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Asymptotic formula for balanced words

Shigeki Akiyama
Abstract.

We give asymptotic formulas for the number of balanced words whose slope α\alpha and intercept ρ\rho lie in a prescribed rectangle. They are related to uniform distribution of Farey fractions and Riemann Hypothesis. In the general case, the error term is deduced using an inequality of large sieve type.

Key words and phrases:
Irrational rotation, Balanced words, Sturmian words, Farey Fraction, Riemann Hypothesis, Large Sieve

1. Introduction

Let 𝒜={0,1}{\mathcal{A}}=\{0,1\} and we denote by 𝒜{\mathcal{A}}^{*} the monoid generated by 𝒜{\mathcal{A}} by concatenation, where the empty word λ\lambda is its identity. The length of x𝒜x\in{\mathcal{A}}^{*} is |x||x|. We denote |x|1|x|_{1} the cardinality of 11 in x𝒜x\in{\mathcal{A}}^{*}. 𝒜{\mathcal{A}}^{{\mathbb{N}}} (resp. 𝒜{\mathcal{A}}^{{\mathbb{Z}}}) is the set of right infinite (resp. bi-infinite) words. If y𝒜y\in{\mathcal{A}}^{*} is a factor (a subword) of x𝒜𝒜𝒜x\in{\mathcal{A}}^{*}\cup{\mathcal{A}}^{{\mathbb{N}}}\cup{\mathcal{A}}^{{\mathbb{Z}}}, we write yxy\prec x. A word in x𝒜𝒜𝒜x\in{\mathcal{A}}^{*}\cup{\mathcal{A}}^{{\mathbb{N}}}\cup{\mathcal{A}}^{{\mathbb{Z}}} is balanced if ||u|1|v|1|1||u|_{1}-|v|_{1}|\leq 1 holds for any u,vxu,v\prec x with |u|=|v||u|=|v|. An infinite word w𝒜w\in{\mathcal{A}}^{{\mathbb{N}}} is sturmian if Card{uw||u|=n}=n+1\text{Card}\{u\prec w\ |\ |u|=n\}=n+1 for all nn\in{\mathbb{N}}. Morse and Hedlund characterized a sturmian word as an aperiodic balanced word in 𝒜{\mathcal{A}}^{{\mathbb{N}}}. They also characterized sturmian words as a coding of irrational rotation. More precisely a lower mechanical word (sn)𝒜(s_{n})\in{\mathcal{A}}^{{\mathbb{N}}} is defined by

sn(α,ρ)=α(n+1)+ραn+ρs_{n}(\alpha,\rho)=\lfloor\alpha(n+1)+\rho\rfloor-\lfloor\alpha n+\rho\rfloor

with a given slope α[0,1]\alpha\in[0,1] and an intercept ρ[0,1)\rho\in[0,1). An upper mechanical word is similarly defined by replacing \lfloor\cdot\rfloor with \lceil\cdot\rceil, which is denoted by s^n(α,ρ)\hat{s}_{n}(\alpha,\rho). Then a sturmian word is a (lower or upper) mechanical word of an irrational slope α\alpha and vice versa. It is known that every balanced word is a factor of a lower and an upper mechanical word [17]. Indeed, for every balanced word x=x1xnx=x_{1}\dots x_{n} we can find a slope α\alpha and an intercept ρ\rho such that xi=si(α,ρ)=s^i(α,ρ)x_{i}=s_{i}(\alpha,\rho)=\hat{s}_{i}(\alpha,\rho). Note that the choice of α\alpha and ρ\rho is not unique. In fact, a balanced xx word corresponds to (ρ,α)(\rho,\alpha) in a convex polygon in [0,1)×[0,1][0,1)\times[0,1]. This geometric idea for enumeration is found in [1, 25, 14, 3]. Let ϕ(n)\phi(n) be the Euler totient function. The formula for the number of balanced words of length nn is given by

1+k=1n(n+1k)ϕ(k).1+\sum_{k=1}^{n}(n+1-k)\phi(k). (1)

Several different proofs are found in [17, 16, 1, 5, 18].

In this paper we refine the formula (1) and give its asymptotic behavior. Denote by B(n,t,u)B(n,t,u) the cardinality of the set of balanced words of length nn whose slope α[1t,1]\alpha\in[1-t,1]111The statement of Theorem 1 is simpler by this choice than taking [0,t][0,t] because the lines in the proof do not intersect (0,1)×{0}(0,1)\times\{0\}. and its intercept ρ[0,u)\rho\in[0,u) and let (n){\mathcal{B}}(n) be the set of balanced words of length nn. Then we show

Theorem 1.
B(n,t,u)=1+mnA(m,t,u)B(n,t,u)=1+\sum_{m\leq n}A(m,t,u)

with

A(m,t,u)=i<jm,(i,j)=1i/jt,mi/j<u1,A(m,t,u)=\sum_{\begin{subarray}{c}i<j\leq m,\ (i,j)=1\\ i/j\leq t,\ \langle mi/j\rangle<u\end{subarray}}1,

where ii and jj are non negative integers.

Theorem 1 slightly generalizes Yasutomi [25, Proposition 4], shown in a different context. By using Theorem 1, we will derive asymptotic formulas. Hereafter, we use conventional terminology in analytic number theory, i.e., Landau OO, oo symbol and Vinogradov symbol \ll. The symbol ε\varepsilon is reserved as an arbitrary positive constant, and the symbol cc in Landau OO is a suitably chosen positive constant, which may differ among formulas.

Theorem 2.
B(n,t,u)=tuπ2n3+O(n2(logn)15/2).B(n,t,u)=\frac{tu}{\pi^{2}}n^{3}+O\left(n^{2}(\log n)^{15/2}\right).

Moreover, we have

B(n,1,1)=n3+3n2π2+O(n2exp(c((logn)3/5(loglogn)1/5)))B(n,1,1)=\frac{n^{3}+3n^{2}}{\pi^{2}}+O\left(n^{2}\exp\left(-c\left((\log n)^{3/5}(\log\log n)^{-1/5}\right)\right)\right)

and

B(n,t,1)=tn3π2+O(n2).B(n,t,1)=\frac{tn^{3}}{\pi^{2}}+O(n^{2}). (2)

For almost all tt, the estimate (2) can be sharpened with the help of Fujii [9], see the discussion in the end of §3. Denote by χA(x)\chi_{A}(x) the indicator function of the set AA. For a rectangle (a,b]×[c,d)(a,b]\times[c,d) in the unit square [0,1)×[0,1][0,1)\times[0,1], we see

Card{x(n)|(ρ,α)(a,b]×[c,d)}=(ba)(dc)π2n3+O(n2(logn)15/2),\mathrm{Card}\{x\in{\mathcal{B}}(n)\ |\ (\rho,\alpha)\in(a,b]\times[c,d)\}=\frac{(b-a)(d-c)}{\pi^{2}}n^{3}+O\left(n^{2}(\log n)^{15/2}\right),

from

χ(a,b]×[c,d)(x)=χ(0,b]×[c,1)(x)χ(0,a]×[c,1)(x)χ(0,b]×[d,1)(x)+χ(0,a]×[d,1)(x).\chi_{(a,b]\times[c,d)}(x)=\chi_{(0,b]\times[c,1)}(x)-\chi_{(0,a]\times[c,1)}(x)-\chi_{(0,b]\times[d,1)}(x)+\chi_{(0,a]\times[d,1)}(x).

Moreover for a Jordan measurable region WW in the unit square, we have

Corollary 3.
Card{x(n)|(ρ,α)W}=Area(W)π2n3+O(n2(logn)15/2)\mathrm{Card}\{x\in{\mathcal{B}}(n)\ |\ (\rho,\alpha)\in W\}=\frac{\mathrm{Area}(W)}{\pi^{2}}n^{3}+O\left(n^{2}(\log n)^{15/2}\right)

where Area\mathrm{Area} is the 22-dimensional Lebesgue measure, since every Jordan measurable set is well approximated by finite union of rectangles.

Farey series (fm(i))(f_{m}(i)) of order mm is the finite increasing sequence composed of irreducible fractions in [0,1)[0,1) whose denominators are not larger than mm:

0=fm(1)<fm(2)<<fm(Φ(m))<10=f_{m}(1)<f_{m}(2)<\dots<f_{m}(\Phi(m))<1

with Φ(m)=k=1mϕ(k)\Phi(m)=\sum_{k=1}^{m}\phi(k). Clearly A(m,t,1)=max{j|fm(j)t}A(m,t,1)=\max\left\{j\ |\ f_{m}(j)\leq t\right\}. It is well-known that Riemann Hypothesis is equivalent to

i=1Φ(m)exp(2π𝔦fm(i))m1/2+ε,\sum_{i=1}^{\Phi(m)}\exp(2\pi\mathfrak{i}f_{m}(i))\ll m^{1/2+\varepsilon},

a strong uniform distribution property of Farey fractions. As pointed out in [8], Franel [7] already noticed that

01(A(m,t,1)tΦ(m))2𝑑tm1+ε\int_{0}^{1}(A(m,t,1)-t\Phi(m))^{2}dt\ll m^{1+\varepsilon}

is also equivalent to Riemann Hypothesis. One can see

A(m,t,1)tΦ(m)=O(m),A(m,t,1)-t\Phi(m)=O(m),

similarly to (8) and (10) in §3, but we expect it is much smaller in average. See also [21, 8, 15]. We state another equivalent statement directly related to the number of balanced words:

Corollary 4.

The estimate

B(n,1,1)=n3+3n2π2+O(n3/2+ε)B(n,1,1)=\frac{n^{3}+3n^{2}}{\pi^{2}}+O\left(n^{3/2+\varepsilon}\right) (3)

is equivalent to Riemann Hypothesis.

2. Proof of Theorem 1

We elucidate a geometric counting discussion of Yasutomi [25, Proposition 4] in our convenient terminology, which is more straightforward than the one in [1]. Let mm be a fixed positive integer and put X:=[0,1)×[0,1]X:=[0,1)\times[0,1]. The map ψ:(ρ,α)(sn(1α,ρ))n=1m\psi:(\rho,\alpha)\to(s_{n}(1-\alpha,\rho))_{n=1}^{m} gives a natural partition

X:=wB(m)ψ1(w).X:=\bigcup_{w\in B(m)}\psi^{-1}(w). (4)

Then XX is cut into convex cells by segments:

Y:=X{(x,y)|x=ny,n{1,,m},{0,1,,n1}}.Y:=X\setminus\{(x,y)\ |\ x=ny-\ell,n\in\{1,\dots,m\},\ell\in\{0,1,\dots,n-1\}\}.

We obtain essentially the same partition by using s^n\hat{s}_{n}, the difference is seen only on the boundary of YY. In this paper, we use sns_{n} for the partition. Fixing α[0,1]\alpha\in[0,1], the intersections of the line y=1αy=1-\alpha and x=nyx=ny-\ell are written as

{Rn(0)|n{1,,m}}×{1α}={nα(mod1)|n=1,2,m}×{1α}\{R^{-n}(0)\ |\ n\in\{1,\dots,m\}\}\times\{1-\alpha\}=\{-n\alpha\,(\bmod{1})\ |\ n=1,2\dots,m\}\times\{1-\alpha\}

where R:(x,1α)(x+α,1α)R:(x,1-\alpha)\mapsto(x+\alpha,1-\alpha) is the rotation map acting on the unit interval [0,1)×{1α}[0,1)\times\{1-\alpha\} which is identified with the torus 𝕋:=/{\mathbb{T}}:={\mathbb{R}}/{\mathbb{Z}}. The partition of [0,1)×{1α}[0,1)\times\{1-\alpha\} by Rn(0)(n=1,,m)R^{-n}(0)\ (n=1,\dots,m) gives cylinder sets of rotation RR, i.e., the points in the same cylinder share the same coding

(χ[1α,1)×{1α}(Ri1(x)))i=1m.\left(\chi_{[1-\alpha,1)\times\{1-\alpha\}}\left(R^{i-1}(x)\right)\right)_{i=1}^{m}.

They are the m+1m+1 different words which correspond to the words of length mm appear in the sturmian word of slope α\alpha when α\alpha is irrational. Slicing XX by the line y=1αy=1-\alpha, we observe the mm-th level cylinder sets of the rotation RR on 𝕋{\mathbb{T}} of slope α\alpha acting on the unit interval [0,1)×{1α}[0,1)\times\{1-\alpha\}. Considering α\alpha as a variable, we reconstruct the partition (4) of XX that every convex cell corresponds to an element of (m){\mathcal{B}}(m), which is consistent with the cylinder partition of [0,1)×{1α}[0,1)\times\{1-\alpha\} for each α[0,1]\alpha\in[0,1]. In this manner, the partition (4) is seen as a pile of cylinder sets of level mm for all (rational/irrational) rotations. The case m=4m=4 is depicted in Figure 1.

Refer to caption
Figure 1. Partition for m=4m=4

To enumerate B(n,t,u)B(n,t,u), we compute B(m,t,u)B(m1,t,u)B(m,t,u)-B(m-1,t,u) for m2m\geq 2, that is, the increase of number of cells as we add a new slope 1/m1/m. Equivalently, we count the number of intersections in [0,u)×[0,t][0,u)\times[0,t] which appear by adding new segments of slope 1/m1/m, see Figure 2. The intersection points (x,y)[0,u)×[0,t](x,y)\in[0,u)\times[0,t] of x=mybx=my-b and x=yc(<m)x=\ell y-c\ (\ell<m) are in one to one correspondence with the set of their yy-coordinates:

P(m):={bcm|bmbcm<b+1m,bcmt,bmcm<u,0b,<m,0c<}.P(m):=\left\{\frac{b-c}{m-\ell}\ \left|\ \begin{aligned} &\frac{b}{m}\leq\frac{b-c}{m-\ell}<\frac{b+1}{m},\quad\frac{b-c}{m-\ell}\leq t,\\ &\frac{\ell b-mc}{m-\ell}<u,\quad 0\leq b,\ell<m,\quad 0\leq c<\ell\end{aligned}\right.\right\}.

We claim that P(m)P(m) coincides with the set

Q(m):={x[0,1)|xt,mx<u,den(x)m}Q(m):=\left\{x\in{\mathbb{Q}}\cap[0,1)\ \left|\ x\leq t,\quad\langle mx\rangle<u,\quad\mathrm{den}(x)\leq m\right.\right\}

where x=xx\langle x\rangle=x-\lfloor x\rfloor and den(y)\mathrm{den}(y) is the denominator of yy\in{\mathbb{Q}}. In fact P(m)Q(m)P(m)\subset Q(m) is clear. Let x=p/qQ(m)x=p/q\in Q(m) with 1qm1\leq q\leq m, (p,q)=1(p,q)=1, p/qtp/q\leq t and mx<u\langle mx\rangle<u. Put =mq\ell=m-q and choose bb with b/mp/q<(b+1)/mb/m\leq p/q<(b+1)/m and let c=bpc=b-p. Since p/q<(b+1)/mp/q<(b+1)/m, we have pbp\leq b and c0c\geq 0. From the property of Farey fraction, we have (bp)/(mq)<b/m<p/q(b-p)/(m-q)<b/m<p/q, which implies c=bp<mq=c=b-p<m-q=\ell. The inequality 0bmcm<10\leq\frac{\ell b-mc}{m-\ell}<1 follows from bmbcm<b+1m\frac{b}{m}\leq\frac{b-c}{m-\ell}<\frac{b+1}{m}. Thus mx=bmcm\langle mx\rangle=\frac{\ell b-mc}{m-\ell} and therefore xP(m)x\in P(m). The claim is proved. Set A(m,t,u)=Card(Q(m))A(m,t,u)=\mathrm{Card}(Q(m)) for m1m\geq 1. From B(1,t,u)=2B(1,t,u)=2 and A(1,t,u)=1A(1,t,u)=1, we obtain Theorem 1.∎

We illustrate this proof by an

Example 5.

Let t=0.7t=0.7 and u=0.59u=0.59. Then (A(m,t,u))m=18=(1,2,4,4,7,8,10,13)(A(m,t,u))_{m=1}^{8}=(1,2,4,4,7,8,10,13) and B(8,t,u)=50B(8,t,u)=50. In Figure 2, we are counting the number of cells in the shaded region [0,u)×[0,t][0,u)\times[0,t]. Dashed segments are of slope 1/81/8 intersecting at 13 points indicated by dots, which contribute to the increase of the number of cells.

Refer to caption
Figure 2. B(8,0.7,0.59)=50B(8,0.7,0.59)=50

Hereafter we show an asymptotic formula of B(n,t,u)B(n,t,u).

3. Theorem 2 for u=1u=1 and Corollary 4

We show a

Lemma 6.
mx(xm)ϕ(m)=x3π2+O(x2exp(c(logx)3/5(loglogx)1/5)).\sum_{m\leq x}(x-m)\phi(m)=\frac{x^{3}}{\pi^{2}}+O\left(x^{2}\exp\left(-c(\log x)^{3/5}(\log\log x)^{-1/5}\right)\right).
Proof.

This is a variation of the argument to deduce prime number theorem [20, Theorem 6.9]. We use the Mellin inversion formula

mx(xm)ϕ(m)=12π𝔦a𝔦a+𝔦ζ(s1)xs+1ζ(s)s(s+1)𝑑s\sum_{m\leq x}(x-m)\phi(m)=\frac{1}{2\pi\mathfrak{i}}\int_{a-\mathfrak{i}\infty}^{a+\mathfrak{i}\infty}\frac{\zeta(s-1)x^{s+1}}{\zeta(s)s(s+1)}ds

for a>2a>2. Here ζ(s)\zeta(s) is Riemann zeta function with a complex variable s=σ+𝔦ws=\sigma+\mathfrak{i}w and put τ=|w|+4\tau=|w|+4. Since |ζ(s1)|τ1/2logτ|\zeta(s-1)|\ll\tau^{1/2}\log\tau and 1/ζ(s)logτ1/\zeta(s)\ll\log\tau in the required range, shifting the path to σ=1+(logx)1=:a0\sigma=1+(\log x)^{-1}=:a_{0}, we can pick the residue at s=2s=2:

mx(xm)ϕ(m)=x3π2+12π𝔦a0𝔦a0+𝔦ζ(s1)xs+1ζ(s)s(s+1)𝑑s.\sum_{m\leq x}(x-m)\phi(m)=\frac{x^{3}}{\pi^{2}}+\frac{1}{2\pi\mathfrak{i}}\int_{a_{0}-\mathfrak{i}\infty}^{a_{0}+\mathfrak{i}\infty}\frac{\zeta(s-1)x^{s+1}}{\zeta(s)s(s+1)}ds. (5)

Truncating this formula,

mx(xm)ϕ(m)=x3π2+12π𝔦a0𝔦Ta0+𝔦Tζ(s1)xs+1ζ(s)s(s+1)𝑑s+O(x2(logT)2T1/2)\sum_{m\leq x}(x-m)\phi(m)=\frac{x^{3}}{\pi^{2}}+\frac{1}{2\pi\mathfrak{i}}\int_{a_{0}-\mathfrak{i}T}^{a_{0}+\mathfrak{i}T}\frac{\zeta(s-1)x^{s+1}}{\zeta(s)s(s+1)}ds+O\left(\frac{x^{2}(\log T)^{2}}{T^{1/2}}\right)

and shifting the path to the zero-free region

σ>1c(logτ)2/3(loglogτ)1/3=:b\sigma>1-\frac{c}{(\log\tau)^{2/3}(\log\log\tau)^{1/3}}=:b

of ζ(s)\zeta(s) due to Vinogradov-Korobov [24, 11], we get a rectangular contour to be studied. The contribution from horizontal segments is O(x2/T3/2ε)O(x^{2}/T^{3/2-\varepsilon}) and the vertical one at σ=b\sigma=b is O(xb+1)O(x^{b+1}). Lemma 6 is obtained by the choice T=exp(c(logx)3/5(loglogx)1/5)T=\exp(c(\log x)^{3/5}(\log\log x)^{-1/5}). ∎

Let us start with the easiest case t=u=1t=u=1. From

A(m,1,1)=i<jm(i,j)=11=Φ(m)A(m,1,1)=\sum_{\begin{subarray}{c}i<j\leq m\\ (i,j)=1\end{subarray}}1=\Phi(m)

and Theorem 1, we have

B(n,1,1)\displaystyle B(n,1,1) =\displaystyle= 1+m=1nΦ(m)\displaystyle 1+\sum_{m=1}^{n}\Phi(m)
=\displaystyle= 1+j=1n(n+1j)ϕ(j)\displaystyle 1+\sum_{j=1}^{n}(n+1-j)\phi(j)
=\displaystyle= n3+3n2π2+O(n2exp(c(logn)3/5(loglogn)1/5)).\displaystyle\frac{n^{3}+3n^{2}}{\pi^{2}}+O\left(n^{2}\exp\left(-c(\log n)^{3/5}(\log\log n)^{-1/5}\right)\right). (7)

Here we used Lemma 6 and the Mertens formula

Φ(n)=3n2π2+O(nlogn).\Phi(n)=\frac{3n^{2}}{\pi^{2}}+O(n\log n).

The error term E(n)=Φ(n)(3/π2)n2E(n)=\Phi(n)-(3/\pi^{2})n^{2} is well studied in literature [22, 6, 24, 19, 13]. However in (3), the effect of this error term cancels out and we find the second main term in (7). We retrieved the formula (1) as well.

Now we show Corollary 4. If Riemann Hypothesis is valid, then (5) holds with a0=1/2+εa_{0}=1/2+\varepsilon. Thus from

limTa0𝔦Ta0+𝔦T|ζ(s1)ζ(s)s(s+1)|𝑑s<,\lim_{T\to\infty}\int_{a_{0}-\mathfrak{i}T}^{a_{0}+\mathfrak{i}T}\left|\frac{\zeta(s-1)}{\zeta(s)s(s+1)}\right|ds<\infty,

we get the estimate (3). Conversely Mellin transformation shows

1(nx(xn)ϕ(n)x3π2)xs2𝑑x=ζ(s1)ζ(s)s(s+1)1π2(s2)\int_{1}^{\infty}\left(\sum_{n\leq x}(x-n)\phi(n)-\frac{x^{3}}{\pi^{2}}\right)x^{-s-2}dx=\frac{\zeta(s-1)}{\zeta(s)s(s+1)}-\frac{1}{\pi^{2}(s-2)}

for σ>2\sigma>2. If (3) is valid, then the parenthesis in the integrand is O(x3/2+ε)O(x^{3/2+\varepsilon}). This gives the holomorphic continuation of the right side to σ>1/2+ε\sigma>1/2+\varepsilon, which finishes the proof.

Let us discuss a small counting issue. Since i=0i=0 implies j=1j=1 in the sum A(m,t,u)A(m,t,u), we have

A(m,t,u)=1+i<jm,i/jtmi/j<u,(i,j)=11A(m,t,u)=1+\sum_{\begin{subarray}{c}i<j\leq m,\,i/j\leq t\\ \langle mi/j\rangle<u,\,(i,j)=1\end{subarray}}1

where ii and jj are positive integers hereafter. Therefore

B(n,t,u)=1+n+i<jmn,i/jtmi/j<u,(i,j)=11.B(n,t,u)=1+n+\sum_{\begin{subarray}{c}i<j\leq m\leq n,\,i/j\leq t\\ \langle mi/j\rangle<u,\,(i,j)=1\end{subarray}}1.

If t<1t<1 then, it is the same as

B(n,t,u)=1+n+i,jmn,i/jtmi/j<u,(i,j)=11.B(n,t,u)=1+n+\sum_{\begin{subarray}{c}i,j\leq m\leq n,\,i/j\leq t\\ \langle mi/j\rangle<u,\,(i,j)=1\end{subarray}}1.

In the case t=1t=1, i=ji=j happens only when i=j=1i=j=1, and we may write

B(n,1,u)=1+i,jmnmi/j<u,(i,j)=11.B(n,1,u)=1+\sum_{\begin{subarray}{c}i,j\leq m\leq n\\ \langle mi/j\rangle<u,\,(i,j)=1\end{subarray}}1.

Let μ\mu be the Möbius function and assume u=1u=1 and t<1t<1. With positive integers a,ba,b, we have

B(n,t,1)1n\displaystyle B(n,t,1)-1-n =\displaystyle= i,jmni/jtk(i,j)μ(k)=kmnμ(k)bm/ka/bt1\displaystyle\sum_{\begin{subarray}{c}i,j\leq m\leq n\\ i/j\leq t\end{subarray}}\,\sum_{k\mid(i,j)}\mu(k)=\sum_{k\leq m\leq n}\mu(k)\sum_{b\leq m/k}\,\sum_{a/b\leq t}1
=\displaystyle= kmnμ(k)bm/kbt\displaystyle\sum_{k\leq m\leq n}\mu(k)\sum_{b\leq m/k}\lfloor bt\rfloor
=\displaystyle= tkmnμ(k)bm/kbkbmnμ(k)bt.\displaystyle t\sum_{k\leq m\leq n}\mu(k)\sum_{b\leq m/k}b-\sum_{kb\leq m\leq n}\mu(k)\langle bt\rangle. (8)

While t=1t=1, the same computation gives

B(n,1,1)1=kmnμ(k)bm/kb,B(n,1,1)-1=\sum_{k\leq m\leq n}\mu(k)\sum_{b\leq m/k}b,

we see

B(n,t,1)=1t+n+tB(n,1,1)kbmnμ(k)btB(n,t,1)=1-t+n+tB(n,1,1)-\sum_{kb\leq m\leq n}\mu(k)\langle bt\rangle (9)

for t<1t<1. Since

kbmμ(k)bt=O(m)\sum_{kb\leq m}\mu(k)\langle bt\rangle=O(m) (10)

by Niederreiter [21, (6) and Lemma 3], we have

kbmnμ(k)bt=O(n2).\sum_{kb\leq m\leq n}\mu(k)\langle bt\rangle=O(n^{2}). (11)

Note that the implied constant does not depend on tt. From (7),(9) and (11), we have shown

B(n,t,1)=tn3π2+O(n2).B(n,t,1)=\frac{tn^{3}}{\pi^{2}}+O(n^{2}). (12)

We do not know whether

kbmnμ(k)bt=o(n2)\sum_{kb\leq m\leq n}\mu(k)\langle bt\rangle=o(n^{2})

holds for all tt. If this estimate is valid for a fixed tt, we observe the second main term:

B(n,t,1)=t(n3+3n2)π2+o(n2).B(n,t,1)=\frac{t(n^{3}+3n^{2})}{\pi^{2}}+o(n^{2}).

Fujii [9] elaborated the improvement of (10) by using Hecke’s Dirichlet series [10]:

Zt(s):=b=1bt12bs.Z_{t}(s):=\sum_{b=1}^{\infty}\frac{\langle bt\rangle-\frac{1}{2}}{b^{s}}.

The analytic property of Zt(s)Z_{t}(s) heavily depends on the Diophantine approximation property of tt by rationals. The refinement of (10) in the proofs of [9, Theorem 1 and 2] imply

B(n,t,1)=t(n3+3n2)π2+O(n2exp(c(lognloglogn)1/3))B(n,t,1)=\frac{t(n^{3}+3n^{2})}{\pi^{2}}+O\left(n^{2}\exp\left(-c\left(\log n\cdot\log\log n\right)^{1/3}\right)\right) (13)

for almost all tt, including all algebraic numbers.222The error term of (13) can be replaced by the one in (7), because we need to improve (11) but not necessarily (10). This may be discussed elsewhere. Note that the proofs are rather different between rational tt and algebraic irrational tt, and the implied OO constant could be very sensitive to the choice of tt.

4. Preliminaries

Let 𝐞(x)=exp(2π𝔦x)\mathbf{e}(x)=\exp(2\pi\mathfrak{i}x). Let 𝐝\mathbf{d} be the natural metric on the torus 𝕋=/{\mathbb{T}}={\mathbb{R}}/{\mathbb{Z}}. For a given interval [α,β][0,1)[\alpha,\beta]\subset[0,1) and a positive J>2/(1β+α)J>2/(1-\beta+\alpha) there exists a smooth function VJV_{J} of period 11 such that

VJ(z)=1\displaystyle V_{J}(z)=1 z[α,β]\displaystyle z\in[\alpha,\beta]
VJ(z)=0\displaystyle V_{J}(z)=0 𝐝(zmod,[α,β])1/J\displaystyle\mathbf{d}(z\bmod{{\mathbb{Z}}},[\alpha,\beta])\geq 1/J (14)
0VJ(z)1\displaystyle 0\leq V_{J}(z)\leq 1 otherwise,\displaystyle\mathrm{otherwise},

whose Fourier expansion is

VJ(z)=hvJ(h)𝐞(hz)V_{J}(z)=\sum_{h\in{\mathbb{Z}}}v_{J}(h)\mathbf{e}(hz)

with

|vJ(h)|min(2π|h|,2J(πh)2),|v_{J}(h)|\leq\min\left(\frac{2}{\pi|h|},\frac{2J}{(\pi h)^{2}}\right), (15)

see [23, Chapter 1,Lemma 12]. We shall use a large sieve inequality [12, Theorem 7.2],[2, Lemma 2.4]:

Lemma 7.

For any real numbers xm,ymx_{m},y_{m} with |xm|X|x_{m}|\leq X and |ym|Y|y_{m}|\leq Y and αm,βm\alpha_{m},\beta_{m}\in{\mathbb{C}}, we have

|mnαmβn𝐞(xmyn)|51+XY(|xixj|<1/Y|αiαj||yiyj|<1/X|βiβj|)1/2.\left|\sum_{m}\sum_{n}\alpha_{m}\beta_{n}\mathbf{e}(x_{m}y_{n})\right|\leq 5\sqrt{1+XY}\left(\sum_{|x_{i}-x_{j}|<1/Y}|\alpha_{i}\alpha_{j}|\sum_{|y_{i}-y_{j}|<1/X}|\beta_{i}\beta_{j}|\right)^{1/2}.

We prepare two more lemmas.

Lemma 8.
|1πTTexp(𝔦αx)sinβxx𝑑xδ(βα)|=O(min(1,1T|βα|))\left|\frac{1}{\pi}\int_{-T}^{T}\exp(\mathfrak{i}\,\alpha x)\,\frac{\sin{\beta x}}{x}dx-\delta(\beta-\alpha)\right|=O\left(\min\left(1,\frac{1}{T\left|\beta-\alpha\right|}\right)\right)

for T,α,β>0T,\alpha,\beta>0 with

δ(y)={1y>01/2y=00y<0.\delta(y)=\begin{cases}1&y>0\\ 1/2&y=0\\ 0&y<0.\end{cases}
Proof.

This follows from the sine integral formula

2π0Tsinxx𝑑x=1+O(min(1,1T)),\frac{2}{\pi}\int_{0}^{T}\frac{\sin x}{x}dx=1+O\left(\min\left(1,\frac{1}{T}\right)\right),

c.f. [4, p.166], [12, Lemma 13.11]. ∎

Lemma 9.
H<h2HM<a2M(h,a)=6HMπ2log(2min(H,M))+O(HM)\sum_{\begin{subarray}{c}H<h\leq 2H\\ M<a\leq 2M\end{subarray}}(h,a)=\frac{6HM}{\pi^{2}}\log(2\min(H,M))+O(HM)
Proof.

Putting d=(h,a)d=(h,a), we have

H<h2HM<a2M(h,a)=\displaystyle\sum_{\begin{subarray}{c}H<h\leq 2H\\ M<a\leq 2M\end{subarray}}(h,a)= d2min(H,M)dH/d<i2H/dM/d<j2M/dk(i,j)μ(k)\displaystyle\sum_{d\leq 2\min(H,M)}d\sum_{\begin{subarray}{c}H/d<i\leq 2H/d\\ M/d<j\leq 2M/d\end{subarray}}\,\sum_{k\mid(i,j)}\mu(k)
=\displaystyle= d2min(H,M)dk2min(H,M)dμ(k)H/(dk)<i2H/(dk)M/(dk)<j2M/(dk)1,\displaystyle\sum_{d\leq 2\min(H,M)}d\sum_{k\leq\frac{2\min(H,M)}{d}}\mu(k)\sum_{\begin{subarray}{c}H/(dk)<i\leq 2H/(dk)\\ M/(dk)<j\leq 2M/(dk)\end{subarray}}1,

where the last sum is

HdkMdk+O(Hdk)+O(Mdk)+O(1).\frac{H}{dk}\frac{M}{dk}+O\left(\frac{H}{dk}\right)+O\left(\frac{M}{dk}\right)+O(1).

The proof follows from dXlog(X/d)=X+O(log(X))\sum_{d\leq X}\log(X/d)=X+O(\log(X)). ∎

5. Proof of Theorem 2

Our strategy is to treat B(n,t,u)B(n,t,u) as a function on tt. Then B(n,t,u)B(n,t,u) is a non-decreasing step function having a finite number of rational discontinuities. Every gap between the discontinuities of B(n,s,t)B(n,s,t) as a function on tt is greater than 1/n21/n^{2}. Depending on nn, we will choose t1tt2t_{1}\leq t\leq t_{2} that t2t1t_{2}-t_{1} is small and t1,t2t_{1},t_{2} have a suitable property to estimate error terms. If we get the same asymptotic formulas and their implied constants of the Landau and Vinogradov symbols are independent of this choice of tit_{i}, we obtain the estimate for tt by the non-decreasing property, because it is sandwiched by the same formula. Therefore if we could perturb tt and get the same error term (up to negligible terms), then we are done. Hereafter we shall be cautious on the above implied constants, whether they can be independent of the choices of tit_{i}.

At the cost of an additional error term O(n2)O(n^{2}) which takes care of the case i/j=ti/j=t, we may assume that 0<t<10<t<1, tt is rational and B(n,t,u)B(n,t,u) is continuous at tt, i.e., constant in the neighborhood of tt. Recall that from the middle of §3, ii and jj are positive integers and

B(n,t,u)=1+n+i,jmn,i/jtmi/j<u,(i,j)=11.B(n,t,u)=1+n+\sum_{\begin{subarray}{c}i,j\leq m\leq n,\,i/j\leq t\\ \langle mi/j\rangle<u,\,(i,j)=1\end{subarray}}1.

Taking VJ(x)V_{J}(x) with respect to the interval [0,u][0,u], we have

i,jmn,i/jt(i,j)=1VJ(mij)=B(n,t,u)n+O(n3J).\sum_{\begin{subarray}{c}i,j\leq m\leq n,\,i/j\leq t\\ (i,j)=1\end{subarray}}V_{J}\left(\frac{mi}{j}\right)=B(n,t,u)-n+O\left(\frac{n^{3}}{J}\right). (16)

As (14) implies vJ(0)=u+O(1/J)v_{J}(0)=u+O(1/J), the main term is

i,jmn,i/jt(i,j)=1(u+O(1J))\displaystyle\sum_{\begin{subarray}{c}i,j\leq m\leq n,\,i/j\leq t\\ (i,j)=1\end{subarray}}\left(u+O\left(\frac{1}{J}\right)\right) =\displaystyle= ui,jmn,i/jt(i,j)=11+O(n3J)\displaystyle u\sum_{\begin{subarray}{c}i,j\leq m\leq n,\,i/j\leq t\\ (i,j)=1\end{subarray}}1+O\left(\frac{n^{3}}{J}\right) (17)
=\displaystyle= tun3π2+O(n2)+O(n3J)\displaystyle\frac{tun^{3}}{\pi^{2}}+O(n^{2})+O\left(\frac{n^{3}}{J}\right)

from (12). The remainder is

0hvJ(h)i,jmn,i/jt𝐞(hmij)k(i,j)μ(k)\displaystyle\sum_{0\neq h\in{\mathbb{Z}}}v_{J}(h)\sum_{\begin{subarray}{c}i,j\leq m\leq n,\,i/j\leq t\end{subarray}}\mathbf{e}\left(\frac{hmi}{j}\right)\sum_{k\mid(i,j)}\mu(k) (18)
=\displaystyle= knμ(k)0hvJ(h)a,bm/k,a/btmn𝐞(hmab)\displaystyle\sum_{k\leq n}\mu(k)\sum_{0\neq h\in{\mathbb{Z}}}v_{J}(h)\sum_{\begin{subarray}{c}a,b\leq m/k,\,a/b\leq t\\ m\leq n\end{subarray}}\mathbf{e}\left(\frac{hma}{b}\right)

with positive integers a,ba,b. From (15), our target is to estimate

C(n,k,t,u)\displaystyle C(n,k,t,u) :=\displaystyle:= 0hvJ(h)a,bm/k,a/btmn𝐞(hmab)\displaystyle\sum_{0\neq h\in{\mathbb{Z}}}v_{J}(h)\sum_{\begin{subarray}{c}a,b\leq m/k,\,a/b\leq t\\ m\leq n\end{subarray}}\mathbf{e}\left(\frac{hma}{b}\right) (19)
=\displaystyle= 0|h|HvJ(h)a,bm/k,a/btmn𝐞(hmab)+O(Jn3Hk2).\displaystyle\sum_{0\neq|h|\leq H}v_{J}(h)\sum_{\begin{subarray}{c}a,b\leq m/k,\,a/b\leq t\\ m\leq n\end{subarray}}\mathbf{e}\left(\frac{hma}{b}\right)+O\left(\frac{Jn^{3}}{Hk^{2}}\right).

By Lemma 8 and (15), we have

UUTT1hHvJ(h)a,bn/kmn𝐞(hmab)(ab)𝔦x(bm)𝔦y\displaystyle\int_{-U}^{U}\int_{-T}^{T}\sum_{1\leq h\leq H}v_{J}(h)\sum_{\begin{subarray}{c}a,b\leq n/k\\ m\leq n\end{subarray}}\mathbf{e}\left(\frac{hma}{b}\right)\left(\frac{a}{b}\right)^{\mathfrak{i}x}\left(\frac{b}{m}\right)^{\mathfrak{i}y}
×sin(xlog(t))πxsin(ylog(1/k))πydxdy\displaystyle\hskip 142.26378pt\times\frac{\sin(x\log(t))}{\pi x}\frac{\sin(y\log(1/k))}{\pi y}dxdy
=\displaystyle= 1hHvJ(h)a,bm/k,a/btmn𝐞(hmab)\displaystyle\sum_{1\leq h\leq H}v_{J}(h)\sum_{\begin{subarray}{c}a,b\leq m/k,\,a/b\leq t\\ m\leq n\end{subarray}}\mathbf{e}\left(\frac{hma}{b}\right)
+O(logUlogHa,bn/kmn1T|logtba|)+O(logHa,bn/k,a/btmnmin(1,1U|logbkm|)).\displaystyle+O\left(\log U\log H\sum_{\begin{subarray}{c}a,b\leq n/k\\ m\leq n\end{subarray}}\frac{1}{T|\log\frac{tb}{a}|}\right)+O\left(\log H\sum_{\begin{subarray}{c}a,b\leq n/k,\,a/b\leq t\\ m\leq n\end{subarray}}\min\left(1,\frac{1}{U|\log\frac{bk}{m}|}\right)\right).

It is important to note that the variables h,a,b,mh,a,b,m are independent in the sum in the double integral (5). We may further assume that tt has a denominator pp, which is the smallest prime exceeding n2n^{2}. Then we have |log(1+(tba1))|1/(2np)|\log(1+(\frac{tb}{a}-1))|\geq 1/(2np) and 1T|logtba|4n3T\frac{1}{T|\log\frac{tb}{a}|}\leq\frac{4n^{3}}{T} which implies

O(logUlogHa,bn/k,mn1T|logtba|)=O(n6k2TlogUlogH).O\left(\log U\log H\sum_{a,b\leq n/k,\,m\leq n}\frac{1}{T|\log\frac{tb}{a}|}\right)=O\left(\frac{n^{6}}{k^{2}T}\log U\log H\right). (22)

Since one can show that the case bk=mbk=m is negligible, we have

O(logHa,bn/k,a/btmnmin(1,1U|logbkm|))=O(n4k2UlogH).O\left(\log H\sum_{\begin{subarray}{c}a,b\leq n/k,\,a/b\leq t\\ m\leq n\end{subarray}}\min\left(1,\frac{1}{U|\log\frac{bk}{m}|}\right)\right)=O\left(\frac{n^{4}}{k^{2}U}\log H\right). (23)

To deal with (5), let us apply Lemma 7 to

G(K,L,M,N):=K<m2KL<h2LM<a2MN<b2NvJ(h)a𝔦xb𝔦(yx)m𝔦y𝐞(hmab)G(K,L,M,N):=\sum_{\begin{subarray}{c}K<m\leq 2K\\ L<h\leq 2L\end{subarray}}\sum_{\begin{subarray}{c}M<a\leq 2M\\ N<b\leq 2N\end{subarray}}v_{J}(h)a^{\mathfrak{i}x}b^{\mathfrak{i}(y-x)}m^{-\mathfrak{i}y}\mathbf{e}\left(\frac{hma}{b}\right)

where M,Nn/(2k)M,N\leq n/(2k), Kn/2K\leq n/2 and K,L,M,NK,L,M,N\in{\mathbb{N}}. Define xmx_{m} and yny_{n} by rearranging multi-sets

{ha|L<h2L,M<a2M}\{ha\ |\ L<h\leq 2L,\,M<a\leq 2M\}

and

{mb|K<m2K,N<b2N}\left\{\left.\frac{m}{b}\ \right|\ K<m\leq 2K,\,N<b\leq 2N\right\}

in the non-decreasing order, keeping multiplicity. Since |xm|4LM|x_{m}|\leq 4LM, |yn|2K/N|y_{n}|\leq 2K/N, we have a bound:

|G(K,L,M,N)|2\displaystyle|G(K,L,M,N)|^{2}
25(1+8LMKN)(|h1a1h2a2|<N2K|vJ(h1)vJ(h2)|)(|m/b1m/b2|<14LM1)\displaystyle\leq 25\left(1+\frac{8LMK}{N}\right)\left(\sum_{|h_{1}a_{1}-h_{2}a_{2}|<\frac{N}{2K}}|v_{J}(h_{1})v_{J}(h_{2})|\right)\left(\sum_{|m/b_{1}-m/b_{2}|<\frac{1}{4LM}}1\right)
(1+8LMKN)1L2(N2K+1)(L<h12LM<a22M(h1,a2))KN(N24LMK+1)\displaystyle\ll\left(1+\frac{8LMK}{N}\right)\frac{1}{L^{2}}\left(\frac{N}{2K}+1\right)\left(\sum_{\begin{subarray}{c}L<h_{1}\leq 2L\\ M<a_{2}\leq 2M\end{subarray}}(h_{1},a_{2})\right)KN\left(\frac{N^{2}}{4LMK}+1\right)
(1+LMKN)MNKL(NK+1)(N2LMK+1)logmin(L,M)\displaystyle\ll\left(1+\frac{LMK}{N}\right)\frac{MNK}{L}\left(\frac{N}{K}+1\right)\left(\frac{N^{2}}{LMK}+1\right)\log\min(L,M)

by Lemma 9. If LN/(8MK)L\leq N/(8MK), then

|G(K,L,M,N)|2(N4KL2+N3L2)logmin(L,M)N4L2logmin(L,M).|G(K,L,M,N)|^{2}\ll\left(\frac{N^{4}}{KL^{2}}+\frac{N^{3}}{L^{2}}\right)\log\min(L,M)\ll\frac{N^{4}}{L^{2}}\log\min(L,M). (24)

If N/(8MK)<LN2/(4MK)N/(8MK)<L\leq N^{2}/(4MK), we have

|G(K,L,M,N)|2(MN3L+MN2KL)logmin(L,M).|G(K,L,M,N)|^{2}\ll\left(\frac{MN^{3}}{L}+\frac{MN^{2}K}{L}\right)\log\min(L,M). (25)

When L>N2/(4MK)L>N^{2}/(4MK), we get the estimate

|G(K,L,M,N)|2(M2NK+M2K2)logmin(L,M).|G(K,L,M,N)|^{2}\ll(M^{2}NK+M^{2}K^{2})\log\min(L,M). (26)

Now (24),(25),(26) shows

G(K,L,M,N)n2(logn)1/2kG(K,L,M,N)\ll\frac{n^{2}(\log n)^{1/2}}{k} (27)

for any K,L,M,NK,L,M,N. Set J=max(n,3/(1t))J=\max(n,3/(1-t)) and H=J2H=J^{2}. Summing up (19),(5),(22),(23),(27) and a similar computation for negative hh, we obtain

C(n,k,t,u)n2k2+n6k2T(logU)(logn)+n4k2Ulogn+n2k(logT)(logU)(logn)9/2C(n,k,t,u)\ll\frac{n^{2}}{k^{2}}+\frac{n^{6}}{k^{2}T}(\log U)(\log n)+\frac{n^{4}}{k^{2}U}\log n+\frac{n^{2}}{k}(\log T)(\log U)(\log n)^{9/2}

where the increase of the exponent of the last logn\log n comes from the number of dissections

1bN=j12jN<b2j+1N\sum_{1\leq b\leq N}=\sum_{j\geq 1}\,\sum_{2^{-j}N<b\leq 2^{-j+1}N}

and the similar ones for a,ma,m and hh. Taking U=n2U=n^{2} and T=n4T=n^{4}, we obtain

C(n,k,t,u)=O(n2k(logn)13/2).C(n,k,t,u)=O\left(\frac{n^{2}}{k}(\log n)^{13/2}\right).

Therefore from (16),(17) and (18), we have

B(n,t,u)=tun3π2+O(n2(logn)15/2).B(n,t,u)=\frac{tun^{3}}{\pi^{2}}+O\left(n^{2}(\log n)^{15/2}\right).
Remark 10.

One can apply the same method to A(m,t,u)A(m,t,u). A slightly easier computation gives

A(m,t,u)=3tum2π2+O(m3/2(logm)9/2)A(m,t,u)=\frac{3tum^{2}}{\pi^{2}}+O\left(m^{3/2}(\log m)^{9/2}\right)

which falls short in showing the above bound for B(n,t,u)B(n,t,u).

Acknowledgment. I am particularly grateful to my colleague Hiroshi Mikawa for showing the idea to introduce the large sieve inequality in this setting. Thanks are also due to Shin’ichi Yasutomi and Fujii Akio for inspiring discussion.

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