a ≤ ≤ b ≤ a ( √ e +1) e a − + 1. Then k < ea log an + bb + eb ≤ ea log( a +1)+ e <
13. Since k < ea log an + bb + eb , we have n > ba (cid:0) e a ( k/e − /b ) − (cid:1) . We can check that ba (cid:0) e a ( k/e − /b ) − (cid:1) ≥ (cid:0) e k/e − / − (cid:1) > kp i k for each k with 2 ≤ k ≤
12. Hence by the above claim we know thatthere is a prime p ≥ p i k satisfying nk +1 < p ≤ nk . Since p ≥ p i k , by the aboveobservation, one has ( k + 1) /p ≤ ( k + 1) /p i k < /
2. It then follows that n − ( k + 1) (cid:16) ak + 2 a + 2 bp (cid:17) > b (cid:16) e a ( k/e − /b ) a − a − (cid:17) − a ( k + 1)( k + 2) > (cid:16) e a ( k/e − / − a − (cid:17) − a ( k + 1)( k + 2) . (3.4)We can easily check that the right-hand side of (3.4) is positive if 2 ≤ a ≤ ≤ k ≤
12. Thus p > nk +1 > ak + 2 a + 2 b/p . Therefore by Lemma 2.5, S a.b ( n, k )is not an integer if 2 ≤ a ≤ ≤ b ≤ a ( √ e +1) e a − + 1. This concludes that S a,b ( n, k ) is not an integer if a ≥
2. To finish the proof of Theorem 1.1, one needsonly to handle the remaining case a = 1. In the following we let a = 1.Let b ≥
45. Then k < e log n + bb + eb ≤ e log + e <
25 and n > b ( e k/e − /b − k with 2 ≤ k ≤
24, we have n > b (cid:0) e k/e − /b − (cid:1) ≥ (cid:0) e k/e − / − (cid:1) > kp i k and n − ( k + 1)( ak + 2 a + 2 b/p i k ) > b (cid:0) e k/e − /b − − k +1) /p i k (cid:1) − ( k + 1)( k + 2) >
0. Hence by the above claim, there is a prime p ≥ p i k with nk +1 < p ≤ nk . For such a prime p , one has p > nk +1 > ak + 2 a + 2 b/p i k >ak + 2 a + 2 b/p , which implies that S a,b ( n, k ) is not an integer by Lemma 2.5.Let b ≤
44 and k ≥
24. If b = 1, then by [1], S , ( n, k ) is not an integer. If2 ≤ b ≤
44, then k < e log n + bb + eb ≤ log 60001 + e/ <
32 and n > b ( e k/e − /b − k with 24 ≤ k ≤
31 that b ( e k/e − /b − > e k/e − / − > kp i k and b ( e k/e − /b − − ( k + 2)( k + 1) − b ( k + 1) /p i k > e k/e − /b − − k + 1) /p i k ) − ( k + 2)( k + 1) >
0. So by the above claim there is a prime p suchthat nk ≥ p > nk +1 > b ( e k/e − /b − k +1 > k + 2 + 2 b/p . Hence by Lemma 2.5, S a,b ( n, k )is not an integer in this case.Let b ≤
44 and 2 ≤ k ≤
23. If n ≥ n ≥ p i , then n ≥ kp i k for 2 ≤ k ≤
23. It follows from the above claim that there is a prime p satisfying nk +1 < p ≤ nk . Further, one has n ≥ > ( k + 1)( k + 2 + b ) for any integer k with 2 ≤ k ≤
23, and so p > nk +1 > k + 2 + b ≥ k + 2 + 2 b/p . Therefore one yieldsfrom Lemma 2.5 that S a,b ( n, k ) is not an integer. If n ≤ S a,b ( n, k ) is not an integer except that b = 1 , n = 3 and k = 2, inwhich case S a,b ( n, k ) = 1.This completes the proof of Theorem 1.1. ∗ Appendix
Program 1.
IntTest1:=proc(a) local i,j,C,b,k,n,S;C:=min(27, floor(3275*a*(1.4143*2.7183+1)/(2.7182^a-1)+1));k:=floor(2.7183*log[2.7182](120000*a+1)/a+2.7183); S:=vector(k,0);n:=[0,4437,2086,1397,1143,550,640,588,515,571,627,516];for b from 1 to C do S:=vector(k,0);for i from k to n[a] do S[1]:=S[1]+1/(a*i-a*k+b);for j from 2 to k do S[j]:= S[j]+S[j-1]/(a*i-a*k+a*j-a+b);if type(S[j],integer) then print(i-k+j, j*IsInt) end if;end do; end do; end do; end procfor a from 2 to 12 do IntTest1(a) end do;
Program 2.
IntTest2:=proc(b)local i,j,S; S:=vector(23,0);for i from 23 to 7612 do S[1]:=S[1]+1/(i-23+b);for j from 2 to 23 do S[j]:= S[j]+S[j-1]/(i+j-24+b);if type(S[j],integer) then print(i-23+j, j*IsInt) end if;end do; end do; end procfor b from 1 to 44 do IntTest2(b) end do
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13 (1923),10-15.[6] C. Wang and S. Hong, On the integrality of the elementary symmetric functions of 1 , / , ..., / (2 n − Math. Slovaca , in press.
Mathematical College, Sichuan University, Chengdu 610064, P.R. China
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