Parttons and compostons over fnte felds Muratovć-Rbć Department of Mathematcs Unversty of Saraevo Zmaa od Bosne 33-35, 71000 Saraevo, Bosna and Herzegovna amela@pmf.unsa.ba Qang Wang School of Mathematcs and Statstcs Carleton Unversty Ottawa, K1S 5B6, Canada wang@math.carleton.ca Submtted: Aug 31, 2012; Accepted: Feb 3, 2013; Publshed: Feb 12, 2013 Mathematcs Subect Classfcatons: 11B30, 05A15, 11T06 Abstract In ths paper we fnd an exact formula for the number of parttons of an element z nto m parts over a fnte feld,.e. we fnd the number of nonzero solutons of the euaton x 1 + x 2 + + x m = z over a fnte feld when the order of terms does not matter. Ths s euvalent to countng the number of m-mult-subsets whose sum s z. When the order of the terms n a soluton does matter, such a soluton s called a composton of z. The number of compostons s useful n the study of zeta functons of torc hypersurfaces over fnte felds. We also gve an applcaton n the study of polynomals of prescrbed ranges over fnte felds. 1 Introducton Let n and m be postve ntegers. A composton of n s an ordered lst of postve ntegers whose sum s n. A m-composton of n s an ordered lst of m postve ntegers m parts whose sum s n. It s well known that there s a becton between all m-compostons of n and -subsets of [n 1] = {1, 2,..., n 1} and thus there are n 1 m-compostons of n and 2 n 1 compostons of n. Smlarly, a weak composton of n s an ordered lst of non-negatve ntegers whose sum s n and a weak m-composton of n s an ordered lst of m non-negatve parts whose sum s n. Usng substtuton of varables, we can easly Supported by NSERC of Canada. the electronc ournal of combnatorcs 201 2013, #P34 1
obtan that the number of weak m-compostons of n.e., the number of non-negatve nteger solutons to x 1 + x 2 + + x m = n s eual to the number of m-compostons of n + m.e., the number of postve nteger solutons to x 1 + x 2 + + x m = n + m, whch s n+ = n+ n. The combnatoral nterpretaton of n+ = n+ n s the number of ways n selectng n-multsets from a set M wth m elements, whch s sometmes called n-combnatons of M wth repettons. Dsregardng the order of the summands, we have the concepts of parttons of n nto m parts, parttons of n nto at most m parts, and so on. For more detals we refer the reader to [9]. Let F be a fnte felds of = p r elements. The subset problem over a subset D F s to determne for a gven z F, f there s a nonempty subset {x 1, x 2,..., x m } D such that x 1 + x 2 + + x m = z. Ths subset sum problem s known to be NP -complete. In the study of the subset sum problem over fnte felds, L and Wan [4] estmated the number, Nm, z, D = #{{x 1, x 2,..., x m } D x 1 + x 2 + x m = z}, of m- subsets of D F whose sum s z F. In partcular, exact formulas are obtaned n cases that D = F or F or F \ {0, 1}. Smlarly, we are nterested n the number Sm, z, D = #{x 1, x 2,..., x m D D D x 1 + x 2 + + x m = z}, that s, the number of ordered m-tuples whose sum s z and each coordnate belongs to D F, as well as the number Mm, z, D whch counts the number of m-multsets of D F whose sum s z F. In partcular, when D = F or F, ths motvated us to ntroduce the followng. Defnton 1. A partton of z F nto m parts s a multset of m nonzero elements n F whose sum s z. The m nonzero elements are the parts of the partton. We denote by Mm, z, F or P m z the number of parttons of z nto m parts over F. Smlarly, we denote by Mm, z, F or ˆP m z the number of parttons of z nto at most m parts over F and by P z the total number of parttons of z over fnte feld F. We remark that Nm, z, F s the number of parttons of an element z over fnte feld F such that all summands are dstnct, and Mm, z, F s the number of parttons of an element z nto m parts over fnte feld F, droppng the restrcton that all summands are dstnct. We also remark that n the study of polynomals of prescrbed ranges over fnte felds [5] there has arsen a need as well for countng the number Mm, 0, F of parttons of 0 wth at most m parts over fnte feld F, whch n turn leads us to answer a recent conecture by Gács et al on polynomals of prescrbed ranges over fnte felds [3]. In ths artcle we frst obtan an exact formula for the number of parttons of an element z F nto m parts over F. Theorem 1. Let m be a non-negatve nteger, F be a fnte feld of = p r elements wth prme p, and z F. The number of parttons of z nto m parts over F s gven by P m z = 1 + m 2 + D m z, m the electronc ournal of combnatorcs 201 2013, #P34 2
where D m z = 0, f m 0 mod p and m 1 mod p; 1 /p 1+, f m = p, 0, and z = 0; 1 /p 1+, f m = p + 1, 0, and z = 0; 1 /p 1+, f m = p, 0, and z F ; /p 1+, f m = p + 1, 0, and z F. 1 Smlarly, we have the followng defnton of compostons over fnte felds. Defnton 2. A composton of z F wth m parts s a soluton x 1, x 2,..., x m to the euaton z = x 1 + x 2 + + x m, 1 wth each x F. Smlarly, a weak composton of z F wth m parts s a soluton x 1, x 2,..., x m to Euaton 1 wth each x F. We denote the number of compostons of z havng m parts by Sm, z, F or S m z. The number of weak compostons of z wth m parts s denoted by Sm, z, F. The total number of compostons of z over F s denoted by Sz. A formula for the number of compostons over F p can be found on page 295 n [1]. A general formula for S m z over F for arbtrary and nonzero z can be obtaned usng a remark on the normalzed Jacob sum of the trval character gven n [2] see Remark 1 on page 144. In fact, the numbers S m 1 are the smplest example of the number of ratonal ponts on an affne torc varety over a fnte feld namely a torc hyperplane; see for example [6], [7], and [8]. In order to compare wth the formula for parttons, we only present a recurrence formula for compostons as follows. Proposton 1. Let m > 2, F be a fnte feld of = p r elements wth prme p, and z F. The number of compostons of z wth m parts over F s gven by It follows that and S m z = 1 m 2 2 + S m 2 z. S m 0 = 1m + 1 m 1 S m z = 1m 1 m, f z 0. Usng the fact that addtve group F, + s somorphc to the addtve group F r p, +, we obtan that the numbers of parttons and compostons of elements over F r p are the same as the numbers of parttons and compostons of correspondng elements over F. Fnally, we demonstrate an applcaton of Theorem 1 n the study of polynomals of prescrbed range. Frst let us recall that the range of the polynomal fx F [x] s a multset M of sze such that M = {fx : x F } as a multset that s, not only values, but also multplctes need to be the same. Here and also n the followng sectons we the electronc ournal of combnatorcs 201 2013, #P34 3
abuse the set notaton for multsets as well. In [3], there s a nce connecton between polynomals wth prescrbed ranges and hyperplanes n vector spaces over fnte felds. We refer the reader to ths paper for more detals. In ths paper, we obtan the followng result as an applcaton of Theorem 1. Theorem 2. Let F be a fnte feld of = p r elements. For every l wth l < 3 2 there exsts a multset M wth b M b = 0 and the hghest multplcty l acheved at 0 M such that every polynomal over the fnte feld F wth the prescrbed range M has degree greater than l. We note that Theorem 2 generalzes Theorem 1 n [5] whch dsproves Conecture 5.1 n [3]. In the followng sectons, we gve the proofs of Theorems 1-2 respectvely. 2 Proof of Theorem 1 In ths secton we prove Theorem 1. Frst of all we prove a few techncal lemmas. Lemma 1. Let a F and m be a postve nteger. Then P m a = P m 1. Proof. Let x 1 + x 2 + + x m = 1. The followng mappng between two multsets defned by {x 1, x 2,..., x m } {ax 1, ax 2,..., ax m } for some a F s one-to-one and onto, whch results n ax 1 + ax 2 +... + ax m = a. Thus P m a = P m 1. It s obvous to see that P 1 z = 1 f z F and P 1 0 = 0. However, we can show that P m 0 = P m z f m 0 mod p and m 1 mod p as follows. Lemma 2. Let m be any postve nteger satsfyng m 0 mod p and m 1 mod p. Then P m 0 = P m 1. Proof. Let x 1 + x 2 + + x m = 0 be a partton of 0 nto m parts. Then x 1 + 1 + x 2 + 1 + + x m + 1 = m s a partton of m F wth at most m parts f x = p 1 then x + 1 = 0, but snce x 0 there s no x + 1 = 1. Moreover, there s a bectve correspondence of multsets {x 1,..., x m } {x 1 + 1,..., x m + 1}. Therefore, n order to fnd the number P m 0 of parttons of 0 nto m parts over F, we need to fnd the number of parttons of m wth at most m parts but no element s eual to 1. Ths means these parttons of m can have parts eual to the zero. Let x 1 + x 2 + + x m = m. We assume that the parts eual to 1 f any appear n the begnnng of the lst: x 1, x 2,..., x m. If x 1 = 1 then x 1 + x 2 + + x m = m mples x 2 + + x m = m 1. Conversely, each partton of m nto m 1 parts can generate a partton of m nto m parts wth the frst part eual to 1. So the number of parttons of m nto m parts wth at least one part eual to 1 s eual to the number of parttons of m 1 nto m 1 parts. Let U 0 be the famly of parttons of m nto m parts wthout zero elements and no part s eual to 1. Therefore U 0 = P m m P m 1. the electronc ournal of combnatorcs 201 2013, #P34 4
Let U 1 be the famly of parttons of m wth m parts wth exactly one element eual to 0 and no element eual to 1. Let x 1 +x 2 + +x m = m be a partton n U 1 and x 1 = 0 and x 0, 1 for = 2,..., m. Obvously, t s euvalent to a partton x 2 + +x m = m of m nto m 1 parts wth all parts not eual to 1. Smlarly as n the case for U 0 we have U 1 = P m P m 2 m 1. More generally, let U be the famly of parttons wth m parts wth parts eual to the zero, say x 1 = x 2 =... = x = 0, and x 0, 1 for = + 1,..., m. Then we have a partton of m nto m parts, x +1 + + x m = m, such that no part s eual to 1. Smlarly, we have U = P m m P m 1 m 1. In partcular, for = m 1 there s only one soluton of the euaton x m = m and thus U = P 1 m = 1. We note that these famles of U s are parwse dsont and ther unon s the famly of parttons of m nto m parts wth no part eual to 1. Therefore we have P m 0 = U 0 + U 1 + + U = P m m P m 1 + P m P m 2 m 1 + + P 2 m P 1 m 1 + P 1 m. If m 0 mod p and m 1 mod p, then m 1 and m are both nonzero elements n F. By Lemma 1, we can cancel P m 1 = P m for = 1,..., m 1. Hence P m 0 = P m m = P m 1. Usng the above two lemmas, we obtan the exact counts of P m z when m 0 mod p and m 1 mod p. Lemma 3. If z F and m s any postve nteger satsfyng m 0 mod p and m 1 mod p then we have P m z = 1 + m 2. m Proof. We note that there are 1+ m multsets of m nonzero elements from F n total and the sum of elements n each multset can be any element n F. Usng Lemmas 1 and 2 we have Pm s = P 1 + m 1 m 1 = m s F and therefore + m 2 for every z F. P m z = P m 1 = 1 m In order to consder other cases, we use an nterestng result by L and Wan [4], whch gves the number Nk, b, F of sets wth all dstnct k nonzero elements that sums to b F. Namely, Nk, b, F = 1 1 k+ k/p νb /p 1 + 1, 2 k k/p where νb = 1 f b 0 and νb = 1 f b = 0 see Theorem 1.2 n [4]. Frst we can prove the electronc ournal of combnatorcs 201 2013, #P34 5
Lemma 4. Let Nk, b, F be the number of sets wth k nonzero elements that sums to b F and m > 1 be a postve nteger. Then P m 0 = 1N1, 1, F P 1 + N1, 0, F P 0 1N2, 1, F P m 2 1 + N2, 0, F P m 2 0 +... + 1 1Nm 2, 1, F P 2 1 + Nm 2, 0, F P 2 0 + 1 m 1Nm 1, 1, F + 1 m+1 Nm, 0, F. Proof. Denote by U the famly of all multsets of m nonzero elements that sums to zero,.e. Pm 0 = U. Let B a be the famly of all multsets of m nonzero elements such that a s a member of each multset and the sum of elements of each multset s eual to 0. Namely, B a B a mples s B a s = 0 and a B a. Obvously, U = a F B a. Now we wll use the prncple of ncluson-excluson to fnd the cardnalty of U. For dstnct a 1,..., a k F and k > m, t s easy to see that B a1 B a2... B ak =, because each multset B a1 contans only m nonzero elements. Moreover, f k = m then the number of multsets n the unon of ntersectons s Nm, 0, F. If B B a1 B a2... B ak and k m 1 then B = {a 1, a 2,..., a k, x k+1,..., x m }. Because x k+1 + + x m = a 1 + + a k, the number of elements n the ntersecton B a1 B a2... B ak s the same as the number of parttons of a 1 + + a k nto m k parts,.e. B a1 B a2... B ak = P m k a 1 a k. We note that none of a s = 1,..., k s eual to zero and Nk, b, F = Nk, 1, F for any b F. In partcular, f k < m 1, then the sum a 1 + + a can be any element n F and thus there are 1Nk, 1, F P m k 1 + Nk, 0, F P m k 0 such multsets B B a1 B a2... B ak for all choces of nonzero dstnct a 1,..., a k. If k = m 1 then the sum a 1 + + a can not be eual to the zero, there are n total 1Nm 1, 1, F such multsets contaned n the ntersecton of m 1 famles of B a s. Fnally we combne the above cases and use the prncple of ncluson-excluson to complete the proof. In the seuel we also need the followng dentty whch s a specal nstance of Chu- Vandermonde dentty. Lemma 5. For all postve ntegers s, we have s 1 2 + s 2 + s 1 +1 = s s =1 the electronc ournal of combnatorcs 201 2013, #P34 6
Proof. The result follows from Chu-Vandermonde theorem 2 F 1 s; 1; 2 + s; 1 = 0. Here we also nclude a drect proof. Multplyng 1 + x 1 = 1 1 k=0 x and seres 2 + k 1 1 + x = 1 k=0 k=0 k 1 k x k, We obtan 1 = 1 + x 1 1 1 + x = 1 1 2 + k x 1 k x k = 1 k s 1 2 + s 1 s x s. s s=0 =0 Therefore for s 1 we have s =0 1s 1 2+s s = 0. Ths mples s 1 2 + s 2 + s 1 s +1 = 1 s. s s =1 Fnally multplyng both sdes of the last eualty by 1 s we complete the proof. Next we prove Theorem 1. In order to do so, we let P m z = 1 2 + m + D m z. 3 m Wthout loss of generalty, we can assume > 2. Obvously, by Lemma 3, we have D m z = 0 for any z F f m 0 mod p and m 1 mod p. Further D m z = D m 1 by Lemma 1 for all z 0. Because P m 0 + 1 P m 1 = 2+m m, we have D m 0 + 1D m 1 = 0,.e., D m 1 = 1 1 D m0. 4 Next we use the conventon that P 0 0 = 1 and P 0 1 = 0 so that D 0 0 = 1 and D 0 1 = 1. Smlarly, P1 0 = 0 and P 1 1 = 1 and thus D 1 0 = 1 and D 1 1 = 1. For the rest of ths secton, we only need to compute D m 0 when m = p or m = p + 1 for some postve nteger because of Euaton 4. To do ths, we apply Lemmas 4 and 5, along wth Euatons 2 3, and the followng euaton 1 1Nm, 1, F + Nm, 0, F = m k=0. 5 Let us consder m = up frst. In ths case, by Lemma 4 and Euaton 3, we have: m 2 [ 1 2 + m s P m 0 = 1 s+1 1Ns, 1, F m s + Ns, 0, F s=1 + 1Ns, 1, F D m s 1 + Ns, 0, F D m s 0 ] + 1 m 1Nm 1, 1, F + 1 m+1 Nm, 0, F. the electronc ournal of combnatorcs 201 2013, #P34 7
Usng Euatons 5 and 2, we obtan P m 0 = 1 m 2 1 2 + m s 1 s+1 s m s s=1 m 2 + 1 s+1 1 1 1D m s 1 + D m s 0 s s=1 m 2 + 1 s+1 1 1 s+ s/p 1 /p 1 D m s 1 + D m s 0 s/p s=1 + 1 m 1 1 1 + 1 m+1 1 1 m 1 m + 1 m + /p 1 /p 1 1 1 m 1/p + 1 m+1 1 m+ m/p 1 /p 1 m/p After rearrangng terms, we use Lemma 5, Lemma 3, Euatons 3 and 4 to smplfy the above as follows: = 1 m 1 2 + m s 1 s+1 s m s s=1 /p 1 + 1 s+1 1 s+ s/p D m s 0 s/p 1 s m 2 s 0, 1modp + 1 u 1 1 [ /p 1 /p 1 ] + u 1 u = 1 2 + up + up 1 s up s 0, 1modp 1 1+ s/p /p 1 s/p D up s 0, where we use Lemma 5 and D 0 0 = D 1 0 = 1 to obtan the last eualty. Now let us rewrte ths as P up 0 = 1 2 + up u 1 /p 1 + 1 1+u t D tp 0 up u t t=0 u 1 /p 1 + 1 u t D tp+1 0. 6 u t 1 t=0 the electronc ournal of combnatorcs 201 2013, #P34 8
Smlarly, for m = up + 1, we have P up+1 0 = 1 2 + up + 1 /p 1 + 1 1+ s/p D up+1 s 0 up + 1 s/p 1 s up 1 s 0, 1modp = 1 2 + up + 1 u 1 /p 1 + 1 1+u t D tp 0 + D tp+1 0 D up 0. up + 1 u t t=1 Next we show D up+1 0 = D up 0 for all u 0 by mathematcal nducton. The base case u = 0 holds because D 1 0 = D 0 0 = 1. Assume now D sp0 = D sp+1 0 for all 0 s < u and plug nto the above formula we obtan P up+1 0 = 1 2 + up + 1 D up 0 up + 1 Because P up+1 0 = 1 2+up+1 +Dup+1 0, we conclude that D up+1 up+1 0 = D up 0. Hence t s true for all u 0. Usng ths relaton we smplfy Euaton 6 to P up 0 = 1 2 + up u 1 /p 1 /p 1 + 1 u t+1 + D tp 0 up u t u t 1 t=0 = 1 2 + up u 1 /p + 1 u t+1 D tp 0 7 up u t t=0 and by usng P up 0 = 1 2+up up + Dup 0 we obtan u 1 /p D up 0 = 1 u t+1 D tp 0. 8 u t t=0 Let fx = =0 D p0x be the generatng functon of the seuence {D up 0 : u = 0, 1, 2,...}. Then /p 1 x /p fx = /p 1 l x l D p 0x l l=0 =0 u 1 /p = D 0 0 + 1 u t D tp 0 + D up 0 x u u t u=1 t=0 = D 0 0 + D up 0 + D up 0x u = D 0 0 = 1. Now 1 x /p fx = 1 mples fx = 1 u=1 1 1 x = 1 /p /p 1 + t x t. t the electronc ournal of combnatorcs 201 2013, #P34 9 t=0
Hence D p 0 = 1 /p 1+ for = 0, 1, 2.... Moreover, we use Euaton 4 and D p+1 0 = D p 0 to conclude D p 0 = 1 D p+1 0 = 1 /p 1 + /p 1 + ; D p 1 = 1 ; D p+1 1 = 1 /p 1 + ; /p 1 + Fnally, together wth Lemma 3 we complete the proof of Theorem 1. Fnally we note that t s straghtforward to derve the followng corollary. Corollary 1. Let m be a non-negatve nteger, F be a fnte feld of = p r elements wth prme p, and z F. The number of parttons of z nto at most m parts over F s gven by m ˆP m z = P k z = 1 1 + m + m D m z, where k=0 3 Proof of Theorem 2 D m z = { Dm z, f m 0 mod p; 0, otherwse. Let l = m. The assumpton l < 3 mples that 4 m. As n [5], we 2 2 denote by T the famly of all subsets of F of cardnalty m,.e., T = {T T F, T = m}. Denote by M the famly of all multsets M of order contanng 0 wth the hghest multplcty l = m and the sum of elements n M s eual to 0,.e., M = {M 0 M, multplcty0 = m, b M b = 0}.. We note that the polynomal wth the least degree m such that t sends m values to 0 can be represented by f λ,t x = λ x s, 9 whch unuely determnes a mappng s F \T F : F T M, 10 defned by λ, T rangef λ,t x. the electronc ournal of combnatorcs 201 2013, #P34 10
In Lemma 2 [3] we found an upper bound for the number rangef of the mages of the polynomal wth the least degree m such that t sends m values to 0, when m < p. Usng ths upper bound, we proved that, for every m wth 3 < m mn{p 1, /2}, there exsts a multset M wth b M b = 0 and the hghest multplcty m acheved at 0 M such that every polynomal over F wth the prescrbed range M has degree greater than m Theorem 1, [5]. Ths result dsproved Conecture 5.1 n [3]. In ths secton, we drop the restrcton of m < p and then use the formula obtaned n Theorem 1 to prove Theorem 2, whch generalzes Theorem 1 n [5]. Frst of all, we prove the followng result. Lemma 6. Let be a prme power, m be a postve nteger and d = gcd 1, m 1. 2 Let F : F T M be defned as n Euaton 10. Then rangef 1 2... m + 1 m! where δ = 1 f p m and zero otherwse. + d, >1 1 φ + δ 1 /p, m/p Proof. As n Lemma 2 of [3] we consder the group G of all non-constant lnear polynomals n F [x] actng on the set F T wth acton Φ : cx + b, λ, T c λ, ct + b. All the elements of the same orbt n F T are all mapped to the same range M M. Thus we need to fnd the number N of orbts under ths group acton. Usng the Burnsde s Lemma, we need to fnd the number of fxed ponts F T g n F T under the acton of gx = cx + b. As n Lemma 2 [3], for gx = x there are 1 m elements fxed by gx. Moreover, f gx = cx + b, c 1 then elements are fxed by gx only f = ordc d = gcd 1, m 1 and n ths case we have F T g = 1 1. Under the assumpton m < p n Lemma 2 [3], we don t need to consder gx = x + b, b 0, because t has p-cycles of the form x, x + b,..., x + p 1b and has no fxed elements. However, for arbtrary m, we must consder ths case. In fact, f gx = x + b fxes some subset T of F wth m elements then we must have p m and T conssts of p-cycles. In partcular, there are p m p of such subsets T fxed by gx = x + b for each b F. Varyng λ and b, we therefore obtan F T g = δ 1 2 /p m/p. Now usng Burnsde s Lemma we obtan N = 1 G F T g g G 1 m 1 = 1 = 1 + m >0, d + 1 1 φ + >0, d δ 1 1 /p φ + δ 1 2 m/p /p m/p. the electronc ournal of combnatorcs 201 2013, #P34 11
In order to prove Theorem 2 t s clear that we only need to show 1 2... m + 1 + 1 δ 1 /p φ + < m! m/p P m 0. 11 d, >1 By Theorem 1, t s enough to show 1... m + 1 m! for m = p + 1 and + d, >1 1 2... m + 1 m! 1 φ + 1 1 + p < 1 + m 2. 12 m + d, >1 /p 1 φ < 1 + m 2 m, 13 for all other cases, because 1 /p m/p = 1 1 /p 1+ when m = p and 1. For the cases m = 4 and m = 5, because 2m, we can check drectly that Ineualty 13 holds and thus Ineualty 11 holds. We now show Ineualtes 12 and 13 hold for m > 5 by usng a combnatoral argument. Let G =< a > be a cyclc group of order 1 wth generator a. Let M be the set of all multsets wth m elements chosen from G. Then M = 2+m m. To estmate the left hand sde of Ineualtes 12 and 13 we count now the number of multsets n some subsets of M defned as follows. These subsets of multsets of m elements are defned from subsets of k-subsets of G when k m. Frst of all, let M 0 be the set of all subsets of G wth m elements. So M 0 M and M 0 = 1 m. Let A be the set of all subsets of G wth m 1 elements. For each A = {a u 1, a u 2,..., a u } A where 0 u 1 < u 2 <... < u < 1 we can fnd a multset M = {a u 1, a u 1, a u 2, a u 3,..., a u } correspondng to A n the unue way. We can use notaton s to denote an element s n a multset M wth multplcty. Hence the above multset M can also be denoted by M = {a u 1 2, a u 2, a u 3,..., a u }. The set of all these multsets M, denoted by M 1, has A = 1 elements. Moreover M 0 M 1 =. Now let M 01 = M 1 M 1. Then M 01 = 1 m + 1 = m. For each satsfyng > 2 and d, we let S =< a > be a cyclc subgroup of G wth 1 elements. From each set C of all subsets of S wth elements, we can defne two dsont subclasses of M contanng multsets wth m elements n G correspondng to C. Frst, let B = {a u1, a u2,..., a u } be a subset of S where 0 u 1 < u 2 <... < 1. For each fxed t such that 0 t < and gcd, t = 1, we can construct a multset correspondng to B as follows: M = {a t a u1, a t a u2,..., a t a u, a m } the electronc ournal of combnatorcs 201 2013, #P34 12
where a m s arbtrarly element n G. For each fxed t ths class of multsets formed from C s denote by M t. Then M t = 1 1. Secondly, for B = {a u1, a u2,..., a u } C and each fxed t, we can construct another multset M = {a t+1 a u1, a t a u2,..., a t a u, 1}, correspondng to B. The set of these multsets s denoted by Note that mples M t 2 M t =. Hence we have M = M t M 1 t 1 t< = φ 1 + gcd,t=1 1 M t. Then M t = 1. 1 = φ. Fnally, f m 1 1 then we let M m =. Otherwse, f m 1 1 then we let M t contans all the multsets of the form M = {a t+, a m }, for = 0, 1,..., 1 1, any postve nteger t < m 1 wth gcdm 1, t = 1, and any a m G. Let M t contan all the multsets of the form {a t+ m 2, a 2 }. It s obvous that a a t+. By comparng the multplctes of two multsets we see that M t M t =. Moreover, M = M t M t 1 t< gcd,t=1 = φm 1 1 1 = φm 1. 1 1 + 1 1 Fnally, f m p + 1 for some 1 we let M m =. Otherwse, f m = p + 1 for some 1 we let C = {s 1, s 2,..., s /p } be a subset of G wth /p < 1 elements. For each subset of elements from C we fnd a correspondng multset M n M m from M n the followng way M = {s p 1, s p 2,..., s p, a m } where a m s arbtrary chosen to be an element from G. Thus there are 1 /p+ 1 multsets n M m. Obvously, M m s dsont M where gcdm 1, 1 because the multplcty of at least one of ts element s p 1. Indeed, t could possbly have common elements only wth M but n ths case m 1 = p 1 so M =. Now M m = 1 /p+ 1. the electronc ournal of combnatorcs 201 2013, #P34 13
Defne δ = 0 f m p + 1 for some and δ = 1 f m = p + 1. Then we obtan M LHS := M 01 M Mm = gcd, 1 >1 + φ m d, >1 1 /p + m 1/p 1 + δ 1 m 1/p We note that the multset {1, 1, 1, a, a 2,..., a m 3 } s not ncluded n the M LHS and thus M LHS < M. Dvdng both sdes by, we have 1 + 1 φ + δ 1 /p + m 1/p 1 < 1 + m 2. 14 m m 1/p m d, >1 Hence both Ineualtes 12 and 13 are satsfed. Ths completes the proof of Theorem 2. Acknowledgements We would lke to thank the anonymous referees for helpful suggestons, n partcular, for pontng out the reference [2], and also thank Keth Conrad for useful suggestons and for brngng the papers [6, 7, 8] to our attenton. References [1] B. C. Berndt, R. J. Evans, K. S. Wllams, Gauss and Jacob sums. Canadan Math. Soc. Seres of Monographs and Advanced Texts. John Wley & Sons, New York, 1998. [2] K. Conrad, Jacob sums and Stckelberger s congruence, Ensegn. Math. 41 1995, 141-153. [3] A. Gács, T. Héger, Z. L. Nagy, D. Pálvölgy, Permutatons, hyperplanes and polynomals over fnte felds, Fnte Feld Appl. 16 2010, 301-314. [4] J. L and D. Wan, On the subset sum problem over fnte felds, Fnte Feld Appl., 14 2008, 911-929. [5] A. Muratovć-Rbć and Q. Wang, On a conecture of polynomals wth prescrbed range, Fnte Feld Appl., 18 2012, no. 4, 728 737. [6] D. Wan, Mrror symmetry for zeta functons, In Mrror Symmetry V, AMS/IP Studes n Advanced Mathematcs, Vol.38, 2006, 159-184. [7] D. Wan, Lectures on zeta functons over fnte felds Gottngen Lecture Notes. n Hgher Dmensonal Geometry over Fnte Felds, eds: D. Kaledn and Y. Tschnkel, IOS Press, 2008, 244-268. [8] C. F. Wong, Zeta functons of proectve torc hypersurfaces over fnte felds. Thess Ph.D. Unversty of Calforna, Irvne. 2008, http://arxv.org/pdf/0811.0887 [9] R. P. Stanley, Enumeratve Combnatorcs, Vol I, Cambrdge Unversty Press, 1997.. the electronc ournal of combnatorcs 201 2013, #P34 14