KYUNGPOOK Math. J. 2019; 59(3): 377-389  
The Convolution Sum al+bm=nδlδm for (a,b) = (1,28),(4,7),(1,14),(2,7),(1,7)
Ayșe Alaca∗, Șaban Alaca and Ebén ézer Ntienjem
School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, K1S 5B6, Canada
e-mails : AyseAlaca@cunet.carleton.ca, SabanAlaca@cunet.carleton.ca and Ebenezer.Ntienjem@carleton.ca
* Corresponding Author.
Received: July 18, 2017; Revised: April 6, 2018; Accepted: July 18, 2018; Published online: September 23, 2019.
© Kyungpook Mathematical Journal. All rights reserved.

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Abstract

We evaluate the convolution sum Wa,b(n):=al+bm=nσ(l)σ(m) for (a, b) = (1, 28), (4, 7), (2, 7) for all positive integers n. We use a modular form approach. We also re-evaluate the known sums W1,14(n) and W1,7(n) with our method. We then use these evaluations to determine the number of representations of n by the octonary quadratic form x12+x22+x32+x42+7(x52+x62+x72+x82). Finally we express the modular forms ∆4,7(z), ∆4,14,1(z) and ∆4,14,2(z) (given in [10, 14]) as linear combinations of eta quotients.

Keywords: convolution sums, sum of divisors function, Eisenstein series, modular forms, cusp forms, Dedekind eta function, eta quotients, octonary quadratic forms, representations.
Introduction
roduction

Let ℕ, ℕ0, and ℂ denote the sets of positive integers, nonnegative integers, integers and complex numbers respectively. For k, n ∈ ℕ the sum of divisors function σk(n) is defined by σk(n)=d|ndk, where d runs through the positive divisors of n. If n ∉ ℕ we set σk(n) = 0. We write σ(n) for σ1(n). For a, b ∈ ℕ with ab we define the convolution sum Wa,b(n) by Wa,b(n):=(l,m)2al+bm=nσ(l)σ(m).Set g = gcd(a, b). Clearly Wa,b(n)={Wa/g,b/g(n/g)ifg|n,0ifgn.Hence we may suppose that gcd(a, b) = 1. The convolution sum Wa,b(n) has been evaluated for (a,b)=(1,b)for1b16,18,20,23,24,25,27,32,36,(2,3),(2,5),(2,9),(3,4),(3,5),(3,8),(4,5),(4,9).See, for example, [2, 3, 4, 5, 7, 10, 13, 14, 17, 18].

In this paper we evaluate the convolution sum Wa,b(n) for (a,b)=(1,28),(4,7),(2,7).We use a modular form approach. The sum W1,14(n) has been evaluated by Royer [14], and the sum W1,7(n) has been evaluated by Lemire and Williams [10] and later by Cooper and Toh [4]. We re-evaluate the sums W1,14(n) and W1,7(n) with our method.

For l, n ∈ ℕ let Rl(n) denote the number of representations of n by the octonary quadratic form x12+x22+x32+x42+l(x52+x62+x72+x82), namely Rl(n):=card{(x1,x2,x3,x4,x5,x6,x7,x8)8|n=x12+x22+x32+x42+l(x52+x62+x72+x82)}.Explicit formulas for Rl(n) are known for l = 1, 2, 3, 4, 5, 6, 8, see, for example, [1, 2, 5, 11, 13]. We use the evaluations of the convolution sums W1,28(n), W4,7(n) and W1,7(n) to determine an explicit formula for R7(n).

Finally we express the modular forms ∆4,7(z), ∆4,14,1(z) and ∆4,14,2(z) (given in [10, 14]) as linear combinations of eta quotients.

Preliminary Results
y Results

Let N ∈ ℕ and . Let Γ0(N) be the modular subgroup defined by Γ0(N)={(abcd)|a,b,c,d,adbc=1,c0(modN)}.We write Mk0(N)) to denote the space of modular forms of weight k and level N. It is known (see for example [15, p. 83]) that Mk(Γ0(N))=Ek(Γ0(N))Sk(Γ0(N)),where Ek0(N)) and Sk0(N)) are the corresponding subspaces of Eisenstein forms and cusp forms of weight k with trivial multiplier system for the modular subgroup Γ0(N).

The Dedekind eta function η(z) is the holomorphic function defined on the upper half plane by the product formula η(z)=eπiz/12n=1(1e2πinz).We set q := q(z) = e2πiz. Then we can express the Dedekind eta function η(z) in (2.2) as η(z)=q1/24n=1(1qn).A product of the form f(z)=1δ|Nηrδ(δz),where , not all zero, is called an eta quotient. We define the following nine eta quotients C1(q):=η5(z)η5(7z)η(2z)η(14z),C2(q):=η2(z)η2(2z)η2(7z)η2(14z),C3(q):=η6(z)η6(14z)η2(2z)η2(7z),C4(q):=η6(2z)η6(7z)η2(z)η2(14z),C5(q):=η2(4z)η4(14z)η2(28z),C6(q):=η6(2z)η6(28z)η2(2z)η2(14z),C7(q):=η4(2z)η6(28z)η2(4z),C8(q):=η(z)η(2z)η(7z)η8(28z)η3(14z),C9(q):=η(2z)η(4z)η9(28z)η3(14z),and integers cr(n) (n ∈ ℕ) for r ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9} by Cr(q)=n=1cr(n)qn.

We use the following theorem to determine if a given eta quotient f (z) is in Mk0(N)). See [8, Theorem 5.7, p. 99] and [9, Corollary 2.3, p. 37].

Theorem 2.1

(Ligozat) Let N ∈ ℕ andf(z)=1δ|Nηrδ(δz)be an eta quotient.

Letk=121δ|Nrδands=1δ|Nδrδ. Suppose that the following conditions are satisfied:

  • 1δ|Nδrδ0(mod 24),

  • 1δ|NNδrδ0(mod 24),

  • 1δ|Ngcd(d,δ)2rδδ0for each positive divisor d of N,

  • k is an even integer,

  • s is the square of a rational integer.

Then f(z) is in Mk0(N)).

(iii)′ In addition to the above conditions, if the inequality in (iii) is strict for each positive divisor d of N, then f(z) is in Sk0(N)).

We note that we have used MAPLE [12] to find the above eta quotients Cj(q) for 1 ≤ j ≤ 9 in a way that they satisfy Theorem 2.1 for N = 28 and k = 4.

The Eisenstein series L(q) and M(q) are defined as L(q):=124n=1σ(n)qn,M(q):=1+240n=1σ3(n)qn,respectively. We use Theorem 2.1 and the Eisenstein series M(q) to give a basis for the modular space M40(28)) in the following theorem.

Theorem 2.2

  • {M (qt) | t = 1, 2, 4, 7, 14, 28} is a basis for E40(28)).

  • {Cj(q) | 1 ≤ j ≤ 9} is a basis for S40(28)).

  • {M (qt) | t = 1, 2, 4, 7, 14, 28} ∪ {Cj(q) | 1 ≤ j ≤ 9} is a basis for M40(28)).

Proof
  • Appealing to [8, Theorem 3.8, p. 50] or [15, Proposition 6.1], we see that dim(E40(28))) = 6. Then we see from [15, Theorem 5.9] that {M (qt) | t = 1, 2, 4, 7, 14, 28} is a basis for E40(28)).

  • It follows from Theorem 2.1 that each Cj(q) is in S40(28)) for 1 ≤ j ≤ 9. By [8, Theorem 3.8, p. 50] or [15, Proposition 6.1], we have dim(S40(28))) = 9. One can see that there is no linear relationship among the eta quotients Cj(q) (1 ≤ j 9). Thus, {Cj(q) |1 ≤ j ≤ 9} constitute a basis for S40(28)).

  • The assertion follows from (a), (b) and (2.1).

We note that {C1(q), C2(q), C3(q), C4(q)} is a basis for S40(14)).

We use the Sturm bound S(N) to show the equality of two modular forms in the same modular space. The following theorem gives S(N) for M40(N)), see [8, Theorem 3.13 and Proposition 2.11] for a general case.

Theorem 2.3

Let f(z), g(z) ∈ M40(N)) with the Fourier series expansionsf(z)=n=0anqnandg(z)=n=0bnqn. The Sturm bound S(N) for the modular space M40(N)) is given byS(N)=N3p|N(1+1/p),and so if an = bnfor all nS(N) then f (z) = g(z).

By Theorem 2.3, the Sturm bounds for the modular spaces M40(14)) and M40(28)) are S(14)=8,S(28)=16,respectively. Using (2.16) and Theorem 2.2 we prove Theorem 2.4. We then use Theorem 2.4 to determine explicit formulas for our convolution sums in the next section.

Theorem 2.4

We have (L(q)28L(q28))2=118125M(q)21125M(q2)112125M(q4)343125M(q7)1029125M(q14)+92512125M(q28)1345225C1(q)8600425C2(q)+252C3(q)+4018825C4(q)+40723225C5(q)+685445C6(q)5241625C7(q)+232780825C8(q)+273100825C9(q),(4L(q4)7L(q7))2=7125M(q)21125M(q2)+1888125M(q4)+5782125M(q7)1029125M(q14)5488125M(q28)8364175C1(q)500425C2(q)+324C3(q)+10716175C4(q)2476825C5(q)+282245C6(q)13881625C7(q)+67660825C8(q)+27340825C9(q),(L(q)14L(q14))2=111125M(q)56125M(q2)686125M(q7)+21756125M(q14)460825C2(q)+67225C3(q)+1027225C4(q),(2L(q2)7L(q7))2=14125M(q)+444125M(q2)+5439125M(q7)2744125M(q14)460825C2(q)+1027225C3(q)+67225C4(q),(L(q)7L(q7))2=1825M(q)+88225M(q7)+5765(C1(q)+4C2(q)).

Proof

We prove only the first and fourth equalities as the remaining three can be proven similarly. Let us prove the first equality. By [15, Theorem 5.8] we have L(q) − 28L(q28) ∈ M20(28)), and so (L(q)28L(q28))2M4(Γ0(28)).By Theorem 2.2(c) there exist coefficients x1, x2, x4, x7, x14, x28, y1, y2, . . . , y9 ∈ ℂ such that (L(q)28L(q28))2=x1M(q)+x2M(q2)+x4M(q4)+x7M(q7)+x14M(q14)+x28M(q28)+i=19yiCi(q).Appealing to (2.16), we equate the coefficients of qn for 0 ≤ n ≤ 16 on both sides of (2.17), and have a system of linear equations with 17 equations and 15 unknowns. By using MAPLE we solve this system and find the asserted coefficients.

Let us now prove the fourth equality. We have 2L(q2)7L(q7)=L(q)7L(q7)(L(q)2L(q2)).By [15, Theorem 5.8], we have L(q)7L(q7)M2(Γ0(7))andL(q)2L(q2)M2(Γ0(2)).Thus it follows from (2.18) and (2.19) that 2L(q2) − 7L(q7) ∈ M20(14)), and so (2L(q2)7L(q7))2M4(Γ0(14)).As M40 (14)) ⊂ M40 (28)), we have (2L(q2) − 7L(q7))2M40 (28)). Thus by Theorem 2.2(c) there exist coefficients x1, x2, x4, x7, x14, x28, y1, y2, . . . , y9 ∈ ℂ such that (2L(q2)7L(q7))2=x1M(q)+x2M(q2)+x4M(q4)+x7M(q7)+x14M(q14)+x28M(q28)+i=19yiCi(q).We equate the coefficients of qnfor 0 ≤ n ≤ 16 on both sides of (2.20) to obtain the asserted coefficients. Alternatively, one can use the fact that {Cj(q) | 1 ≤ j ≤ 4} is a basis for S40(14)), and find the asserted cofficients for the formulas of W2,7(n), W1,14(n) and W1,7(n) accordingly.

Evaluating the Convolution Sum Wa,b(n)
talic#!#)

We now present explicit formulas for the convolution sum Wa,b(n) for (a, b) = (1, 28), (4, 7), (1, 14), (2, 7), (1, 7). We make use of Theorem 2.4 and the classical identity L2(q)=1+n=1(240σ3(n)288nσ(n))qn,see for example [6] and [10, Lemma 3.1].

Theorem 3.1

Let n ∈ ℕ. ThenW1,28(n)=12400σ3(n)+1800σ3(n2)+1150σ3(n4)+492400σ3(n7)+49800σ3(n14)+49150σ3(n28)+(124n112)σ(n)+(124n4)σ(n28)+112167200c1(n)+238922400c2(n)1128c3(n)334967200c4(n)101200c5(n)1740c6(n)+13200c7(n)433150c8(n)25475c9(n),W4,7(n)=12400σ3(n)+1800σ3(n2)+1150σ3(n4)+492400σ3(n7)+49800σ3(n14)+49150σ3(n28)+(124n28)σ(n4)+(124n16)σ(n7)+697470400c1(n)+13922400c2(n)9896c3(n)893470400c4(n)+431400c5(n)740c6(n)+2411400c7(n)8811050c8(n)178525c9(n),W1,14(n)=1600σ3(n)+1150σ3(n2)+49600σ3(n7)+49150σ3(n14)+(124n56)σ(n)+(124n4)σ(n14)+2175c2(n)1600c3(n)1074200c4(n),W2,7(n)=1600σ3(n)+1150σ3(n2)+49600σ3(n7)+49150σ3(n14)+(124n28)σ(n2)+(124n8)σ(n7)+2175c2(n)1074200c3(n)1600c4(n),W1,7(n)=1120σ3(n)+49120σ3(n7)+(124n28)σ(n)+(124n4)σ(n7)170c1(n)235c2(n).

Proof

We prove the theorem only for the convolution sum W1,28(n) as the other four sums can be proven similarly. Replacing q by q28 in (3.1), we obtain L2(q28)=1+n=1(240σ3(n28)727nσ(n28))qn.We have L(q)L(q28)=(124n=1σ(n)qn)(124n=1σ(n)q28n)=124n=1σ(n)qn24n=1σ(n28)qn+576n=1W1,28(n)qn.We obtain from (3.1)(3.3) that (L(q)28L(q28))2=L2(q)+784L2(q28)56L(q)L(q28)=729+n=1(240σ3(n)+188160σ3(n28)+32256(124n112)σ(n)+32256(124n4)σ(n28)32256W1,28(n))qn.We equate the coefficients of qn on the right hand sides of (L(q) − 28L(q28))2 in (3.4) and the first part of Theorem 2.4, and solve for W1,28(n) to obtain the asserted formula.

Theorem 3.2

Let n ∈ ℕ. Then R7(n)=8σ(n)32σ(n4)+8σ(n7)32σ(n28)+64W(1,7)(n)+1024W(1,7)(n4)256(W(4,7)(n)+W(1,28)(n)).

Proof

For n ∈ ℕ0 let r4(n) denote the number of representations of n as sum of four squares, namely r4(n)=card{(x1,x2,x3,x4)4|n=x12+x22+x32+x42},so that r4(0) = 1. It is a classical result of Jacobi, see for example [16], that r4(n)=8d|n4dd=8σ(n)32σ(n4)forn.By (1.2) and (3.5) we have R7(n)=(l,m)02l+7m=nr4(l)r4(m)=r4(n)r4(0)+r4(0)r4(n7)+(l,m)2l+7m=nr4(l)r4(m)=8σ(n)32σ(n4)+8σ(n7)32σ(n28)+(l,m)2l+7m=nr4(l)r4(m).We need to determine the last sum in (3.6). Using (3.5) we obtain (l,m)2l+7m=nr4(l)r4(m)=(l,m)2l+7m=n(8σ(l)32σ(l4))(8σ(m)32σ(m4))=64(l,m)2l+7m=nσ(l)σ(m)+1024(l,m)2l+7m=nσ(l4)σ(m4)256(l,m)2l+7m=nσ(l4)σ(m)256(l,m)2l+7m=nσ(l)σ(m4)=64W1,7(n)+1024W1,7(n/4)256(W4,7(n)+W1,28(n)).The assertion now follows from (3.6) and (3.7).

We deduce the following corollary from Theorems 3.1 and 3.2.

Corollary 3.1

Let n ∈ ℕ. ThenR7(n)=825σ3(n)1625σ3(n2)+12825σ3(n4)+39225σ3(n7)78425σ3(n14)+627225σ3(n28)928175c1(n)76825c2(n)+325c3(n)+2272175c4(n)+230425c5(n)+7685c6(n)115225c7(n)+2457625(c8(n)+c9(n)).

Proof

We substitute the formulas of W1,28(n), W4,7(n), W1,7(n) and W1,7(n/4) from Theorem 3.1 into the right hand side of R7(n) in Theorem 3.2 to obtain the formula R7(n)=825σ3(n)1625σ3(n2)+12825σ3(n4)+39225σ3(n7)78425σ3(n14)+627225σ3(n28)68161225c1(n)5696175c2(n)+325c3(n)+162241225c4(n)+21248175c5(n)+7685c6(n)10624175c7(n)+166912175(c8(n)+c9(n))51235(c1(n/4)+4c2(n/4)).By Theorem 2.1, one can see that C1(q4) and C2(q4) are in S40(56)), and so we have C1(q4) + 4C2(q4) ∈ S40(56)). By Theorem 2.2(b), we know that Cj(q) (1 ≤ j ≤ 9) are in S40(28)) ⊂ S40(56)). We want to see if there exist coefficients x1, x2, …, x9 ∈ ℂ such that C1(q4)+4C2(q4)=x1C1(q)+x2C2(q)++x9C9(q).By Theorem 2.3, the Sturm bound for the modular space M40(56)) is S(56) = 32. By using MAPLE we equate the coefficients of qn for 1 ≤ n ≤ 32 on both sides of (3.9) and have a system of linear equations with 32 equations and nine unknowns. We then solve this system to obtain the identity C1(q4)+4C2(q4)=156C1(q)18C2(q)18C3(q)+156C4(q)+2C5(q)C7(q)2C8(q)2C9(q).We deduce from (3.10) that, for n ∈ ℕ, c1(n/4)+4c2(n/4)=156c1(n)18c2(n)18c3(n)+156c4(n)+2c5(n)c7(n)2c8(n)2c9(n).The asserted expression for R7(n) now follows by substituting (3.11) into (3.8).

Expressing ∆4,7(z), ∆4,14,1(z) and ∆4,14,2(z) as Linear Combinations of Eta Quotients
Quotients

In this section we express the modular forms ∆4,7(z), ∆4,14,1(z) and ∆4,14,2(z) as linear combinations of the eta quotients Ci(q) (1 ≤ i ≤ 4). We refer the reader to [14, Remark 1.2 and Tables 6 and 7] for the definitions of ∆4,7(z), ∆4,14,1(z) and ∆4,14,2(z).

We first express the cusp form ∆4,7(z) as a linear combination of two eta quotients. The sum W1,7(n) has been given by Lemire and Williams [10, Theorem 2] as W1,7(n)=1120σ3(n)+49120σ3(n7)+(124n28)σ(n)+(124n4)σ(n7)170u(n),where u(n) is given by (with our notation) n=1u(n)qn=(η16(z)η8(7z)+13η12(z)η12(7z)+49η8(z)η16(7z))1/3=Δ4,7(z)(withthenotationin[14,Remark1.2])=n=1τ4,7(n)qn.Equating the right hand sides of (4.1) and W1,7(n) in Theorem 3.1, we obtain u(n)=c1(n)+4c2(n)forn,where c1(n) and c2(n) are given by (2.13). Then, by (4.3), (2.4) and (2.5), we obtain Δ4,7(z)=n=1τ4,7(n)qn=n=1u(n)qn=C1(q)+4C2(q).

As for ∆4,14,1(z) and ∆4,14,2(z), which are in S40(14)) (see [14, p. 236]), we use the values of τ4,14,1(n) and τ4,14,2(n) in [14, Tables 6 and 7] and the Sturm bound given in (2.16) to obtain Δ4,14,1(z):=n=1τ4,14,1(n)qn=C3(q)+C4(q),Δ4,14,2(z):=n=1τ4,14,2(n)qn=4C2(q)+C3(q)+C4(q).

The sum W1,14(n) has been given by Royer [14, Theorem 1.7] as W1,14(n)=1600σ3(n)+1150σ3(n2)+49600σ3(n7)+49150σ3(n14)+(124n56)σ(n)+(124n4)σ(n14)3350τ4,7(n)6175τ4,7(n/2)184τ4,14,1(n)1200τ4,14,2(n),where τ4,7(n) and τ4,7(n/2) are given by (4.4), and the values of τ4,14,1(n) and τ4,14,2(n) are given in (4.5) and (4.6), respectively. Finally, equating the right hand sides of (4.7) and W1,14(n) in Theorem 3.1, we obtain 2175c2(n)1600c3(n)1074200c4(n)=3350τ4,7(n)6175τ4,7(n/2)184τ4,14,1(n)1200τ4,14,2(n)forn,from which, by (4.4)(4.6) and (2.13), we obtain the identity 4C1(q2)+16C2(q2)=C1(q)3C2(q)+C3(q)+C4(q).

Acknowledgements

The authors are grateful to Professor Emeritus Kenneth S. Williams for helpful discussions throughout the course of this research. The authors would also like to thank the anonymous referee for helpful comments on the original manuscript.

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