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# On correlation of the 3-fold divisor function with itself

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## Abstract

Let $\zeta^k(s) = \sum_{n=1}^\infty \tau_k(n) n^{-s}, \Re s > 1$. We present three conditional results on the additive correlation sum $$\sum_{n\le X} \tau_3(n) \tau_3(n+h)$$ and give numerical verifications of our method. The first is a conditional proof for the full main term of this correlation sum for the case $h=1$, on assuming an averaged level of distribution for the three-fold divisor function $\tau_3(n)$ in arithmetic progressions to level two-thirds. The second is a derivation for the leading term asymptotics of this correlation sum, valid for any composite shift $h$. The third result gives a complete expansion of the polynomial for the full main term also for the case $h=1$ but from the delta-method, showing that our answers match. We also refine an unconditional result of Heath-Brown on the classical correlation of the usual divisor function and numerically study the error terms, showing square-root cancellation supporting our approach. Our method is essentially elementary, especially for the $h=1$ case, uses congruences, and, as alluded to earlier, gives the same answer as in prior prediction of Conrey and Gonek, previously computed by Ng and Thom, and heuristic probabilistic arguments of Tao. Our procedure is general and works to give the full main term with a power-saving error term for any correlations of the form $\sum_{n\le X} \tau_k(n) f(n+h)$, to any composite shift $h$, and for any arithmetic function $f(n)$, such as $f(n) = \tau_\ell(n), \Lambda(n),$ et cetera.
ON CORRELATION OF THE 3-FOLD DIVISOR FUNCTION WITH
ITSELF
DAVID T. NGUYEN
Abstract. Let ζk(s) = P
n=1 τk(n)ns,<s > 1. We present three conditional results on
nX
τ3(n)τ3(n+h)
and give numerical veriﬁcations of our method. The ﬁrst is a conditional proof for the full
main term of this correlation sum for the case h= 1, on assuming an averaged level of
distribution for the three-fold divisor function τ3(n) in arithmetic progressions to level two-
thirds. The second is a derivation for the leading term asymptotics of this correlation sum,
valid for any composite shift h. The third result gives a complete expansion of the polynomial
for the full main term also for the case h= 1 but from the delta-method, showing that our
answers match. We also reﬁne an unconditional result of Heath-Brown [13] on the classical
correlation of the usual divisor function and numerically study the error terms, showing
square-root cancellation supporting our approach. Our method is essentially elementary,
especially for the h= 1 case, uses congruences, and, as alluded to earlier, gives the same
answer as in prior prediction of Conrey and Gonek [5], previously computed by Ng and
Thom [21], and heuristic probabilistic arguments of Tao [26]. Our procedure is general and
works to give the full main term with a power-saving error term for any correlations of the
form PnXτk(n)f(n+h), to any composite shift h, and for any arithmetic function f(n),
such as f(n) = τ(n),Λ(n),et cetera.
Contents
1. Introduction and statements of results 2
2. Lemmata 9
3. Proof of Theorem 1 11
4. Conditional proof of the asymptotic for the correlation sum PnXτ3(n)τ3(n+ 1) 13
5. General case of mixed correlations and composite shifts 18
6. Comparison with a conjectural formula of Conrey and Gonek 26
7. Numerical evidence for Conjecture 1: Square-root cancellation in the error term
of the classical correlation PnXτ(n)τ(n+ 1) 31
Acknowledgments 39
References 39
Appendix: Proof of Corollary 1 40
Date: Tuesday 14th June, 2022
Research supported in part by an NSF grant (Focus Research Group DMS-1854398) and the American
Institute of Mathematics.
1
arXiv:2206.05877v1 [math.NT] 13 Jun 2022
1. Introduction and statements of results
For k1 let
ζk(s) =
X
n=1
τk(n)
ns,(<s > 1).
(1.1) Dk,(X, h) = X
nX
τk(n)τ(n+h)
of higher divisor functions τk(n) are instrumental in the study of moments of L-functions,
dating back to 1918 from G. Hardy and J. Littlewood in their pioneering work on the Second
moment of the magnitude of the Riemann zeta function on the vertical line with real part
one-half, corresponding to the case k== 2. Despite its importance, no one to this
day has been able to rigorously prove even an asymptotic formula for this correlation when
both kand are three or larger, though it is widely believed (see, e.g., [21, Conjecture
1.1], [26, Conjecture 1], [5, Conjecture 3], and [18, Conjecture 1.1 (ii)]), that
(1.2) X
nX
τ3(n)τ3(n+ 1) 1
4Y
p14
p2+4
p31
p4Xlog4X,
as X . More generally, the additive divisor correlation problem asks for an asymptotic
of the form
X
nX
τ(n)τk(n+ 1) = M,k(X) + E,k (X),
where M,k(X) is a main term of order exactly X(log X)+k2and E,k (X) is an error term
of order strictly smaller than M,k(X). In Table 1we summarize results on the error term
E,k(X) for various and k.
An approach to the shifted convolution τk(n)τ(n+h) is through what is called a “level of
distribution”. It is a folklore conjecture that τk(n) all have a level of distribution up to 1,
for any > 0. Some known level, or exponent, of distribution for τk(n) was summarized
in [22, Table 1, p. 33]. One of our purpose is to provide a conditional proof for the full
asymptotic expansion for (1.2), on assuming the following upper bound for the averaged
level of distribution of τ3(n) in arithmetic progressions up to level 2/3 for k== 3, and to
indicate the barrier in the correlation problem. This obstacle is summarized in the following
Conjecture 1. Let > 0. Then, for any k1, we have, uniformly in 1hXk1
k, the
upper bound
(1.3) X
qXk1
k
X
nX
nh( mod q)
τk(n)1
ϕq
(h,q)X
nX
(n, q
(h,q))=1
τk(n)
X1
2+,
as X , where the implied constant is independent of hand only depends on .
Remark 1. Numerical evidence for this conjectural upper bound is provided in the last
section, where we numerically determine an upper bound for the exponent of the error term
and also the size of the implied constant for the two error terms E2,2(X, 1) and E3,3(X, 1).
2
Table 1. Progress on the error term E,k (X) in the asymptotic
PnXτ(n)τk(n+ 1) = M,k (X) + E,k (X), as X , where E,k (X) is
of order strictly smaller than X(log X)+k2.
 k References E,k(X)
2 2 Ingham [16, (8.5) p. 205] (1927) Xlog X
Estermann [9, p. 173] (1931) X11/12(log X)17/6
Heath-Brown [13, Theorem 2, p. 387] (1979) X5/6+
Deshouillers & Iwaniec [6, Theorem, p. 2] (1982) X2/3+
2 3 Hooley [15, Theorem 1, p. 412] (1957) X(log Xlog log X)2
Friedlander & Iwaniec [11, p. 320] (1985) X1δ(δ > 0)
Heath-Brown [14, Theorem 3, p. 32] (1986) X11
102 +
Bykovskii, Vinogradov [4, p. 3004] (1987) X8/9+
24 Linnik [17, Teopema 3, p. 961] [17] (1958) X(log X)k1(log log X)4
Bredikhin [3, Teopema, p. 778] (1963) X(log X)k1(log log X)4
Motohashi [19, Theorem 1, p. 43] (1980) X(log log X)c(k)(log X)1
Fouvry, Tenenbaum [10, Theoreme 1, p. 44] (1985) Xexp (c(k)(log X)1/2)
Bykovskii, Vinogradov [4, p. 3004] (1987) X11
2k+
Drappeau [7, Theorem 1.5, p. 687] (2017) X1δ/k (δ > 0)
Topacogullari [27, Theorem 1.1, p. 7682] (2018) X14
15k9++X11
57 +
3 3 Open–no unconditional bound on E,k(X) is known.
Our ﬁrst result gives the full main term for the shifted convolution D3,3(X, 1).
Corollary 1. Assume Conjecture 1. Let > 0. Then, we have, as X , the asymptotic
equality
D3,3(X, 1) = M3,3(X, 1) + E3,3(X, 1),
where, with at least sixty-eight digits accuracy in the coeﬃcients,
M3,3(X, 1) = X(0.054444679154884094580751878529861703282699438750338984412069100
(1.4)
8809066227780631551394813609558909414229584839437008 log4X
+0.710113929053644747553958926673505372958197119463757504939845715359739 log3X
+2.02119605787987777943324240784753809467091508369917789267040603543881 log2X
+0.677863310832980388541571083062733656003222322704135348688102425159897 log X
+0.287236647746619417221664617814645950166036274397222249618913907447198) + O(X),
and the error term satisﬁes
E3,3(X, 1) X1/2+.
3
Remark 2. The coeﬃcients of (1.4)can be computed to any degree of accuracy–see the proof
of Corollary 1in the Appendix 7for more.
For comparison with our method, in Section 6, we explicitly work out all the main terms
in full details from a previously conjectured formula of Conrey and Gonek [5, Conjecture 3]
for the speciﬁc case k= 3 and h= 1, showing complete agreement in our answers to at least
68 digits down to the constant term. This is
Theorem 3. Let > 0. Let m3(X, 1) be deﬁned via the delta method by (6.2). Then, we
have, as X , with at least 71 digits accuracy in the coeﬃcients,
m3(X, 1)(1.5)
= 0.05444467915488409458075187852986170328269943875033898441206910088090
66227780631551394813609558909414229584839437008Xlog4(X)
+ 0.710113929053644747553958926673505372958197119463757504939845715359
739076661971842253983213149206Xlog3(X)
+ 2.0211960578798777794332424078475380946709150836991778926704060354
3880548628848354775122568369734Xlog2(X)
+ 0.67786331083298038854157108306273365600322232270413534868810242
515989727867201461267995359769Xlog(X)
+ 0.287236647746619417221664617814645950166036274397222249618913
90744731664345218868780687078219X+O(X).
A numerical computation provided by B. Conrey shows that the prediction (1.5) gives, for
X= 109,
m3(109,1) 17.243 395 216 318 ×1012
which compares extremely well with the data
D3,3(109,1) = 17,243,358,889,275,
with an error of size 3.706 ×107, which is just 0.000210% of D3,3(109,1). A graphical
comparison is provided in Figure 1, showing great alignment. In Figure 3, a plot of the error
term E3,3(X, 1), or diﬀerence between the data D3,3(X, 1) and the prediction M3,3(X, 1) is
shown.
Corollary 1is derived from the following theorem with the help of Mathematica1to carry
out the residues computations.
Theorem 1. Assume Conjecture 1. Let > 0. We have, as X ,
(1.6) D3,3(X, 1) = M3,3(X, 1) + E3,3(X, 1),
1Mathematica ﬁles available at https://aimath.org/dtn/papers/correlations/
4
200 000 400 000 600 000 800 000 1×106
1×109
2×109
3×109
4×109
Figure 1. A plot of the three functions D3,3(X, 1) (solid blue), M3,3(X, 1) in
(1.4) (dotted red), and 1
4Qp14
p2+4
p31
p4Xlog4X(large dash in green),
for X106.
where
M3,3(X, 1) = 3 Res
s=1
w1=w2=0 X1
3(w1+2w2+3s)
sw1w2
ζ3(s)ζ(w1+w2+ 1)ζ(w2+ 1)A1(s, w1, w2)!
(1.7)
3 Res
s=1
w2=1,w1=0 X1
3(w1+2w2+s)
sw1w2
ζ3(s)ζ(w1+w2+ 1 s)ζ(w2+ 1 s)A2(s, w1, w2)!
+ Res
s=1
w1=w2=1 X1
3(w1+w2+s)
sw1w2
ζ3(s)ζ(w1+ 1 s)ζ(w2+ 1 s)A3(s, w1, w2)!+O(X),
where
A1(s, w1, w2) = Y
p11
pw1+w2+1 11
pw2+1
×
1 + 11
ps3
11
p1
pw1+w2+1 1+1
pw2+1 1+1
(pw1+w2+1 1)(pw2+1 1)
,
5
A2(s, w1, w2) = Y
p11
pw1+w2+1s11
pw2+1s
×
1 + 11
ps3
11
p1
pw1+w2+1s1+1
pw2+1s1+1
(pw1+w2+1s1)(pw2+1s1)
,
and
A3(s, w1, w2) = Y
p11
pw1+1s11
pw2+1s
(1.8)
×
1 + 11
ps3
11
p1
pw1+1s1+1
pw2+1s1+1
(pw1+1s1)(pw2+1s1)
,
and the error term satisﬁes
E3,3(X, 1) X1
2+.
Remark 3. Our method applies equally to correlations between the von Mangoldt function
Λ(n)and τk(n)of the form
(1.9) Pk(X, h) = X
nX
τk(n)Λ(n+h).
In particular, by assuming the Elliott-Halberstam conjecture for Λ(n), the full main-term
for the prime correlation (1.9)can be derived and numerically tested, similar to the case for
D2,2(X, 1) and D3,3(X, 1) demonstrated here. In this sense, Conjecture 1can be seen as an
Elliott-Halberstam conjecture, but for the k-fold divisor function τk(n).
When the shift his larger than one, we work out in our third result the leading order main
term in Mk,(X, h), showing our answer matches previous computations of Ng and Thom [21]
and Tao [26].
Corollary 2. Assume Conjecture 1. We have, for any k,  2and composite shift h,
(1.10) Dk,(X, h)Ck,fk,(h)
(k1)!(1)!Xlogk+2X,
where
Ck, =Y
p 11
pk1
+11
p1
11
pk+2!,
and fk,(h)is given by equation (5.11)below. In particular, for k== 3, we have
(1.11) X
nX
τ3(n)τ3(n+ 1) 1
4Y
p14
p2+4
p31
p4f3,3(h)Xlog4X,
6
where
f3,3(h) = Y
p|hνp(h)2(p1)2(p+ 1) + pνp(h)+2 + 4pνp(h)+3
+pνp(h)+4 +νp(h)4p3+ 6p24p35p2+ 4p1
/pνp(h)(p1)2p2+ 2p1,
with νp(h)the highest power of pthat divides h.
Remark 4. The conditional asymptotic (1.11)conﬁrms a recent Conjecture in [21, Conjec-
ture, page 35] for k== 3 and 1hX2/3.
Corollary 2is derived from the following unconditional
Theorem 2. We have, as X ,
X
1X1/k X
2X(k1)/k
1
· · · X
k1X(k1)/k
1···k2X
kX
1···k1
1
ϕ(q1)Res
s=1
(X/δ)s
sX
(n,q1=1)
τ()
ns
(1.12)
Ck,fk,(h)
k!(1)! Xlogk+2X,
where q=1· · · k1,δ= (h, q), and q1=q/δ.
We give an elementary proof, essentially, for (1.12) for the case k== 3 and h= 1
in Section 4. For the general situation, it turns out to be more robust to use generating
functions. Moreover, our procedure also works to replace the leading order term asymptotic
(1.10) by the full main term, as in (1.6). We give a more detailed explanation of this process
in Section 5, where the asymptotic (1.12) for h > 1 is derived.
In the last Section 7, we provide further numerical evidence for Conjecture 1for the case
k= 2. More precisely, we reﬁne an unconditional result of Heath-Brown [13, Theorem 2] on
the shifted correlation D2,2(X, h) of the usual divisor function, giving
Theorem 4. Let > 0. We have, uniformly for all 1hX1/2, the asymptotic equality
X
nX
τ(n)τ(n+h) = M2,2(X, h) + E2,2(X, h),
where
M2,2(X, h) = Xc2(h) log2X+c1(h) log X+c0(h),
with
c2(h) = 6
π2X
d|h
1
d,
c1(h) = (4γ2)fh(1,0) + 2f(0,1)
h(1,0) + f(1,0)
h(1,0),
and
c0(h)=2f(0,1)
h(1,0) + γ2f(0,1)
h(1,0) + f(1,0)
h(1,0) fh(1,0)+f(1,1)
h(1,0) + 2γ2fh(1,0)
+f(1,0)
h(1,0) + 2(γ1)fh(1,0),
7
with the constants fh,f(0,1)
h,f(1,0)
h, and f(1,1)
hat (1,0) depending only on hgiven in Lemmas
11 and 12, and with the error term satisfying
E2,2(X, h)X5/6+.
As a consequence of this result, we obtain the following
Corollary 3. We have, for any > 0, with at least 148 digits accuracy in the coeﬃcients,
M2,2(X, 1) = X6
π2log2(X)
+1.5737449203324910789070569280484417010544014980534581993991047787172106559673
1173018329789033856157663793482022187619702084359231966550508901828044158 log(X)
0.5243838319228249988207213304174247109766097340170991428485246582967458363611
4606090215515124475866524185215534024889460792901985996741204565400064583) + O(X).
For example, our M2,2(X, 1) given above for the main term of D2,2(X, 1) for X= 20,220,000
yields
M2,2(20.22 ×106,1) 4,003,240,490,
which is just 25 parts-per-billion of the answer
(1.13) X
n20,220,000
τ(n)τ(n+ 1) = 4,003,240,588;
whereas the corresponding leading order asymptotic
6
π2(20,220,000) log2(20,220,000) 3.478542795 ×109
is far from (1.13).
A graph of the error term E2,2(X, 1) is plotted in Figure 2. In Figure 4, a log-log-plot of
this error term is shown, numerically suggesting that this error is bounded by |E2,2(X, 1)|
7X0.51, which is in favor of the conjectural bound (1.3).
Remark 5. Unconditional lower bounds for the additive divisor sum Dk,(X, h)have been
sharpened from Ng and Thom [21] by Andrade and Smith [1], who approximate, in our
notation, the general divisor function τk(n)by partial divisor functions
τ(n, A) = X
q|n:qnA
τ1(q)
parametrized by A(0,1].
Remark 6. A similar quantity to the left side of (1.3)was investigated for a special set
of moduli d=rq in [22, Theorem 1, p. 35] using the method of [28] with d < X 1
2+1
584
for a ﬁxed residue class nh(d). This is one approach towards bounding this error term
E,k(X)–maybe a weaker form of (1.3)is suﬃcient for certain applications.
8
200 000 400 000 600 000 800 000 1×106
-5000
5000
Figure 2. A plot of the error term E2,2(X, 1) in solid blue, and ±7X0.51 in
dashed red, for Xup to one million.
Remark 7. It would be interesting to also sum over hand investigate the variance of divisor
sums, such as
X
hHX
nX
τ3(n)τ3(n+h)M3,3(X, h)
2
,
with M3,3(X, h)given by (1.7)and with H=Xcfor various ranges of c. An analogous
variance, but of the k-fold divisor function in arithmetic progressions, was studied in, e.g,
[24].
2. Lemmata
We start by ﬁrst generalizing a combinatorial Lemma of Hooley [15, Lemma 4, p. 405] for
τk(n).
Lemma 1. For any nX, we have
(2.1) τk(n) = kΣk(n) + O(E(n)),
where
Σk(n) = X
12···k=n
12···k1X(k1)/k;1X1/k
1
and
E(n) = X
12···k=n
12···k1X(k1)/k;kX1/k
1.
9
Proof. This follows from the identity
X
1···k=n
1,··· ,kX1/k
=X
1···k=n
1X
1ikX
1···k=n
i>X1/k
1 + X
1i1<i2kX
1···k=n
i1,i2>X1/k
1 + · · ·
+ (1)jX
1i1<···<ijkX
1···k=n
i1,··· ,ij>X1/k
1 + · · · + (1)kX
1···k=n
1,··· ,k>X1/k
1.
Lemma 2. For any h1, we have
(2.2)
X
n=1
τk(nh)
ns=ζk(s)Ah(s),(σ > 1),
where
(2.3) Ah(s) = Y
p|h11
pskk+νp(h)1
k12F1(1, k +νp(h); 1 + νp(h); ps),
where 2F1is a hypergeometric function.
Proof. By multiplicativity and Euler products, we have
X
n=1
τk(nh)
ns=Y
p|h
X
j=0
τk(pj+νp(h))
pjs !Y
p-h
X
j=0
τk(pj)
pjs !
=Y
p|h
X
j=0 k+j+νp(h)1
k11
pjs
X
j=0
τk(pj)
pjs Y
p
X
j=0
τk(pj)
pjs !.
By a hypergeometric relation, we have
X
j=0 k+j+νp(h)1
k11
pjs =k+νp(h)1
k12F1(1, k +νp(h); 1 + νp(h); ps).
This, together with
Y
p
X
j=0
τk(pj)
pjs !=ζk(s),
give (2.2).
Lemma 3. For any h1, we have
X
(n,h)=1
τk(n)
ns=ζk(s)Y
p|h11
psk
,(σ > 1).
10
Proof. Going to Euler products gives
X
(n,h)=1
τk(n)
ns=Y
p-h
X
j=0
τk(pj)
pjs =ζk(s)Y
p|h11
psk
.
3. Proof of Theorem 1
Lemma 4. For any nX, we have
(3.1) τ3(n) = 1(n)2(n)+Σ3(n),
where
Σ1(n) = X
123=n
12X2/3;1X1/3
1,
Σ2(n) = X
123=n
12X2/3;1,3X1/3
1,
Σ3(n) = X
123=n
1,2,3X1/3
1.
Substituting (3.1) in for τ3(n) in D3,3(X, h), we have
D3,3(X, h)(3.2)
= 3 X
nX
τ3(n+h1(n)3X
nX
τ3(n+h2(n) + X
nX
τ3(n+h3(n)
= 11(X)21(X)+Σ31(X),
say. Interchanging the order of summations in Σ11(X), we have
Σ11(X) = X
1X1/3X
2X2/3
1X
3X
12
τ3(123+h).
Making a change of variables in the 3sum, we get
(3.3) Σ11(X) = X
1X1/3X
2X2/3
1X
nX+h
nh(12)
τ3(n).
Similarly, we obtain
(3.4) Σ21(X) = X
1X1/3X
2X2/3
1X
n12X1/3+h
n1(12)
τ3(n)
and
(3.5) Σ31(X) = X
1X1/3X
3X1/3X
n13X1/3+h
nh(13)
τ3(n).
11
By (1.3), we have
X
nY
nh(q)
τ3(n)1
ϕ(q)Res
s=1
Ys
sζ3(s)fq(s),
where
(3.6) fq(s) = Y
p|q11
ps3
.
Thus, by (1.3), (3.3), (3.4), and (3.5), D3,3(X, h) becomes
D3,3(X, h) = 3Res
s=1
Xs
sζ3(s)X
1X1/3X
2X2/3
1
f12(s)
ϕ(12)
(3.7)
3Res
s=1
ζ3(s)
sX
1X1/3X
2X2/3
1
f12(s)
ϕ(12)(12X1/3+h)s
+ Res
s=1
ζ3(s)
sX
1X1/3X
2X1/3
f12(s)
ϕ(12)(12X1/3+h)s
.
We treat the double sums by Perron’s formula. By Perron’s formula, we have
X
1X1/3X
2X2/3
1
f12(s)
ϕ(12)=1
(2πi)2Z(2) Z(2)
Xw1/3
w1
X2w2/3
w2
X
1,2=1
f12(s)
ϕ(12)
1
w1+w2
1
1
w2
2
dw1dw2.
By multiplicativity and Euler products, we write the ’s sums as
(3.8)
X
1,2=1
f12(s)
ϕ(12)
1
w1+w2
1
1
w2
2
=Y
pX
j1,j2
fpj1+j2(s)
ϕ(pj1+j2)
1
pj1(w1+w2)+j2w2.
By (3.6) and deﬁnition of ϕ(n), the j’s sums become
X
j1,j2
fpj1+j2(s)
ϕ(pj1+j2)
1
pj1(w1+w2)+j2w2= 1 + 11
ps3
11
pX
j1,j21
1
pj1(w1+w2)
1
pj2w2
(3.9)
= 1 + 11
ps3
11
p1
pw1+w2+1 1+1
pw2+1 1+1
(pw1+w2+1 1)(pw2+1 1).
Thus, by (3.8) and (3.9), we get
X
1X1/3X
2X2/3
1
f12(s)
ϕ(12)=1
(2πi)2Z(2) Z(2)
Xw1/3
w1
X2w2/3
w2
ζ(w1+w2+ 1)ζ(w2+ 1)A1(s, w1, w2)dw1dw2,
12
where A1(s, w1, w2) is given as in (1.8). Thus, with this, the ﬁrst term on the right side of
(3.7) becomes
3 Res
s=1
w1=w2=0 X1
3(w1+2w2+3s)+hs
sw1w2
ζ3(s)ζ(w1+w2+ 1)ζ(w2+ 1)A1(s, w1, w2)!.
This gives the ﬁrst term on the right side of (1.7). Similarly, we obtain the remaining two
terms in (1.7). This completes the proof of Theorem 1.
4. Conditional proof of the asymptotic for the correlation sum
PnXτ3(n)τ3(n+ 1)
Let h= 1. Recall that Σ11(X), Σ21 (X), and Σ31(X) are given by (3.3), (3.4), and (3.5),
respectively. We will evaluate Σ11(X) asymptotically, and give bounds for Σ21 (X) and
Σ31(X).
4.1. Using level of distribution for τ3(n)in AP’s to evaluate the sum Σ11(X).We
treat the most inner sum in (3.3) using an averaged level of distribution for τ3(n).
The main term in (1.3) is explicit.
Lemma 5. For any q1, we have
1
ϕ(q)X
nX
(n,q)=1
τ3(n) = Xa1(q) log2X+a2(q) log X+a3(q)
(4.1)
+Oτ(q)X2/3log X
ϕ(q),
where
a1(q) = 1
2
ϕ(q)2
q3,(4.2)
a2(q) = ϕ(q)2
q3
3γ7
6+7
3X
p|q
log p
p1
,(4.3)
a3(q) = ϕ(q)2
q3
3γ23γ+ 3γ1+X
p|q
log p
p1
4γ3 + X
p|q
log p
p1
.
Proof. See, e.g, [23, Lemma 51, p. 153].
Thus, assuming Conjecture 1with k= 3, we get, by (1.3), (4.1), and (3.3), that
Σ11(X)(X+ 1) b1(X) log2(X+ 1) + b2(X) log(X+ 1) + b3(X)+O(E(X)),(4.4)
13
where
b1(X) = X
1X1/3X
2X2/3
1
a1(12),(4.5)
b2(X) = X
1X1/3X
2X2/3
1
a2(12),(4.6)
b3(X) = X
1X1/3X
2X2/3
1
a3(12),
and
(4.7) E(X)=(X+ 1)2/3log(X+ 1) X
1X1/3X
2X2/3
1
τ(12)
ϕ(12).
We will evaluate b1(X) asymptotically and estimate b2(X), b3(X), and E(X) below.
4.1.1. Evaluation of b1(X).By (4.5) and (4.2), we have
(4.8) b1(X) = 1
2X
1X1/3X
2X2/3
1
ϕ(12)2
(12)3.
We evaluate b1(X) in the following lemma.
Lemma 6. There are computable constants c1and c2such that
(4.9) b1(X) = 1
12 Y
p14
p2+4
p31
p4log2(X) + c1log X+c2+OX2
3+.
Proof. We apply Perron’s formula twice to (4.8), ﬁrst to the 2sum, then to the 1sum. Let
(4.10) f(n) = Y
p|n11
p2
and
(4.11) gd(n) = f(nd)
f(d).
The functions f(n) and gd(n) are both multiplicative in n. By (4.8), deﬁnition of ϕ(n),
(4.10), and (4.11), we have
(4.12) b1(X) = 1
2X
1X1/3
f(1)
1
Σ(X, 1),
where
(4.13) Σ(X, 1) = X
nX2/3
1
g1(n)
n.
14
By Euler products, we have
X
n=1
g1(n)
ns+1 =ζ(s+ 1)A(s)B1(s),(σ > 0),
where
(4.14) A(s) = Y
p12
ps+2 +1
ps+3 ,(σ > 1),
(4.15) Bn(s) = Y
p|n11
p2
12
ps+2 +1
ps+3
,(σ > 1),
and A(s) and B1(s) are convergent in the larger region σ > 1. Thus, by (4.13) and Perron’s
formula, we have
Σ(X, 1) = A(0)B1(0) log X2/3
1+ (AB1)0(0) + γA(0)B1(0) + O X2/3
11!.
Hence, by (4.12) and the above, we have
(4.16) b1(X) = b11(X) log X+b12 (X) + b13(X) + OX1
3+,
where
b11(X) = 2
3A(0) X
nX1/3
Bn(0)
n,(4.17)
b12(X) = A(0) X
nX1/3
Bn(0)
nlog n,(4.18)
b13(X) = ((AB1)0(0) + γA(0)B1(0)) X
nX1/3
1
n.(4.19)
We now evaluate the b’s. By the deﬁnition (4.15) and Euler products, we have
(4.20)
X
n=0
Bn(0)
ns+1 =ζ(s+ 1)B(s),(σ > 0),
where
(4.21) B(s) = Y
p
11
ps+1 +1
ps+1 11
p2
12
p2+1
p3
,(σ > 1).
By (4.14) and (4.21), we have
(4.22) A(0)B(0) = Y
p14
p2+4
p31
p4.
15
Thus, by (4.17), Perron’s formula, (4.20), and (4.22), we have
b11(X) = 2
9Y
p14
p2+4
p31
p4log X(4.23)
+2
3A(0) (B0(0) + γB(0)) + OX1
3+.
Next, by (4.18), partial summation, and the above, we have
b12(X) = 1
18 Y
p14
p2+4
p31
p4log2X(4.24)
1
2A(0) (B0(0) + γB(0)) log X+OX1
3+.
Lastly, we have, from (4.19)
(4.25) b13(X) = ((AB1)0(0) + γA(0)B1(0)) 1
3log X+γ+O1
X.
Therefore, combining (4.16), together with (4.23), (4.24), and (4.25), the estimate (4.9)
follows.
4.1.2. Bounds for b2(X),b3(X), and E1(X).
Lemma 7. We have, as X ,
b2(X)log2X,
b3(X)log2X,
E(X)X2
3+.
Proof. We have
X
p|q
log p
p11.
Thus, by (4.6), (4.3), the above, and (4.5), we have
b2(X)X
1X1/3X
2X2/3
1
ϕ(12)2
(12)3b1(X)log2X,
by (4.9). Similarly, we get
b3(X)log2X.
We now estimate E1(X). We have
X
1X1/3X
2X2/3
1
τ(12)
ϕ(12)XX
1X1/3
1
1X
2X2/3
1
1
2
X.
Hence, by (4.7) and the above, we get
E(X)X2
3+.
16
Therefore, combining (4.4), Lemmas 6and 7, we have, on assuming Conjecture 1we obtain
the following
Proposition 1. Assume Conjecture 1for k= 3. Then, we have, as X ,
Σ11(X)1
12 Y
p14
p2+4
p31
p4(X+ 1) log2(X+ 1) log2X,
with Σ11(X)deﬁned in (3.2)and given in (3.3).
4.2. Applying Shiu’s bound to estimate the remaining sums Σ21(X)and Σ31(X).
We apply Shiu’s bound below to treat the last two sums Σ21(X) and Σ31(X).
Lemma 8 (Shiu’s bound).Suppose that 1N < N 0<2X,N0N > Xd, and (a, d) = 1.
Then for j, ν 1we have
(4.26) X
NnN0
na(d)
τj(n)νN0N
ϕ(d)(log X)jν1.
The implied constants depending on , j, and νat most.
Proof. See [25, Theorem 2].
This is
Proposition 2. We have
Σ21(X)Xlog3Xlog log X,
Σ31(X)Xlog3Xlog log X.
Proof. We treat Σ21(X) ﬁrst. By Shiu’s bound (4.26), the most inner sum over nin Σ21(X)
is
(4.27) 1
ϕ(13)(3X2/3+ 1) log2(3X2/3+ 1).
Thus, by (3.4) and (4.27),
Σ21(X)X2/3log2XX
1X1/3X
3X1/3
3
ϕ(13)Xlog3Xlog log X.
Similarly, we have, from (4.26), that
Σ31(X)Xlog2Xlog log X.
Therefore, on assuming Conjecture 1, we obtain, by (3.2), Propositions 1and 2, the
asymptotic (1.11) for h= 1.
17
5. General case of mixed correlations and composite shifts
In this section we derive the asymptotic (1.12) and describe the general procedure to
extract the leading order main term of the mixed correlation sum Dk,(X, h) in (1.1) with
composite shifts h.
Write
h=Y
p
pνp(h).
We replace τk(n) in (1.1) by Hooley’s identity (2.1), giving
Dk,(X, h)kX
nX
τ(n+h)X
12···k=n
12···k1X(k1)/k;1X1/k
1
=kX
1X1/k X
2X(k1)/k
1X
3X(k1)/k
12
· · · X
k1X(k1)/k
1···k2X
kX
1···k1
τ(1· · · k+h).
Making a change of variables n=1· · · k+hin the most inner ksum, the above becomes
kX
1X1/k X
2X(k1)/k
1X
3X(k1)/k
12
· · · X
k1X(k1)/k
1···k2X
nX+h
nh(1···k1)
τ(n).
By the bound (1.3), the error term is negligible and the above is in turns asymptotic to
kX
1X1/k X
2X(k1)/k
1X
3X(k1)/k
12
· · · X
k1X(k1)/k
1···k2
1
ϕ1···k1
(h,1···k1)X
nX+h
n, 1···k1
(h,1···k1)=1
τ(n).
Thus, by Perron’s formula, we obtain that
Dk,(X, h)kRes
s=1
w1=···=wk1=0 Xs+w1
k+k1
k(w2+···wk1)
sw1· · · wk1
Tk,(s, w1,· · · , wk1;h)!,(5.1)
where
Tk,(s, w1,· · · , wk1;h) =
X
1,...,k1=1
1
(h, 1· · · k1)s
1
ϕ1···k1
(h,1···k1)
×X
n, 1···k1
(h,1···k1)=1
τ(n(h, 1· · · k1))
ns
k1
Y
j=1
Pk1
i=jwi
j.
By multiplicativity and Euler products, the above generating function Tk, can be written as
(5.2) Tk,(s, w1,· · · , wk1;h) = Y
p|h
Ap(s;w1,· · · , wk1;h)
Bp(s;w1,· · · , wk1)Y
p
Bp(s;w1,· · · , wk1),
18
where
Ap(s;w1,· · · , wk1;h)(5.3)
=X
j1,··· ,jk1
1
pmin(j1+···+jk1p(h))s
1
ϕ(pj1+···+jk1min(j1+···+jk1p(h)))
×X
(n,pj1+···+jk1min(j1+···+jk1p(h))=1
τ(npmin(j1+···+jk1p(h))
ns
1
pPk1
i=1 jiPk1
κ=iwκ,
and
(5.4) Bp(s;w1,· · · , wk1) = ζ(s)
1 + 11
p
11
p
k1
X
j=1 X
σΞj,k1
j
Y
i=1
1
pwσ(i)+1 1
(we have used a nonstandard notation here, Ξj,n ={(α1· · · αj)Sn:α1<· · · < αj}and
σ(i) to mean αi, where Snis the usual symmetric group on nletters). From (5.4), we can
further factor out a product of zetas from Bpas
Y
p
Bp(s;w1,· · · , wk1) = ζ(s)ζ(w1+w2+· · · wk1+ 1)ζ(w2+· · · wk1+ 1) × · · ·(5.5)
×ζ(wk1+ 1) Y
p
BBp(s;w1,· · · , wk1),
where
(5.6)
BBp(s;w1,· · · , wk1) =
k1
Y
i=1 11
pwi+···+wk1+1
1 + 11
p
11
p
k1
X
j=1 X
σΞj,k1
j
Y
i=1
1
pwσ(i)+1 1
.
The product QpBBp(s;w1,· · · , wk1) converges in a wider region than QpBpsince we have
factored out all the poles from the latter. Similarly, from Lemmas 2and 3, the local Euler
factors can be written as
Ap(s;w1,· · · , wk1;h) = ζ(s)AAp(s;w1,· · · , wk1;h)(5.7)
with AAp(s;w1,· · · , wk1;h) a nice Euler product converging in a larger region. From (5.7)
and (5.5), the factor ζ(s) cancels out in the ratio Ap
Bp, and that the generating function
Tk,(s, w1,· · · , wk1;h) (5.2) can thus be written as
Tk,(s, w1,· · · , wk1;h) = ζ(s)ζ(w1+w2+· · · wk1+ 1)ζ(w2+· · · wk1+ 1) × · · ·(5.8)
×ζ(wk1+ 1) Y
p|h
Ap(s;w1,· · · , wk1;h)
Bp(s;w1,· · · , wk1)Y
p
BBp(s;w1,· · · , wk1),
and, hence, we conclude that Tk,(s, w1,· · · , wk1;h) has poles at s= 1 and w1=· · · =
wk1= 0. Therefore, by (5.8) above, we obtain from (5.1) that
(5.9) Dk,(X, h)Ck,fk,(h)
(k1)!(1)!X(log X)k+2,
19
where
(5.10) Ck, =Y
p
BBp(0; ~
0) = Y
p11
pk1 1 + 11
p1k1
X
j=1 k1
j
(p1)j!
and
(5.11) fk,(h) = Y
p|h
Ap(1;~
0; h)
Bp(1;~
0) ,
where we have abbreviated ~
0 for 0,· · · ,0k1 times.
Lastly, we show that the constant Ck, from (5.10) above matches the predicted global
constant from equation (1.6) of [21].
Proposition 3. We have
(5.12) Ck, =Y
p 11
pk1
+11
p1
11
pk+2!.
Proof. We have the identity
k1
X
j=1 k1
j
(p1)j=1 + p
p1k
1
pp
p1k
.
Substituting the above into the right side of (5.10) and simplifying then give the right side
of (5.12).
5.1. The case k= 3 and , h 1.In this subsection, we demonstrate how to apply our
general method developed above to extract the leading order main term for the case k= 3
and , h 1, showing that our answers match with previously conjectured values.
Let k= 3 and ﬁx , h 1. The procedure from previous subsection gives that
X
nX
τ3(n)τ(n+h)3X
1X1/3X
2X2/3
1X
nX+h
nh(12)
τ(n)(5.13)
3 Res
s=1
w1=w2=0 X1
3w1
w1
X2
3w2
w2
(X+h)s
sT(s, w1, w2;h)!,
with
T(s, w1, w2;h) =
X
1,2=1
1
(h, 12)s
1
ϕ(12/(h, 12))
×X
n, 12
(h,12)=1
τ(n(h, 12))
ns
1
w1+w2
1
1
w2
2
=Y
p|h
Ap(s;w1, w2;h)
Bp(s;w1, w2)Y
p
Bp(s;w1, w2),
20
with the global Euler factor Bp(s;w1, w2) given in (5.4) with k= 3, and local factor
Ap(s, w1, w2;h) = X
j1,j2
1
pmin(j1+j2p(h))s
1
ϕ(pj1+j2min(j1+j2p(h)))
×X
(n,pj1+j2min(j1+j2p(h))=1
τ(npmin(j1+j2p(h))
ns
1
pj1(w1+w2)+j2w2.
Thus, (5.13) predicts that
X
nX
τ3(n)τ(n+h)1
4Y
p14
p2+4
p31
p4Y
p|h
f3,(h)Xlog4X.
with
(5.14) f3,(h) = Ap(1; 0,0; h)
Bp(1; 0,0) .
We ﬁrst evaluate f3,(h) in (5.14) for = 3 and hprime.
5.2. Prime shifts.
Proposition 4. Let hbe a prime. We have
(5.15) f3,3(h) = h3+ 6h2+ 3h4
h(h2+ 2h1) .
In particular, assuming the bound (1.3)for k= 3, we have, for hprime,
D3,3(X, h)1
4Y
p14
p2+4
p31
p4h3+ 6h2+ 3h4
h(h2+ 2h1) Xlog4X.(5.16)
Proof. By Perron’s formula and (1.3), we have
X
1X1/3X
2X2/3
1X
nX+h
nh(12)
τ3(n)(5.17)
1
(2πi)3Z(2) Z(2) Z(2)
X1
3w1
w1
X2
3w2
w2
Xs
sT3(s, w1, w2)dw1dw2ds,
where
T3(s, w1, w2) =
X
1,2=1
1
ϕ(q1)
1
δsX
(n,q1)=1
τ3()
ns
1
w1+w2
1
1
w2
2
with
δ= (h, 12)
and
q1=1`2
δ.
21
By Euler products, we can write this function as
T3(s, w1, w2) = Y
p|h
Ap(s;w1, w2;h)
Bp(s;w1, w2)Y
p
Bp(s;w1, w2),
where
Bp(s;w1, w2) = X
n
τ3(n)
ns+X
j1,j2
j1j26=0
1
ϕ(pj1+j2)X
(n,pj1+j2)=1
τ3(n)
ns
1
pj1(w1+w2)+j2w2
(5.18)
and
(5.19) Ap(s;w1, w2;h) = X
n
τ3(n)
ns+X
j1,j2
j1j26=0
1
ϕ(pj1+j21)
1
hsX
(n,pj1+j21)=1
τ3(nh)
ns
1
pj1(w1+w2)+j2w2.
We split the jisums in (5.18) into
X
j1,j2
j1j26=0
=X
j11
j2=0
+X
j1=0
j21
+X
j11
j21
.
We have
X
j11
j2=0
1
ϕ(pj1+j2)X
(n,pj1+j2)=1
τ3(n)
ns
1
pj1(w1+w2)+j2w2
=
X
j1=1
1
pj111
pX
(n,p)=1
τ3(n)
ns
1
pj1(w1+w2)
=ζ3(s)11
ps3
11
p
X
j1=1
1
pj1(w1+w2+1)
=ζ3(s)11
ps3
11
p
1
pw1+w2+1 1,
X
j1=0
j21
1
ϕ(pj1+j2)X
(n,pj1+j2)=1
τ3(n)
ns
1
pj1(w1+w2)+j2w2
=ζ3(s)11
ps3
11
p
1
pw2+1 1,
22
and
X
j11
j21
1
ϕ(pj1+j2)X
(n,pj1+j2)=1
τ3(n)
ns
1
pj1(w1+w2)+j2w2
=ζ3(s)11
ps3
11
p
1
pw1+w2+1 1
1
pw2+1 1.
Thus,
Bp(s;w1, w2) = ζ3(s)
1 + 11
ps3
11
p1
pw1+w2+1 1
(5.20)
+1
pw2+1 1+1
pw1+w2+1 1
1
pw2+1 1
and, hence,
Y
p
Bp(s;w1, w2) = ζ3(s)ζ(w1+w2+ 1)ζ(w2+ 1)BB(s;w1, w2),
where
BB(s;w1, w2) = Y
p11
pw1+w2+1 11
pw2+1
×
1 + 11
ps3
11
p1
pw1+w2+1 1+1
pw2+1 1
+1
pw1+w2+1 1
1
pw2+1 1.
We have that
BB(1; 0,0) = Y
p11
p2 1 + 11
p21
p1+1
p1+1
(p1)2!
(5.21)
=Y
p14
p2+4
p31
p4.
We evaluate the dw2integral in (5.17) ﬁrst, picking up a double pole at w2= 0, then perform
the dw1integral, collecting the triple pole at w1= 0, and ﬁnally the ds integral, with a triple
pole at s= 0. Thus, the left side of (5.16) is asymptotic to
(5.22) BB(1; 0,0)A(1; 0,0; h)
12 Xlog4X.
We next evaluate (5.19).
23
Because of the exponent j1+j21 in (5.19) being non-negative, we split the jsums in
(5.19) into
X
j1,j2
j1j26=0
=X
j1=1
j2=0
+X
j1=0
j2=1
+X
j12
j2=0
+X
j1=0
j22
+X
j11
j21
.
We have
X
j1=1
j2=0
1
ϕ(pj1+j21)
1
hsX
(n,pj1+j21)=1
τ3(nh)
ns
1
pj1(w1+w2)+j2w2
=1
hsX
n
τ3(nh)
ns
1
pw1+w2=1
hs
1
pw1+w2ζ3(s)Ah(s),
X
j1=0
j2=1
1
ϕ(pj1+j21)
1
hsX
(n,pj1+j21)=1
τ3(nh)
ns
1
pj1(w1+w2)+j2w2
=1
hsX
n
τ3(nh)
ns
1
pw2=1
hs
1
pw2ζ3(s)Ah(s),
X
j12
j2=0
1
ϕ(pj1+j21)
1
hsX
(n,pj1+j21)=1
τ3(nh)
ns
1
pj1(w1+w2)+j2w2
=
X
j1=2
1
ϕ(pj11)
1
hsX
(n,h)=1
τ3(h)τ3(n)
ns
1
pj1(w1+w2)
=1
hsτ3(h)ζ3(s)Y
p|h11
ps3
X
j1=2
1
pj1111
p
1
pj1(w1+w2)
=3
ps1ζ3(s)11
ps3
11
p
X
j1=2
1
pj1(w1+w2+1)
=3
ps1ζ3(s)11
ps3
11
p
1
pw1+w2+1
1
pw1+w2+1 1,
24
X
j1=0
j22
1
ϕ(pj1+j21)
1
hsX
(n,pj1+j21)=1
τ3(nh)
ns
1
pj1(w1+w2)+j2w2
=
X
j2=2
1
ϕ(pj21)
1
hsX
(n,h)=1
τ3(h)τ3(n)
ns
1
pj2w2
=3
ps1ζ3(s)11
ps3
11
p
1
pw2+1
1
pw2+1 1,
and
X
j11
j21
1
ϕ(pj1+j21)
1
hsX
(n,pj1+j21)=1
τ3(nh)
ns
1
pj1(w1+w2)+j2w2
=X
j11
j21
1
ϕ(pj1+j21)
1
hsX
(n,h)=1
τ3(h)τ3(n)
ns
1
pj1(w1+w2)+j2w2
=3
ps1ζ3(s)11
ps3
11
p
1
pw1+w2+1 1
1
pw2+1 1.
Thus, the local Euler product Ap(s;w1, w2;h) of T3(s;w1, w2) is equal to
ζ3(s)
1 + 1
psAp(s)1
pw1+w2+1
pw2+3
ps111
ps3
11
p1
pw2+1
1
pw2+1 1
+1
pw1+w2+1
1
pw1+w2+1 1+1
pw1+w2+1 1
1
pw2+1 1.
Thus, by this, (5.19) and (5.20),
Ap(s;w1, w2;h)
=
1 + 1
psAp(s)1
pw1+w2+1
pw2+3
ps1
(11
ps)3
11
p1
pw2+1 1
pw2+11+1
pw1+w2+1 1
pw1+w2+11+1
pw1+w2+11
1
pw2+11
1 + (11
ps)3
11
p1
pw1+w2+11+1
pw2+11+1
pw1+w2+11
1
pw2+11.
Now, by (2.3) with k= 3, we have
Ap(1) = 3 3
p+1
p2.
25
Hence,
Ap(1; 0,0; h) = p3+ 6p2+ 3p4
p(p2+ 2p1) .
This, together with (5.22) and (5.21), give the right side of (5.16).
5.3. Composite shift h.Similarly, for any hcomposite, Mathematica calculations2give
f3,3(h) = Y
p|hνp(h)2(p1)2(p+ 1) + pνp(h)+2 + 4pνp(h)+3
(5.23)
+pνp(h)+4 +νp(h)4p3+ 6p24p35p2+ 4p1
/pνp(h)(p1)2p2+ 2p1,
f3,4(h) = Y
p|h
pνp(h)3(p+ 1)(p1)3νp(h)27p2+ 6p4(p1)2
(5.24)
+νp(h)16p4+ 33p222p+ 5+ 2 pνp(h)+2 + 5pνp(h)+3
+5pνp(h)+4 +pνp(h)+5 6p49p3+ 9p25p+ 1
/2pνp(h)(p1)3p3+ 2p23p+ 1,
and
f3,5(h) = Y
p|hνp(h)4(p+ 1)(p1)4νp(h)311p2+ 8p7(p1)3
(5.25)
νp(h)244p3+ 31p250p+ 17(p1)2νp(h)76p5+p4
200p3+ 200p294p+ 17+ 6 6pνp<