On sums of coefficients of Borwein type polynomials over arithmetic progressions
aa r X i v : . [ m a t h . N T ] J u l THE BORWEIN CONJECTURES OVER ARITHMETIC PROGRESSIONS
JIYOU LI AND XIANG YU
Abstract.
We obtain asymptotic formulas for sums of coefficients over arithmetic progressions of polyno-mials related to the Borwein conjectures. Let a i denote the coefficient of q i in the polynomial Q nj =1 Q p − k =1 (1 − q pj − k ) s , where p is an odd prime, and n, s are positive integers. In this note, we prove that (cid:12)(cid:12)(cid:12) X i = b mod 2 pn a i − ( p − p sn − n (cid:12)(cid:12)(cid:12) ≤ p sn/ , if b is divisible by p , and (cid:12)(cid:12)(cid:12) X i = b mod 2 pn a i + p sn − n (cid:12)(cid:12)(cid:12) ≤ p sn/ , if b is not divisible by p . This improves a recent result of Goswami and Pantangi [6]. Introduction
Let p and s be two positive integers. For a positive integer n , let the sequence ( a i ) be defined by n Y j =1 p − Y k =1 (1 − q pj − k ) s = sn ( p − p/ X i =0 a i q i . (1.1)In 1990, Peter Borwein discovered some intriguing sign patterns of the coefficients a i for three differentcases ( p, s ) = (3 , , (3 , , (5 , − − , + − − and + − − − − respectively. Equivalently, the sign of a i is determined by i mod p . These conjectures were formalized byAndrews in 1995 [1], which are stated as follows. Conjecture 1.1 (First Borwein conjecture) . For the polynomials A n ( q ) , B n ( q ) and C n ( q ) defined by n Y j =1 (1 − q j − )(1 − q j − ) = A n ( q ) − qB n ( q ) − q C n ( q ) , each has non-negative coefficients. Conjecture 1.2 (Second Borwein conjecture) . For the polynomials α n ( q ) , β n ( q ) and γ n ( q ) defined by n Y j =1 (1 − q j − ) (1 − q j − ) = α n ( q ) − qβ n ( q ) − q γ n ( q ) , each has non-negative coefficients Conjecture 1.3 (Third Borwein conjecture) . For the polynomials v n ( q ) , φ n ( q ) , χ n ( q ) , ψ n ( q ) and ω n ( q ) defined by n Y j =1 (1 − q j − )(1 − q j − )(1 − q j − )(1 − q j − ) = v n ( q ) − qφ n ( q ) − q χ n ( q ) − q ψ n ( ) − q ω n ( q ) , each has non-negative coefficients. All these conjectures had been open for many years. In a recent paper [2], Wang gave an analytic proofof the first Borwein conjecture using the saddle point method and a formula discovered by Andrews [[1],Theorem 4.1] for the polynomials A n ( q ) , B n ( q ) and C n ( q ). It is not clear if his method can be applied toother conjectures. Even for the first Borwein conjecture, an algebraic proof would be very interesting. Instead of evaluating a i directly, it is natural to consider the Borwein conjectures on average overarithmetic progressions. Let d be an integer divisible by p and b be an integer with 0 ≤ b ≤ d −
1. If wedefine S d,b := X i = b mod d a i , then the positivity (negativity) part of the Borwein conjectures follows from the positivity (negativity,respectively) of S d,b for sufficiently large d , say d ≥ sn ( p − p/
2. Please note that here S d,b should be S p,s,n,d,b . For notational simplicity the subscripts p , s and n are omitted when there is no confusion.Using estimates of exponential sums, Zaharescu [3] first studied S d,b for a large classes of d . He provedthe following theorem. Theorem 1.4 (Zaharescu) . Let p, q be two distinct odd primes with q ≤ n , and let b be an integer with ≤ b ≤ pq − . Then (cid:12)(cid:12)(cid:12) S pq,b − ( p − p sn − q (cid:12)(cid:12)(cid:12) ≤ ( p − q − p s ⌊ n/q ⌋− sq ( p − n −⌊ n/q ⌋ q ) q , if b is divisible by p , and (cid:12)(cid:12)(cid:12) S pq,b + p sn − q (cid:12)(cid:12)(cid:12) ≤ ( p − q − p s ⌊ n/q ⌋− sq ( p − n −⌊ n/q ⌋ q ) q , if b is not divisible by p , where ⌊ x ⌋ denote the greatest integers bounded by x . For instance, when ( p, s ) = (3 , (cid:12)(cid:12)(cid:12) S q,b − · n − q (cid:12)(cid:12)(cid:12) ≤ q − ⌊ n/q ⌋− n ( n −⌊ n/q ⌋ q ) q , for b divisible by 3. Note that to insure this bound is nontrivial, q must be a prime bounded by n . Thusa new question naturally arises. Problem 1.5.
For larger d , give a reasonable bound for S d,b . In the case ( p, s ) = (3 , q is a prime and in fact obtained an estimatewith a very small error bound. He showed that Theorem 1.6 (Li) . Let p = 3 , s = 1 , and b be an integer with ≤ b ≤ n − . Then (cid:12)(cid:12)(cid:12)(cid:12) S n,b − · n − n (cid:12)(cid:12)(cid:12)(cid:12) ≤ n , if b is divisible by , and (cid:12)(cid:12)(cid:12)(cid:12) S n,b + 3 n − n (cid:12)(cid:12)(cid:12)(cid:12) ≤ n , if b is not divisible by . Goswami and Pantangi [6] generalized this bound to general cases ( p, s ) and d = pn following Li’sargument and Li-Wan’s sieving argument. They proved the following theorem. Theorem 1.7 (Goswami and Pantangi) . Let p be an odd prime and b be an integer with ≤ b ≤ pn − .Then (cid:12)(cid:12)(cid:12) S pn,b − ( p − p sn − n (cid:12)(cid:12)(cid:12) ≤ p sn/ , if b is divisible by p , and (cid:12)(cid:12)(cid:12) S pn,b + p sn − n (cid:12)(cid:12)(cid:12) ≤ p sn/ , if b is divisible by p . In this note, we improve the result of Goswami and Pantangi to arithmetic progressions with a largercommon difference of 2 pn . We proved the following result: HE BORWEIN CONJECTURES OVER ARITHMETIC PROGRESSIONS 3
Theorem 1.8.
Let p be an odd prime and b be an integer with ≤ b ≤ pn − . Then (cid:12)(cid:12)(cid:12) S pn,b − ( p − p sn − n (cid:12)(cid:12)(cid:12) ≤ p sn/ , if b is divisible by p , and (cid:12)(cid:12)(cid:12) S pn,b + p sn − n (cid:12)(cid:12)(cid:12) ≤ p sn/ , if b is not divisible by p . Notation.
The congruence notion a = b mod n means a − b is divisible by n . We use ord( χ ) to denotethe order of the character χ and | E | to denote the cardinality of the set E . If S is a statement, we use 1 S to denote the indicator function of S , thus 1 S = 1 when S is true and 1 S = 0 when S is false.2. Reduction to a subset-sum type problem
As in [6], we first reduce the problem to a subset-sum type problem over the (additive) group of integersmodulo 2 pn . The starting point is the following equality(1 − q j ) = − q j (1 − q − j ) , which allows us to write the polynomial (1.1) as n Y j =1 p − Y k =1 (1 − q pj − k ) s = n Y j =1 p − Y k =( p +1) / (( − s q s ( pj − k ) (cid:0) − q − ( pj − k ) ) s ) ( p − / Y k =1 (1 − q pj − k (cid:1) s = ( − sn ( p − / q sn ( p − pn +1 − p ) / n Y j = − ( n −
1) ( p − / Y k =1 (1 − q pj − k ) s . Let b i denote the coefficient q i in the Laurent polynomial Q nj = − ( n − Q ( p − / k =1 (1 − q pj − k ) s . Then the aboveequation implies a i = ( − sn ( p − / b i − sn ( p − pn +1 − p ) / . In particular, we have S pn,b = X i = b mod 2 pn a i = ( − sn ( p − / X i = b − sn ( p − pn +1 − p ) / pn b i . (2.1)Thus to prove Theorem 1.8, it suffices to consider the sum P i = b mod 2 pn b i .Let D denote set { pj − k : − ( n − ≤ j ≤ n, ≤ k ≤ ( p − / } . Given integers 0 ≤ m i ≤ | D | , 1 ≤ i ≤ s and 0 ≤ b ≤ pn −
1, we define N ( m , m , . . . , m s ; b ) to be cardinality of the set N D ( m , m , . . . , m s , b ) := { ( V , V , . . . , V s ) : V i ⊂ D, | V i | = m i , ≤ i ≤ s, s X i =1 X x ∈ V i x = b mod 2 pn } . That is, N D ( m , m , . . . , m s , b ) is the number of ordered s -tuples of subsets of D with prescribed cardi-nalities m i which sum to b . In the subset-sum problem, we count the number of subsets (equivalently,1-tuples of subsets) with prescribed cardinality which sum to a given element. Thus this problem canbe viewed as a variant of the subset-sum problem. We also define N D ( b ) to be the alternating sum of N D ( m , m , . . . , m s , b ) N D ( b ) = X ≤ m i ≤| D | , ≤ i ≤ s ( − P si =1 m i N D ( m , m , . . . , m s , b ) . (2.2)From the definitions of b i and N D ( b ), it is not hard to see that N D ( b ) = X i = b mod 2 pn b i . (2.3)The problem is now reduced to counting N D ( b ) and thus to counting N D ( m , m , . . . , m s , b ), which canbe viewed as a subset-sum type problem over the group of integers modulo 2 pn . JIYOU LI AND XIANG YU Li-Wan sieve and some combinatorial formulas
For the purpose of the proof, we briefly introduce the Li-Wan sieve [5] and present some combinatorialformulas.Let A be finite set and let A m be the m -th fold Cartesian product of A . Let X be a subset of A m . Let X denote the elements in X with distinct coordinates X = { ( x , x , . . . , x m ) ∈ X : x i = x j , ∀ i = j } . Let S m be the symmetric group on the set { , , . . . , m } . Given a permutation τ ∈ S m , we can write it adisjoint cycle product τ = C C · · · C c ( τ ) , where c ( τ ) denote the number of disjoint cycles of τ . We definethe signature of τ to be sign( τ ) = ( − k − c ( τ ) . We also define the set X τ to be X τ = { ( x , x , . . . , x m ) ∈ X : x i are equal for i ∈ C j , ≤ j ≤ c ( τ ) } . In other words, X τ is the set of elements in X fixed under the action of τ defined by τ ◦ ( x , x , . . . , x m ) :=( x τ (1) , x τ (2) , . . . , x τ ( m ) ). The Li-Wan sieve gives a formula for calculating sums over X via sums over X τ . Theorem 3.1 ([5], Theorem 2.6) . Let f : X → C be a complex-valued function defined over X . Then wehave X x ∈ X f ( x ) = X τ ∈ S m sign( τ ) X x ∈ X τ f ( x ) . A permutation τ ∈ S m is said to be of type ( c , c , . . . , c m ) if it has exactly c i cycles of length i ,1 ≤ i ≤ m . Let N ( c , c , . . . , c m ) denote the number of permutations of type ( c , . . . , c m ). It is well-known[7] that N ( c , c , . . . , c m ) = m !1 c c !2 c c ! · · · m c m c m ! . (3.1)If we define an m -variate polynomial Z m via Z m ( t , t , . . . , t m ) = 1 m ! X P ic i = m N ( c , c , . . . , c m ) t c t c · · · t c m m , then it follows from (3.1) that Z m satisfies the generating function X m ≥ Z m ( t , t , . . . , t m ) u m = exp( t u + t u t u · · · ) . (3.2)We give some combinatorical lemmas that will be used later. Lemma 3.2 ([4], Lemma 2.3) . If t i = a for e | i and t i = 0 otherwise, then we have Z m ( t , t , . . . , t m ) = Z m (0 , . . . , | {z } e − , a, , . . . , | {z } e − , a, . . . ) = [ u m ](1 − u e ) − a/e . Lemma 3.3.
Let B be a finite set of complex numbers. If t i = P b ∈ B b i a for e | i and t i = 0 otherwise.Then we have Z m ( t , t , . . . , t m ) = Z m (0 , . . . | {z } e − , X b ∈ B b e a, , . . . , | {z } e − , X b ∈ B b e a, . . . ) = [ u m ] Y b ∈ B (1 − b e u e ) − a/e . Proof.
Substituting the values of t i into (3.2), we see that Z m ( t , t , . . . , t m ) = [ u m ] exp( a ∞ X i =1 P b ∈ B b ie u ie ie ) = [ u m ] exp( − ae X b ∈ B log(1 − b e u e )) = [ u m ] Y b ∈ B (1 − b e u e ) − a/e . (cid:3) HE BORWEIN CONJECTURES OVER ARITHMETIC PROGRESSIONS 5 Proof of the main result
Now we prove Theorem 1.8. In view of (2.1) and (2.3), we have S pn,b = ( − sn ( p − / N D ( b − sn ( p − pn + 1 − p ) / . (4.1)Thus we need to estimate the quantity N D ( b ) and thus to estimate the quantity N D ( m , . . . , m s , b ). Asin [4], we use character sums to estimate it. Let G = Z / pn Z be the cyclic group of integers modulo2 pn . Let X i = D m i denote the m i -th fold Cartesian product of D = { pj − k : − ( n − ≤ j ≤ n, ≤ k ≤ ( p − / } and X i denote the set of elements in X i with distinct coordinates. For an ordered k -tuple x = ( x , x , . . . , x m ) ∈ D m , where m is a positive integer, let σ ( x ) := P mi =1 x m denote the sumof its coordinates. Using the fact that | G | P χ ∈ G χ ( x ) is 1 if x = 0 and is 0 otherwise, we can express N D ( m , m , . . . , m , b ) as N D ( m , m , . . . , m s , b ) = 1 m ! m ! · · · m s ! X ( x ,x ,...,x s ) ∈ X × X ×···× X s σ ( x )+ σ ( x )+ ··· + σ ( x s )= b = 1 m ! m ! · · · m s ! X ( x ,x ,...,x s ) ∈ X × X ···× X s | G | X χ ∈ b G χ ( σ ( x ) + σ ( x ) + · · · + σ ( x s ) − b )= 1 | G | X χ ∈ b G χ ( b ) X ( x ,x ,...,x s ) ∈ X × X ×···× X s s Y i =1 (cid:16) m i ! χ ( σ ( x i ) (cid:17) = 1 | G | X χ ∈ b G χ ( b ) s Y i =1 (cid:16) m i ! X x i ∈ X i χ ( σ ( x i )) (cid:17) . (4.2)Thus we need to evaluate the character sums of the form S m ( χ ) := 1 m ! X x ∈ X χ ( σ ( x )) = 1 m ! X ( x ,x ,...,x m ) ∈ X χ ( x ) χ ( x ) · · · χ ( x m ) , (4.3)where X = D m and X τ consists of elements in X fixed by τ .Evaluating S m ( χ ) a distinct coordinate counting problem that can be handled by the Li-Wan sieve.Applying Theorem 3.1, we can write S m ( χ ) as S m ( χ ) = 1 m ! X τ ∈ S k sign( τ ) X ( x ,x ,...,x m ) ∈ X τ χ ( x ) χ ( x ) · · · χ ( x m ) . (4.4)Let τ = C · · · C c ( τ ) be a disjoint cycle product of τ . Then from the definition of X τ , we have X ( x ,x ,...,x m ) ∈ X τ χ ( x ) χ ( x ) · · · χ ( x m ) = c ( τ ) Y i =1 ( X x ∈ D χ ℓ i ( x )) , (4.5)where ℓ i denotes the length of the cycle C i , 1 ≤ i ≤ c ( τ ). Thus we have to determine character sums overthe set D = { pj − k : − ( n − ≤ j ≤ n, ≤ k ≤ ( p − / } .Let [ D ] denote the image of D under the quotient map q : Z → G that sends a to a + 2 pn Z . We observethat [ D ] is a disjoint union of translations of the subgroup pG , where pG = { pg : g ∈ G } . Precisely, wehave [ D ] = U ( p − / k =1 ( pG − k ). Thus X x ∈ D χ ( x ) = X x ∈ [ D ] χ ( x ) = ( p − / X k =1 X x ∈ pG χ ( x − k ) = ( ( p − / X k =1 χ ( k )) X x ∈ pG χ ( x )The sum P x ∈ pG χ ( x ) vanishes unless χ is a trivial character on pG for which ord( χ ) = 1 or p . This impliesthat • P x ∈ D χ ( x ) = 0 if ord( χ ) = 1 , p ; • P x ∈ D χ ( x ) = ( P ( p − / k =1 χ ( k )) | pG | = ( P ( p − / k =1 χ ( k )) | G | /p if ord( χ ) = 1 or p . Note that in thecase ord( χ ) = 1, the formula can be further simplified as P x ∈ D χ ( x ) = | D | . JIYOU LI AND XIANG YU
Now suppose that χ is a the character of order e . From the above discussion, we see that for thecase p ∤ e , P x ∈ D χ i ( x ) = | D | if e | i and P x ∈ D χ i ( x ) = 0 otherwise; for the case p | e , P x ∈ D χ i ( x ) =( P ( p − / k =1 χ i ( k )) | G | p if ep | i and P x ∈ D χ i ( x ) = 0 otherwise. Thus we have two cases. Case 1: p ∤ e . In this case, we have P x ∈ D χ i ( x ) = | D | e | i . Then according to (4.4) and (4.5), wededuce that S m ( χ ) = 1 m ! X τ ∈ S m sign( τ ) ℓ Y i =1 ( X x ∈ D χ ℓ i ( x ))= 1 m ! X P ic i = m N ( c , c , . . . , c m )( − m − P mi =1 c i m Y i =1 ( | D | e | i ) c i = ( − m Z m (0 , . . . , | {z } e − , −| D | , , . . . , | {z } e − , −| D | , . . . )= ( − m [ u m ](1 − u e ) | D | /e . The last step is due to Lemma 3.2.
Case 2: p | e . In this case, we have P x ∈ D χ i ( x ) = ( P ( p − / k =1 χ i ( k )) | G | p ep | i . A similar calculation as inCase 1 shows that the sum S m ( χ ) equals S m ( χ ) = ( − m Z m (0 , . . . , | {z } ep − , ( p − / X k =1 χ ep ( k ) | G | p , , . . . , | {z } dp − , ( p − / X k =1 χ ep ( k ) | G | p , . . . )= ( − m [ u m ] ( p − / Y k =1 (1 − χ e/p ( k ) u e/p ) | G | /e , where we used Lemma 3.3.To sum up, we have S m ( χ ) = ( ( − m [ u m ](1 − u ord( χ ) ) | D | ord( χ ) , if p ∤ ord( χ );( − m [ u m ] Q ( p − / k =1 (1 − χ ord( χ ) p ( k ) u ord( χ ) p ) | G | ord( χ ) , if p | ord( χ ) . Now we are ready to estimate N D ( b ). In view of (2.2) and (4.2), we have N D ( b ) = X ≤ m i ≤| D | , ≤ i ≤ s ( − P si =1 m i N D ( m , m , . . . , m s , b )= 1 | G | X ≤ m i ≤| D | , ≤ i ≤ s ( − P si =1 m i X χ ∈ b G χ ( b ) s Y i =1 S m i ( χ ) . Using (4.3) and the above results for S m ( χ ), we conclude that N D ( b ) = 1 | G | X χ ∈ b G : p | ord( χ ) χ ( b ) X ≤ m i ≤| D | , ≤ i ≤ s s Y i =1 [ u m i ] ( p − / Y k =1 (1 − χ ord( χ ) p ( k ) u ord( χ ) p ) | G | ord( χ ) + 1 | G | X χ ∈ b G : p ∤ ord( χ ) χ ( b ) X ≤ m i ≤| D | , ≤ i ≤ s s Y i =1 [ u m i ](1 − u ord( χ ) ) | D | ord( χ ) = 1 | G | X χ ∈ b G : p | ord( χ ) χ ( b ) s Y i =1 (cid:0) | D | X m i =0 [ u m i ] ( p − / Y k =1 (1 − χ ord( χ ) p ( k ) u ord( χ ) p ) | G | ord( χ ) (cid:1) + 1 | G | X χ ∈ b G : p ∤ ord( χ ) χ ( b ) s Y i =1 (cid:0) | D | X m i =0 [ u m i ](1 − u ord( χ ) ) | D | ord( χ ) (cid:1) HE BORWEIN CONJECTURES OVER ARITHMETIC PROGRESSIONS 7 = 1 | G | X χ ∈ b G : p | ord( χ ) χ ( b ) ( p − / Y k =1 (1 − χ ord( χ ) p ( k )) s | G | ord( χ ) + 1 | G | X χ ∈ b G : p ∤ ord( χ ) χ ( b )(1 − s | D | ord( χ ) = 1 | G | X χ ∈ b G :ord( χ )= p χ ( b ) ( p − / Y k =1 (1 − χ ord( χ ) p ( k )) s | G | p + 1 | G | X χ ∈ b G : p | ord( χ ) , ord( χ ) >p χ ( b ) ( p − / Y k =1 (1 − χ ord( χ ) p ( k )) s | G | ord( χ ) = 1 | G | X χ ∈ b G :ord( χ )= p χ ( b ) ( p − / Y k =1 (1 − χ ( k )) s | G | p + O ( max χ ∈ b G : p | ord( χ ) , ord( χ ) >p (cid:12)(cid:12)(cid:12) ( p − / Y k =1 (1 − χ ord( χ ) p ( k )) (cid:12)(cid:12)(cid:12) s | G | p ) . Note that the implied constant in the big O can be 1.Since G = Z / pn Z , we have | G | = 2 pn . A substitution into the above equation yields N D ( b ) = 12 pn X χ ∈ b G :ord( χ )= p χ ( b ) ( p − / Y k =1 (1 − χ ( k )) sn + O ( max χ ∈ b G : p | ord( χ ) , ord( χ ) >p (cid:12)(cid:12)(cid:12) ( p − / Y k =1 (1 − χ ord( χ ) p ( k )) (cid:12)(cid:12)(cid:12) sn ) . (4.6)For a character χ of order p , let P ( χ ) denote the product P ( χ ) := ( p − / Y k =1 (1 − χ ( k )) . To determine the value of P ( χ ), we evaluate its modulus and argument separately.We first compute the modulus of P ( χ ). By definition, we have | P ( χ ) | = P ( χ ) P ( χ ) = ( p − / Y k =1 (1 − χ ( k )) p − / Y k =1 (1 − χ ( − k )) = ( p − / Y k =1 (1 − χ ( k )) p − / Y k =1 (1 − χ ( p − k )) = p − Y k =1 (1 − χ ( k )) = p . where we used the fact that { χ ( k ), 1 ≤ k ≤ p − } gives a complete list of primitive p -roots of unity. Thisgives | P ( χ ) | = p . Note that for a character χ with p | ord( χ ) and ord( χ ) > p , the character χ ord( χ ) p is oforder p . Thus the above equality also implies that | Q ( p − / k =1 (1 − χ ord( χ ) p ( k )) | = √ p and consequently theerror term in (4.6) is O ( p sn/ )Then we compute the argument of P ( χ ). To this end, we need the explicit form of χ . Since χ is acharacter of order p , it must be of the form χ ( k ) = e πiak/p for some integer a with 1 ≤ a ≤ p −
1. Fromthe definition of P ( χ ), we havearg( P ( χ )) = ( p − / X k =1 arg((1 − χ ( k )) ) mod 2 π. From the equality (1 − e iθ ) = 2(1 − cos( θ )) e i ( θ + π ) , we have arg((1 − χ ( k ) ) = arg((1 − e πiak/p ) ) =2 πak/p + π mod 2 π . This implies thatarg( P ( χ )) = ( p − / X k =1 (2 πak/p + π ) mod 2 π = π ( p − p a + p −
12 ) mod 2 π, Combining the results of the modulus of argument of P ( χ ) together, we conclude that P ( χ ) = ( p − / Y k =1 (1 − χ ( k )) = pe πi ( p − p a + p − ) JIYOU LI AND XIANG YU for χ defined by χ ( k ) = e πiak/p . Substituting this into (4.6), we see that N D ( b ) = p sn pn p − X a =1 e πi bp a e πisn ( p − p a + p − ) + O ( p sn/ ) . (4.7)The above sum is a geometric series with common ratio of e πi ( bp + sn p − p ) . When it is 1, that is, b + sn p − =0 mod p , we have N D ( b ) = p sn pn e πisn p − ( p −
1) + O ( p sn/ ) = ( − sn ( p − / ( p − p sn pn + O ( p sn/ ) . Otherwise we have N D ( b ) = p sn pn e πi (2 bp + sn ( p − p + p − )) − e πisn ( p − + p − ) − e πi ( bp + sn p − ) + O ( p sn/ )= p sn pn e πisn p − e πi (2 bp + sn p − p ) − − e πi ( bp + sn p − ) + O ( p sn/ )= − ( − sn ( p − / p sn pn + O ( p sn/ ) . where we used p − p .In summary, we have N D ( b ) = ( ( − sn ( p − / p − p sn pn + O ( p sn/ ) , if b + sn p − = 0 mod p ; − ( − sn ( p − / p sn pn + O ( p sn/ ) , othewise . (4.8)It then follows from (4.1) and (4.8) that S pn,b = ( p − p sn pn + O ( p sn/ ) if( b − sn ( p − pn + 1 − p )8 ) + sn p −
18 = 0 mod p, (4.9)and S pn,b = − p sn pn + O ( p sn/ ) otherwise. A direct simplification shows that the condition (4.9) is equivalentto b − s n ( n − p p −
12 = 0 mod p, which is equivalent to p | b , since n ( n − is always an integer and p − S pn,b = ( ( p − p sn − pn + O ( p sn/ ) , if p | b ; − p sn − pn + O ( p sn/ ) , if p ∤ b. This completes the proof. (cid:3)
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Department of Mathematics, Shanghai Jiao Tong University, Shanghai, P.R. China
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