There is a body of work in the literature on various restricted sums of the number of divisors of an integer function including that described in [2-9, 11] and summarised in Section 2 below

Broughan, Kevin A.

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Kevin A. Broughan

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Restricted divisor sums

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ACTA ARITHMETICA101.2 (2002)Restricted divisor sumsbyKevin A. Broughan (Hamilton)1. Introduction. There is a body of work in the literature on variousrestricted sums of the number of divisors of an integer function includingthat described in [2–9, 11] and summarised in Section 2 below. In this paperasymptotic expressions are derived for sums of the form∑1≤n≤xdα(f(n))where the function f(n) is n, n2 or n2 + n + (p + 1)/4, where p ≡ 3 mod 4is a rational prime, and wheredα(n) = #{d : d |n and 1 ≤ d ≤ α}for real α ≥ 1.Motivation for considering these sums comes from an expression whichis derived for the class number of a quadratic field with discriminant −p, interms of a certain restricted divisor sum. This sum is currently too difficultto estimate, in that the restrictions on divisors depend on the summationvariable n.In deriving asymptotic expressions for the sum∑1≤n≤xdα(n2)it is natural to introduce two so-called integer square root functions r+(n)and r−(n). Both are multiplicative and take integer values. Their Dirichletseries are expressible in a compact rational form in terms of the Riemannzeta function.2. The class number. Define, for positive integral n and real a, b witha < b, the restricted divisor function:(1) d(n, (a, b)) = #{d : d |n and a < d < b}.2000 Mathematics Subject Classification: 11A25, 11A51, 11M06, 11N25, 11N37.Key words and phrases: Dirichlet series, divisor sum, class number, integer squareroot.[105]106 K. A. BroughanTheorem 2.1. Let p be a rational prime with p > 3 and p ≡ 3 mod 4.Then the class number for quadratic forms with discriminant −p can beexpressed in the form(2) h(−p)= 1 + 2∑0≤n≤ 12√p3−1d(n2 + n+p+ 14,(2n+ 1,√n2 + n+p+ 14)).Proof. If f(x, y) = ax2 + bxy+ cy2 in Z[x, y] is a quadratic form then itis primitive if(i) 0 ≤ b ≤ a = c or(ii) −a < b < a < c or(iii) 0 < b = a < c.The discriminant is(3) −p = b2 − 4ac,so b 6= 0.Corresponding to each triple (a, b, c) satisfying (i), (ii) or (iii) there is aform, and different triples correspond to inequivalent forms.In case (i) p = (2a)2− b2 = (2a+ b)(2a− b) so 2a+ b = p and 2a− b = 1.Therefore a = (p+ 1)/4 and b = (p− 1)/2 so there is one solution at most.But b ≤ a implies p ≤ 3, so there are no solutions in case (i).In case (ii) assume 1 ≤ b ≤ a < c, since if b is a solution so is −b andvice versa.By (3), b is odd. Since also p ≡ 3 mod 4,ac =b2 + p4is an integer. Hence a | (b2 +p)/4 and a < c therefore a <√(b2 + p)/4 hence4a2 − p < b2 < a2 so a <√p/3 and b ≤√p/3− 1.Similarly a < c if and only if a <√(b2 + p)/4.In case (iii), 1 ≤ b = a < c so p = a(4c − a). The relation a = p isimpossible, since then 4c− a = 1 and c = (p+ 1)/4 < p = a.If a = 1 and 4c− a = p we obtain c = (p+ 1)/4, leading to the so-calledprincipal solution(a, b, c) = (1, 1, (p+ 1)/4).Conversely, if a | (p+b2)/4 and b < a <√(p+ b2)/4 with 1 ≤ b ≤√p/3−1,then −p = b2 − 4ac and 1 ≤ b < a < c.For each odd value of b satisfying 1 ≤ b ≤√p/3−1 we count the numberof divisors a of (b2 +p)/4 satisfying b < a <√(b2 + p)/4, double to accountRestricted divisor sums 107for each solution (a,−b, c), and add 1 for the principal solution to obtainh(−p) = 1 + 2∑1≤b≤√p/3−1b oddd(p+ b24,(b,√p+ b24)).Finally, let b = 2n+ 1 to obtain formula (2).Corollary 2.1. The class number h(−p) is odd.Corollary 2.2. For all primes p ≡ 3 mod 4 with p > 3,h(−p) ≥ d(1 + p4)− 1since (1 + p)/4 is never a square.Corollary 2.3. The following upper bound is an immediate conse-quence:h(−p) ≤(1− 12√p3)+∑0≤n≤ 12√p3−1d(n2 + n+p+ 14).Example 2.1. If p= 59 a set of inequivalent representatives is {(1, 1, 15),(3, 1, 5), (3,−1, 5)} and h(−59) = 3.If p = 151 thend(38, (1, 6]) = 1, d(40, (3, 6]) = 2, d(44, (5, 6]) = 0, d(50, (7, 7]) = 0,so h(−151) = 1 + 2(1 + 2) = 7.3. Existing results. To begin with there is Dirichlet’s famous divisorsum theorem of 1850, and its improvements due to Voronoi in 1903 and vander Corput in 1922. We haveD(x) =∑n≤xd(n) = x log x+ (2γ − 1)x+O(f(x))where a more recent improvement is f(x) = xa with a = 12/37 + ε, due toKolesnik [7], and f(x) = xa logb x with a = 23/73 and b = 461/146, due toHuxley [5], is the best known published result.In 1952 Erdo˝s [2] showed that if f is a polynomial with integer coeffi-cients, then there are positive constants A1 and A2 such thatA1x log x <∑n≤xd(f(n)) < A2x log xwhere the constantsAi depend on the coefficients (and hence also the degree)of f .108 K. A. BroughanIn [12] Scourfield quoting a result of Bellman–Shapiro states that if f isan irreducible quadratic polynomial with integral coefficients, then∑n≤xd(f(n)) = Ax log x+O(x log log x),where the constant A depends on the coefficients of f . This result was im-proved by McKee in [8–10] who derived an error bound of O(x).In 1963 Hooley [4] considered the special case of∑n≤x d(n2 + a) andfound asymptotic expressions for the cases a = −k2 and a 6= −k2.Other results for restricted divisor sums include divisors in short intervals[3, 11] and a number of results for divisors in arithmetic progressions. Themonograph [3] covers in depth a range of related concepts.In this article we begin the task of analysing the class number divisorsum derived in Theorem 2.1 above by looking at the sums where the divisorsare restricted in size, independent of the summation range. This is done firstfor f(n) = n then f(n) = n2 and finally f(n) = n2 + n+ (p+ 1)/4. In eachcase asymptotic expressions are derived.4. Sums restricted by divisor sizeTheorem 4.1. Let 1 ≤ α ≤ x be real numbers. ThenD(x, α) =∑1≤n≤xdα(n) = x logα+ xγ +O(x/α) +O(α).Proof. Simply count the lattice points below the curve uv = x and abovethe interval [1, α] (see for example [1]):D(x, α) =∑1≤j≤α⌊xj⌋=∑1≤j≤α(xj−{xj})= x(logα+ γ +O(1/α)) +O(α)= x logα+ xγ +O(x/α) +O(α)where γ is Euler’s constant.5. Integer square roots. In this section we will derive an asymptoticexpression for the restricted divisor sumD2(x, α) =∑1≤n≤xdα(n2)by first expressing it in terms of the integer square root function r+(n)defined as follows: If n is a positive integer, r+(n) |n and n | r+(n)2, and ifd is such that d |n with n | d2 then r+(n) | d. This defines r+(n) uniquely.Restricted divisor sums 109Let r−(n) = n/r+(n). Then ifn =m∏i=1pαiiwe haver+(n) =m∏i=1pdαi/2ei and r−(n) =m∏i=1pbαi/2ci .Note that r+(n)2 = n if and only if n is a perfect square, that for allprimes p, if p |n then p | r+(n), r+(n)r−(n) = n and√n ≤ r+(n) ≤ n. Also(r+(n), r+(m)) = r+((n,m)) where (n,m) is the greatest common divisor.Finally, both r+ and r− are multiplicative, but not completely multiplica-tive.We will develop four Dirichlet series for these functions:ψ±(s) =∞∑n=1r±(n)ns, φ±(s) =∞∑n=11/r±(n)ns.Theorem 5.1. For σ = <(s) sufficiently large, the Dirichlet series sat-isfy the following :φ+(s) =ζ(2s+ 1)ζ(s+ 1)ζ(2s+ 2)(σ > 0),φ−(s) =ζ(2s+ 1)ζ(s)ζ(2s)(σ > 1),ψ+(s) =ζ(2s− 1)ζ(s− 1)ζ(2s− 2) (σ > 2),ψ−(s) =ζ(2s− 1)ζ(s)ζ(2s)(σ > 1).Proof. If p is a prime then r+(pα) = pdα/2e soφ+(s) =∏p(1 +1r+(p)ps+1r+(p2)p2s+ . . .)=∏p(1 +1pps+1pp2s+1p2p3s+1p2p4s+ . . .)=∏p(1 +ps(pp2s)1+(1pp2s)1+ps(pp2s)2+(1pp2s)2+ . . .)=∏p(11− 1/(pp2s) +ps/(pp2s)1− 1/(pp2s))=∏p(1− 1p2s+1)−1(1 +1ps+1).110 K. A. BroughanButζ(2s)ζ(s)=∏p(1 +1ps)−1.Henceφ+(s) =ζ(2s+ 1)ζ(s+ 1)ζ(2s+ 2).Finally σ + 1 > 1, 2σ + 1 > 1 and 2σ + 2 > 1 if σ > 0.Next, to derive the expression for ψ−(s) useψ−(s) =∞∑n=1r−(n)ns=∞∑n=11/r+(n)ns−1= φ+(s− 1).The other two derivations follow in a similar manner.Lemma 5.1.D2(x, α) =∑1≤j≤α⌊xr+(j)⌋.Proof. If 1 ≤ j ≤ α and jm = n2 for some n ≤ x, let j0 be such thatjj0 = n20 is the smallest multiple of j which is a square: ifj =m∏i=1pαiithenj0 =m∏i=1pαi mod 2iand (αi + αi mod 2)/2 = dαi/2e so r+(j) = n0.ThenD2(x, α) =∑1≤j≤α∑ji=n2n≤x1 =∑1≤j≤α∑jj0m2=n2n≤x1.But jj0m2 = n2 ⇔ n20m2 = n2 ⇔ n0m = n, n ≤ x⇔ m ≤ x/n0. HenceD2(x, α) =∑1≤j≤α⌊xr+(j)⌋.Example 5.1. An elementary derivation leads to an asymptotic formulafor the partial sums of the squarefree reciprocals:F (x) =∑1≤n≤xn squarefree1n=log xζ(2)+O(1).Restricted divisor sums 111To see this, observe that∑1≤n≤xµ(n)2 =∑1≤n≤x1∑d2|nµ(d) =∑d≤√xµ(d)⌊xd2⌋= x∑d≤√xµ(d)d2+O(√x)= x∞∑d=1µ(d)d2+O(√x) =6pi2x+O(√x).The result now follows by Abel’s Theorem for partial summation [1].A more precise result was found by Suryanarayana [13]. It is this formwhich we use in the restricted divisor derivation below, so we state it as alemma.Lemma 5.2 (Suryanarayana).F (x) =1ζ(2)(log x+ γ − 2ζ′(2)ζ(2))+O(1√x).Theorem 5.2. For x→∞ and 1 ≤ α ≤ x,D2(x, α) =x log2 α4ζ(2)+x logαζ(2)[3γ2− ζ′(2)ζ(2)]+O(x) +O(α).Proof. Using Lemma 5.1, we obtainD2(x, α) =∑n≤α⌊xr+(n)⌋= x∑n≤α1r+(n)+∑n≤α{xr+(n)}= xS(α) +O(α),whereS(x) =∑n≤x1r+(n).Now let x = y2 and for d = 1, 2, . . . let Sd be the set of positive integersn ≤ x with largest squared factor d2. Note that if Q(x) is defined to bethe set of squarefree integers less than or equal to x then Sd has |Q(y2/d2)|elements. Note also that if n is squarefree, then r+(nd2) = nd. ThereforeS(y2) =∑d≤y∑n∈Q(y2/d2)1r+(nd2)=∑d≤y∑n∈Q(y2/d2)1nd=∑d≤y1d∑n∈Q(y2/d2)1n=∑d≤y1dF(y2d2).Now letβ = γ − 2ζ′(2)ζ(2)112 K. A. Broughanand apply Lemma 5.2:S(x) =∑n≤√x1nF(xn2)=1ζ(2)[(log x+ β)( ∑n≤√x1n− 2∑n≤√xlog nn)]+O(1√x∑1≤√x1)=1ζ(2)(log x+ β)(log√x+ γ +O(1√x))− 2ζ(2)(12log2√x+ A+O(log x√x))+O(1)=14ζ(2)log2 x+log xζ(2)(3γ2− ζ′(2)ζ(2))+O(1)for some constant A. ThereforeD2(x, α) = x[log2 α4ζ(2)+logαζ(2)(3γ2− ζ′(2)ζ(2))+O(1)]+O(α)=x log2 α4ζ(2)+1ζ(2)x logα[3γ2− ζ′(2)ζ(2)]+O(x) +O(α).6. Bounds for restricted sums for quadratic formsTheorem 6.1. Let f(n) be an irreducible polynomial with f(n) > 0 forn = 1, 2, . . . If 1 ≤ α ≤ x are real then there exist positive constants c1 andc2 such thatc1x logα ≤∑1≤n≤xdα(f(n)) ≤ c2x logα.Proof. Define the three functions θ, % and N byθ(i, j) ={1 if i | f(j),0 if i - f(j),%(i) = #{j : i | f(j), 1 ≤ j ≤ i},N(x, i) = #{j : i | f(j), 1 ≤ j ≤ x}.Then, since i | f(j)⇔ i | f(i+ j) we have(4)⌊xi⌋%(i) ≤ N(x, i) ≤(⌊xi⌋+ 1)%(i)and sox2i%(i) ≤ N(x, i) ≤ 2xi%(i).Restricted divisor sums 113Therefore ∑1≤i≤αx2i%(i) ≤∑1≤i≤α∑1≤j≤xθ(i, j) ≤∑1≤i≤α2xi%(i)and sox2∑1≤i≤α%(i)i≤∑1≤j≤x∑1≤i≤αθ(i, j) ≤ 2x∑1≤i≤α%(i)iorx2R(α) ≤∑1≤j≤xdα(f(j)) ≤ 2xR(α)whereR(x) =∑1≤i≤x%(i)i.By Lemma 9 and Section 5 of [2],2c1 log x ≤ R(x) ≤ 12c2 log xand the conclusion of the theorem follows directly.In case f is quadratic, the previous result can be strengthened to givean asymptotic formula, using the results of McKee [8–10].Theorem 6.2. Let f(n) = an2 + bn+ c be an irreducible quadratic poly-nomial with f(n) > 0 for n = 1, 2, . . . Let ∆ = b2− 4ac < 0 not be a perfectsquare. If 1 ≤ α ≤ x are real then there exists a positive constant Af suchthat ∑1≤n≤xdα(f(n)) = Afx logα+O(x)whereAf =6H(∆)pi√−∆∏p|a(1− 1p+ 1),H(∆) is the weighted class number (namely , the number of primitive andimprimitive forms Ax2 +Bxy+Cy2 with B2−4AC = ∆, giving weight onehalf to forms proportional to x2 + y2, and one third to those proportional tox2 + xy + y2).Proof. Using the same notation as in the previous theorem, it followsdirectly from (4) thatN(x, i) = x%(i)i+O(%(i)).The same argument as that used in the theorem shows that∑1≤n≤xdα(f(n)) = x∑1≤i≤α%(i)i+O( ∑1≤i≤α%(i)).114 K. A. BroughanUsing the results of [10] given in Lemma 7 and Lemma 8 for the two sumsin this formula, we obtain the result of the theorem with the given value forthe constant Af .Acknowledgments. This work was done in part while the author wason study leave at Columbia University. The support of the Departmentof Mathematics at Columbia University and the valuable discussions heldwith Patrick Gallagher are warmly acknowledged. Helpful comments froma referee, especially those resulting in Theorem 6.2, are also gratefully ac-knowledged.References[1] T. M. Apostol, Introduction to Analytic Number Theory , Springer, New York, 1976.[2] P. Erdo˝s, On the sum∑xk=1 d(f(k)), J. London Math. Soc. 27 (1952), 7–15.[3] R. R. Hall and G. Tenenbaum, Divisors, Cambridge Univ. Press, 1988.[4] C. Hooley, On the number of divisors of quadratic polynomials, Acta Math. 110(1963), 97–114.[5] M. N. Huxley, Area, Lattice Points and Exponential Sums, Oxford Univ. Press,1996.[6] A. E. Ingham, Some asymptotic formulae in the theory of numbers, J. London Math.Soc. (3) 2 (1927), 202–208.[7] G. A. Kolesnik, An improvement in the remainder term in the divisor problem,Mat. Zametki 6 (1969), 545–554 (in Russian); English transl.: Math. Notes 6 (1969),784–791.[8] J. F. McKee, On the average number of divisors of quadratic polynomials, Math.Proc. Cambridge Philos. Soc. 117 (1995), 389–392.[9] —, The average number of divisors of an irreducible quadratic polynomial , ibid. 126(1999), 17–22.[10] —, A note on the number of divisors of quadratic polynomials, in: Sieve Methods,Exponential Sums, and their Application to Number Theory, London Math. Soc.Lecture Note Ser. 237, Cambridge Univ. Press, 1997, 275–281.[11] Y. Motohashi, On the sum of the number of divisors in a short segment , Acta Arith.52 (1970), 249–253.[12] E. J. Scourfield, The divisors of a quadratic polynomial , Proc. Glasgow Math. Assoc.5 (1961), 8–21.[13] D. Suryanarayana, Asymptotic formula for∑n≤x µ2(n)/n, Indian J. Math. 9(1967), 543–545.Department of MathematicsUniversity of WaikatoHamilton, New ZealandE-mail: kab@math.waikato.ac.nzReceived on 11.4.2000and in revised form on 15.12.2000

Restricted divisor sums

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