The floor and ceiling functions appear often in mathematics and manipulating sums involving floors and ceilings is a subtle game. Fortunately, the well-known textbook Concrete Mathematics provides a nice introduction with a number of techniques explained and a number of single or double sums treated as exercises. For two such double sums we provide their single-sum analogues. These closed-form identities are given in terms of a dual partition of the multiset (regarded as a partition) of all b-ary digits of a nonnegative integer. We also present the double- and single-sum analogues involving the fractional part function and the shifted fractional part function

Podrug, Luka

Svrtan, Dragutin

English

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MATHEMATICAL COMMUNICATIONS 303Math. Commun. 28(2023), 303–316Some refinements of formulae involving floor and ceilingfunctionsLuka Podrug1,∗and Dragutin Svrtan21 Faculty of Civil Engineering, University of Zagreb, Kačićeva 26, HR-10 000 Zagreb2 Department of Mathematics, Faculty of Science, University of Zagreb, Bijenička cesta30, HR-10 000 ZagrebReceived April 1, 2023; accepted September 15, 2023Abstract. The floor and ceiling functions appear often in mathematics and manipulatingsums involving floors and ceilings is a subtle game. Fortunately, the well-known textbookConcrete Mathematics provides a nice introduction with a number of techniques explainedand a number of single or double sums treated as exercises. For two such double sums weprovide their single-sum analogues. These closed-form identities are given in terms of adual partition of the multiset (regarded as a partition) of all b-ary digits of a nonnegativeinteger. We also present the double- and single-sum analogues involving the fractional partfunction and the shifted fractional part function.AMS subject classifications: 11A99Keywords: sum, floor function, ceiling function, fractional part function1. IntroductionIn this paper, we are concerned with computing some sums involving floor and ceilingfunctions, presenting several new formulae that refine previously known results. Forthe reader’s convenience, we start by stating some basic definitions.Let x be any real number. The floor of x is defined as the greatest integer lessthan or equal to x, that is, bxc = max {n | n ≤ x, n ∈ Z}. The ceiling of x is theleast integer greater than or equal to x, that is, dxe = min {n | n ≥ x, n ∈ Z}. Forevery n ∈ Z and x ∈ R we have bn+ xc = n+ bxc and dn+ xe = n+ dxe. Finally,the fractional part of x is {x} = x− bxc. We sometimes call bxc the integer part ofx since x = bxc + {x} with 0 ≤ {x} < 1. For every n ∈ Z and 0 ≤ x < 1 we have{n+ x} = x.Throughout the paper we use the Iverson notation [4], which encloses a true-or-false statement P in brackets, whose result is 1 if the statement is true, and 0 if thestatement is false. For example,[j = k] ={1, if j = k;0, if j 6= k.∗Corresponding author. Email addresses: luka.podrug@grad.unizg.hr (L. Podrug),dragutin.svrtan@gmail.com (D. Svrtan)https://www.mathos.unios.hr/mcc©2023 School of Applied Mathematics and Computer Science, University of Osijek304 L.Podrug and D. SvrtanOur main result is a closed-form expression for the sum∑k≥1⌊n+ jbk−1bk⌋,where b ≥ 2, 0 ≤ j < b, and n is any nonnegative integer. The sum is infinite, butonly finitely many of its terms are nonzero, and thus it is a well-defined integer. Thenew formula is a refinement of the known identity∑k≥1∑0<j<b⌊n+jbk−1bk⌋= n [3, Ex.44, p. 44]. In the second part of the paper, we obtain the closed-form expression forthe single-sum ceiling analogue∑0≤k≤logb x⌈x+ jbkbk+1⌉,and in Section 4, for the following double- and single-sum fractional analogues:∑0≤k≤logb n∑0<j<b{n+ jbkbk+1},∑0≤k≤logb n{n+ jbkbk+1}.2. Sums involving the floor functionWe start with a problem which appears as an exercise in the book Concrete Mathe-matics [2, Ex. 22, p. 97], but which can be traced back to a much older collection ofthe 400 best problems [5, Problem 4346, p. 48]. This problem concerns the evalua-tion of the sum S(n) =∑k≥1⌊n2k+ 12⌋and we state two most common ways of dealingwith it.In the first approach, for a fixed integer k, let fk(n) =⌊n2k+ 12⌋. The functionx → x2k+ 12 is clearly a continuous, monotonically increasing function. Ifn2k+ 12 =m ∈ N for some n, then we have fk(n− 1) < fk(n). For what value(s) of n does thishappen?n+ 2k−12k= m =⇒ n = 2k−1 (2m− 1)︸ ︷︷ ︸odd.Hence, fk(n−1) < fk(n), when n = b ·2k−1, with b odd. For given n, there is exactlyone b with that property. So, for fixed n, let b be odd and such that n = b · 2kn−1for some kn ≥ 1. We haveS(n) =∑k≥1fk(n) and S(n− 1) =∑k≥1fk(n− 1),Formulae involving floor and ceiling functions 305and fk(n) = fk(n−1) for every k 6= kn. For k = kn we have fkn(n) = fkn(n−1)+1,so S(n) = S(n− 1) + 1 and thus by induction S(n) = n.The second approach to evaluating S(n) (which appears in the original solution)uses binary expansions. Suppose n’s binary expansion isn = (cmcm−1 · · · c1c0)2,i.e.,n = cm2m + cm−12m−1 + · · ·+ c121 + c0,where each ci is either 0 or 1 and the leading bit cm is 1.We claim that only digits cs for s ≥ k − 1 contribute. To show this, we consider⌊n2k+12⌋=∑s≥0cs2s2k+12=∑s≥0cs2s−k +12= ∑0≤s≤k−1cs2s−k +∑s≥kcs2s−k +12=∑s≥kcs2s−k + ∑0≤s≤k−1cs2s−k +12=∑s≥kcs2s−k + ck−1.By summing over k ≥ 1 we get∑k≥1⌊n2k+12⌋=∑k≥0ck +∑k≥1∑s≥kcs2s−k=∑k≥0ck +∑s≥1∑1≤k≤scs2s−k=∑k≥0ck +∑s≥1cs2s − 12− 1=∑k≥0ck +∑s≥12scs −∑s≥1cs=∑s≥02scs= n.More generally, for every base b we have the following identity: [3, Ex. 44, p. 44]∑k≥1∑0<j<b⌊n+ jbk−1bk⌋= n. (1)306 L.Podrug and D. SvrtanTo prove it, we can use the following replication identity [2, 3.26, p. 85]:bxc+⌊x+1m⌋+⌊x+2m⌋+ · · ·+⌊x+m− 1m⌋= bmxcwhich leads to ∑k≥1∑0<j<b⌊n+ jbk−1bk⌋=∑k≥1∑0<j<b⌊nbk+jb⌋=∑k≥1(⌊ nbk−1⌋−⌊ nbk⌋)= n.To move toward a new refined result (stated in Theorem 1 below) we need morenotation. For every base b ≥ 2 let n’s b-ary expansion be n = (cmcm−1 · · · c0)b, thatis,n = cmbm + cm−1bm−1 + · · ·+ c1b+ c0,where ci ∈ {0, 1, . . . , b− 1}, m = blogb nc, and where the leading b-ary digit cm isnonzero. Consider the multiset of b-ary digits {c0, c1, . . . , cm} and let sb(n) denotethe sum of all b-ary digits of n, that is,sb(n) = sbblogb nc∑s=0csbs = blogb nc∑s=0cs.Let λ = λ(n, b) = (λ1 ≥ λ2 ≥ · · · ≥ λm) be the partition containing the b-ary digits{c0, c1, . . . , cm} in descending order and let λ′ = λ′(n, b) =(λ′1 ≥ λ′2 ≥ · · · ≥ λ′b−1)be the transpose of the partition λ. This follows Macdonald’s notation [7]. Notethat λ′j is the number of b-ary digits greater than or equal to j and, since there areno digits greater than b−1, we set λ′b to be 0 . For example, let n = 1024 and b = 3.We have 1024 = (1101221)3. Then s3(1024) = 8, λ = (2 ≥ 2 ≥ 1 ≥ 1 ≥ 1 ≥ 1 ≥ 0)and λ′ = (6 ≥ 2).Remark 1. Observe that both λ and λ′ are partitions of the number sb(n), i.e.,sb(n) = |λ| =∑0≤j≤mλj =∑0<j<bλ′j .Theorem 1. For every base b ≥ 2, fixed j, 0 ≤ j < b, and for every nonnegativeinteger n we have the following identity:∑k≥1⌊n+ jbk−1bk⌋=n− sb(n)b− 1+ λ′b−j . (2)For j = 0 and b = p prime, we obtain the right equality of the well-known numbertheoretic formula:νp(n!) =∑k≥1⌊npk⌋=n− sp(n)p− 1,where νp(m) is the multiplicity of p in the factorization of m.First, we state and prove an auxiliary result.Formulae involving floor and ceiling functions 307Lemma 1. For every base b, let n’s b-ary expansion be n = (cmcm−1 · · · c0)b with0 ≤ cs ≤ b− 1 for every s and 0 ≤ j < b. Then for every k⌊ck + jb+ck−1b2+ · · ·+ c0bk+1⌋= [ck + j ≥ b] .Proof of Lemma 1. Since⌊ck + jb+ck−1b2+ · · ·+ c0bk+1⌋≤⌊2b− 2b+b− 1b2+ · · ·+ b− 1bk+1⌋=⌊2− 2b+1b− 1b2+1b2− 1b3+ · · ·+ 1bk− 1bk+1⌋=⌊2− 1b− 1bk+1⌋< 2,the integer part⌊ck+jb +ck−1b2 + · · ·+c0bk+1⌋is either 0 or 1.For ck + j ≥ b we have⌊ck+jb +ck−1b2 + · · ·+c0bk+1⌋= 1, and if ck + j < b, we haveck + j ≤ b− 1, thus0 ≤⌊ck + jb+ck−1b2+ · · ·+ c0bk+1⌋≤⌊b− 1b+b− 1b2+ · · ·+ b− 1bk+1⌋=⌊1− 1b+1b− 1bk+1⌋=⌊1− 1bk+1⌋= 0.Proof of Theorem 1. By substituting n with the b-ary expansion of n and byusing Lemma 1 we obtain∑k≥1⌊n+ jbk−1bk⌋=∑k≥1⌊nbk+jb⌋=∑k≥1∑s≥0csbsbk+jb=∑k≥1∑s≥kcsbs−k +ck−1 + jb+ck−2b2+ · · ·+ c0bk=∑k≥1∑s≥kcsbs−k +∑k≥1⌊ck−1 + jb+ck−2b2+ · · ·+ c0bk⌋308 L.Podrug and D. Svrtan=∑k≥1∑s≥kcsbs−k +∑k≥1[ck−1 + j ≥ b] (by Lemma 1)=∑s≥1cs(b0 + b1 + · · ·+ bs−1)+∑s≥0[cs ≥ b− j]=∑s≥1cs ·bs − 1b− 1+ λ′b−j=1b− 1∑s≥1csbs + c0 − c0 −∑s≥1cs+ λ′b−j=1b− 1∑s≥0csbs −∑s≥0cs+ λ′b−j=n− sb(n)b− 1+ λ′b−j .The double-sum formula (1) then follows immediately:Corollary 1. For every base b ≥ 2 and for every nonnegative integer n we have thefollowing identity: ∑k≥1∑0<j<b⌊n+ jbk−1bk⌋= n.Proof. By changing the order of summation the left-hand side is equal to:∑k≥1∑0<j<b⌊n+ jbk−1bk⌋=∑0<j<b∑k≥1⌊n+ jbk−1bk⌋=∑0<j<b(n− sb(n)b− 1+ λ′b−j)(by Theorem 1)= n− sb(n) + (λb−1 + · · ·+ λ1)= n.3. Sums involving ceiling functionNow we turn our attention to the well-known ceiling analogue of Formula (1). Adouble sum [2, Ex. 39, p. 99] ∑0≤k≤logb x∑0<j<b⌈x+ jbkbk+1⌉is equal to (b − 1) (blogb xc+ 1) + dxe − 1 for every real number x ≥ 1 and everyinteger b ≥ 2.Formulae involving floor and ceiling functions 309More generally, we state the following refinement of this formula. Again, letb ≥ 2 be any base, let n := dxe and let n = (cmcm−1 · · · c0)b be n’s b-ary expansion,λ = λ(n, b) and λ′ = λ′(n, b), as before, partition of the multiset of the b-ary digitsof n and its transpose. Note that m = blogb nc is the position of the leading digit inn’s b-ary expansion. We also define νb(n) := max{k | bk\n}. We shall evaluate thesum for every fixed j, 0 < j < b.Theorem 2. For every base b ≥ 2, fixed j, 0 < j < b, and for every real x ≥ 1 wehave the following identity:∑0≤k≤logb x⌈x+ jbkbk+1⌉=n− sb (n)b− 1+m+ λ′b−j +[cνb(n) 6= b− j], (3)where n = dxe and m = blogb nc is the position of the leading digit in n’s b-aryexpansion.Proof. Let b ≥ 2, 0 < j < b and 0 ≤ k ≤ logb x be integers and f(x) =x+jbkbk+1.Then f is a continuous, monotonically increasing function. Suppose f(x) = z ∈ Z.Then we have x = zbk+1− jbk ∈ Z and, since f satisfies the necessary conditions [2,(3.10), p. 71], we can conclude that df(x)e = df (dxe)e for all x ∈ R. Then∑0≤k≤logb x⌈x+ jbkbk+1⌉=∑0≤k≤logb x⌈n+ jbkbk+1⌉.Suppose n’s b-ary expansion isn = (cmcm−1 · · · c1c0)b,i.e.,n = cmbm + cm−1bm−1 + · · ·+ c1b1 + c0,where each cs ∈ {0, 1, . . . , b− 1} and the leading digit cm is nonzero.∑0≤k≤logb x⌈x+ jbkbk+1⌉=∑0≤k≤logb x⌈nbk+1+jb⌉=∑0≤k≤logb x∑s≥0csbsbk+1+jb=∑0≤k≤logb x∑s≥k+1csbsbk+1+∑0≤s≤kcsbsbk+1+jb=∑0≤k≤logb x∑s≥k+1csbs−k−1+∑0≤k≤logb x⌈ck + jb+ck−1b2+ · · ·+ c0bk+1⌉.310 L.Podrug and D. SvrtanBy changing the order of summation, the first (double) sum is equal to:∑0≤k≤logb x∑s≥k+1csbs−k−1 =∑1≤s≤logb x+1csbs−1∑0≤k≤s−1b−k=∑1≤s≤logb x+1csbs−1 · 1− b−s1− b−1=∑1≤s≤logb x+1cs ·bs−1 − b−11− b−1=∑1≤s≤logb x+1cs ·bs − 1b− 1=1b− 1 ∑1≤s≤logb x+1csbs −∑1≤s≤logb x+1cs=n− sb (n)b− 1.The second (single) sum is bounded by⌈ck + jb+ck−1b2+ · · ·+ c0bk+1⌉≤⌈2b− 2b+b− 1b2+ · · ·+ b− 1bk+1⌉=⌈2− 2b+1b− 1b2+1b2− 1b3+ · · ·+ 1bk− 1bk+1⌉=⌈2− 1b− 1bk+1⌉≤ 2.Hence,⌈ck+jb +ck−1b2 + · · ·+c0bk+1⌉is either 1 or 2.For ck + j > b we have⌈ck+jb +ck−1b2 + · · ·+c0bk+1⌉= 2. For ck + j < b we haveck + j ≤ b− 1 and1 ≤⌈ck + jb+ck−1b2+ · · ·+ c0bk+1⌉≤⌈b− 1b+b− 1b2+ · · ·+ b− 1bk+1⌉=⌈1− 1b+1b− 1b2+1b2− 1b3+ · · ·+ 1bk− 1bk+1⌉=⌈1− 1bk−1⌉= 1.For ck + j = b and ck−1 = ck−2 = · · · = c0 = 0 we have that bk exactly divides n, soνb(n) = k and ⌈ck + jb+ck−1b2+ · · ·+ c0bk+1⌉= 1.Formulae involving floor and ceiling functions 311Furthermore, for ck + j = b and at least one cs > 0 with 0 ≤ s ≤ k − 1 we have⌈ck+jb +ck−1b2 + · · ·+c0bk+1⌉= 2. So:∑0≤k≤logb x⌈ck + jb+ck−1b2+ · · ·+ c0bk+1⌉= m+ 1 + λ′b−j −[cνb(n) = b− j]= m+ λ′b−j +[cνb(n) 6= b− j].Finally, we have∑0≤k≤logb x⌈x+ jbkbk+1⌉=n− sb (n)b− 1+m+ λ′b−j +[cνb(n) 6= b− j].Corollary 2. For every base b ≥ 2 and for every real x ≥ 1 we have the followingidentity: ∑0≤k≤logb x∑0<j<b⌈x+ jbkbk+1⌉= (b− 1) (m+ 1) + n− 1,where n = dxe and m = blogb nc is the position of the leading digit in n’s b-aryexpansion.Proof.∑0≤k≤logb x∑0<j<b⌈x+ jbkbk+1⌉=∑0<j<b∑0≤k≤logb x⌈x+ jbkbk+1⌉=∑0<j<b(n− sb (n)b− 1+m+ 1 + λ′b−j −[cνb(n) = b− j])=n− sb (n) + (b− 1) (m+ 1)+∑0<j<b(λ′b−j −[cνb(n) = b− j]).It remains to prove that∑0<j<b(λ′b−j −[cνb(n) = b− j])= sb (n)− 1.By Remark 1,∑0<j<bλ′b−j = sb (n). We can also see that∑0<j<b[cνb(n) = b− j]= 1.This is true because there is exactly one k that νb(n) = k. Since cνb(n) > 0 andj ranges from 1 to b − 1, the required digit that makes expression[cνb(n) = b− j]equal to 1 must also occur and also exactly once.312 L.Podrug and D. SvrtanFinally, we have∑0≤k≤logb x∑0<j<b⌈x+ jbkbk+1⌉=n− sb (n) + (b− 1) (m+ 1)+∑0<j<b(λ′b−j −[cνb(n) = b− j])=n− sb (n) + (b− 1) (m+ 1) + sb (n)− 1=(b− 1) (m+ 1) + n− 1.4. Sums involving fractional part functionIn the following theorem, we obtain the fractional analogue of the floor and ceilingformulae (2) and (3). Esser, Tao, Totaro, and Wang [1] considered signed fractionalfunction g(x) = x+⌊12 − x⌋which takes values in〈− 12 ,12], and proved thatr∑j=0g(k2j)+ 2g(k2r+1)= 1.Here we give a formula for a sum where a standard fractional part function is used.Theorem 3. For every base b ≥ 2, fixed j, 0 < j < b, and for every nonnegativeinteger n we have the following identity:∑0≤k≤logb n{n+ jbkbk+1}=1b− 1(sb(n)−nbm+1)+ (m+ 1)jb− λ′b−j ,where m = blogb nc is the position of the leading digit in n’s b-ary expansion.Proof. Consider n’s b-ary expansion as before.∑0≤k≤logb n{n+ jbkbk+1}=∑0≤k≤logb n∑s≥0csbsbk+1+jb=∑0≤k≤logb n ∑s≥k+1csbsbk+1+∑0≤s≤kcsbsbk+1+jb=∑0≤k≤logb n ∑0≤s≤kcsbsbk+1+jb.Formulae involving floor and ceiling functions 313The inner sum is bounded byck + jb+ck−1b2+ · · ·+ c0bk+1≤ 2b− 2b+b− 1b2+ · · ·+ b− 1bk+1= 2− 1b− 1bk+1< 2.So,{ ∑0≤s≤kcsbsbk+1+ jb}is either∑0≤s≤kcsbsbk+1+ jb or∑0≤s≤kcsbsbk+1+ jb−1. For ck+j ≥ bwe have∑0≤s≤kcsbsbk+1+ jb ≥ 1 and{ ∑0≤s≤kcsbsbk+1+ jb}=∑0≤s≤kcsbsbk+1+ jb−1. For ck+j < bwe have ck + j ≤ b− 1 and{ck + jb+ck−1b2+ · · ·+ c0bk+1}≤{b− 1b+b− 1b2+ · · ·+ b− 1bk+1}= 1− 1bk−1< 1.We conclude that{ ∑0≤s≤kcsbsbk+1+ jb}=∑0≤s≤kcsbsbk+1+ jb .Therefore,∑0≤k≤logb n{n+ jbkbk+1}=∑0≤k≤logb n ∑0≤s≤kcsbsbk+1+jb− [ck ≥ b− j]=∑0≤s≤logb ncs∑s≤k≤logb nbs−k−1 + (blogb nc+ 1)jb− λ′b−j=∑0≤s≤logb ncs ·b−1 − bs−blogb nc−21− 1b+ (m+ 1)jb− λ′b−j=∑0≤s≤logb ncs ·1− bs−m−1b− 1+ (m+ 1)jb− λ′b−j=∑0≤s≤logb ncsb− 1−∑0≤s≤logb ncsbsbm+1(b− 1)+ (m+ 1)jb− λ′b−j=1b− 1(sb(n)−nbm+1)+ (m+ 1)jb− λ′b−j .Finally, by simple summation of the formula in Theorem 3 over 0 < j < b, theclosed-form formula for the double-sum fractional analogue reads:314 L.Podrug and D. SvrtanCorollary 3. For every base b ≥ 2 and for every nonnegative integer n we have thefollowing identity: ∑0≤k≤logb n∑0<j<b{n+ jbkbk+1}= (m+ 1)b− 12− nbm+1,where m = blogb nc is the position of the leading digit in n’s b-ary expansion.Proof. By changing the order of summation the left-hand side is equal to:∑0<j<b∑0≤k≤logb n{n+ jbkbk+1}=∑0<j<b(1b− 1(sb(n)−nbm+1)+ (m+ 1)jb− λ′b−j)= sb(n)−nbm+1+ (m+ 1)b(b− 1)2b− sb(n)= (m+ 1)b− 12− nbm+1.Yet another among 400 best problems [5, Problem 4375, p. 50] concerns thesummation of the following function:((x)) ={x− bxc − 12(= {x} − 12), if x is not an integer;0, if x is an integer.This function is basic for the definition of the well-known Dedekind sum [6, Def.5.1, p. 92]. Here we can easily obtain the closed-form analogues of single and doublesums for this function.Remark 2. For 0 ≤ k ≤ logb n and fixed j, 0 < j < b, only one among the fractionsn+ jbkbk+1is an integer.Proof. If νb(n) = k, then ck 6= 0 and n =m∑s=kcsbs, where m = blogb nc. Thenn+ jbkbk+1=ck + jb+ ck+1 + ck+2b+ · · ·+ cmbm−k−1.Then ck = b− j gives the only integer value of the fraction above.Theorem 4. For every base b ≥ 2, fixed j, 0 < j < b, and for every nonnegativeinteger n we have the following identity:∑0≤k≤logb n((n+ jbkbk+1))=1b− 1(sb(n)−nbm+1)+ (m+ 1)(jb− 12)− λ′b−j +12[cνb(n) = b− j],where m = blogb nc is the position of the leading digit in n’s b-ary expansion.Formulae involving floor and ceiling functions 315Proof. Consider n’s b-ary expansion as before.∑0≤k≤logb n((n+ jbkbk+1))=∑0≤k≤logb n{n+ jbkbk+1}−∑0≤k≤logb n12+12[cνb(n) = b− j]=1b− 1(sb(n)−nbm+1)+ (m+ 1)jb− λ′b−j −m+ 12+12[cνb(n) = b− j]=1b− 1(sb(n)−nbm+1)+ (m+ 1)(jb− 12)− λ′b−j +12[cνb(n) = b− j].Corollary 4. For every base b ≥ 2 and for every nonnegative integer n we have thefollowing identity:∑0≤k≤logb n∑0<j<b((n+ jbkbk+1))=12− nbm+1,where m = blogb nc is the position of the leading digit in n’s b-ary expansion.Proof. Note that by Remark 2, the sum∑0<j<b12[cνb(n) = b− j]is equal to 12 .∑0≤k≤logb n∑0<j<b((n+ jbkbk+1))=∑0<j<b(1b− 1(sb(n)−nbm+1)+ (m+ 1)(jb− 12)− λ′b−j +12[cνb(n) = b−j])= sb(n)−nbm+1+ (m+ 1)(b(b− 1)2b− b− 12)− sb(n) +∑0<j<b12[cνb(n) = b−j]=12− nbm+1.We hope that our summation techniques may be applied further in number the-ory, approximation theory or elsewhere.References[1] L. Esser, T. Tao, B. Totaro, C. Wang, Optimal Sine and Sawtooth Inequalities,J. Fourier Anal. Appl., Paper No. 14, 28(2002).[2] R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics, Addison-Wesley, 1994.316 L.Podrug and D. Svrtan[3] D. E. Knuth, The Art of Computer Programming, Addison-Wesley, 1997.[4] D. E. Knuth, Two notes on notation, Amer. Math. Monthly 99 (1992), 403–422.[5] H. Eves, E. P. Starke, The Otto Dunkel Memorial Problem Book, Supplement tothe Amer. Math. Monthly 64(1957).[6] F. Hirzebruch, D. Zagier, The Atiyah-Singer Theorem and Elementary NumberTheory, Publish or Perish Inc., 1974.[7] I. G. Macdonald, Symmetric Functions and Hall Polynomials, Oxford UniversityPress Inc., 1995.

Some Refinements of Formulae Involving Floor and Ceiling Functions

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