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ON CORRELATION OF THE 3-FOLD DIVISOR FUNCTION WITH

ITSELF

DAVID T. NGUYEN

Abstract. Let ζk(s) = P∞

n=1 τk(n)n−s,<s > 1. We present three conditional results on

the additive correlation sum X

n≤X

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 Pn≤Xτk(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 Pn≤Xτ3(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 Pn≤Xτ(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 k≥1 let

ζk(s) =

∞

X

n=1

τk(n)

ns,(<s > 1).

The additive correlation sums

(1.1) Dk,`(X, h) = X

n≤X

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

n≤X

τ3(n)τ3(n+ 1) ∼1

4Y

p1−4

p2+4

p3−1

p4Xlog4X,

as X→ ∞. More generally, the additive divisor correlation problem asks for an asymptotic

of the form

X

n≤X

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)`+k−2and 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 k≥1, we have, uniformly in 1≤h≤Xk−1

k, the

upper bound

(1.3) X

q≤Xk−1

k

X

n≤X

n≡h( mod q)

τk(n)−1

ϕq

(h,q)X

n≤X

(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

Pn≤Xτ`(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)`+k−2.

` 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) X1−δ(δ > 0)

Heath-Brown [14, Theorem 3, p. 32] (1986) X1−1

102 +

Bykovskii, Vinogradov [4, p. 3004] (1987) X8/9+

2≥4 Linnik [17, Teopema 3, p. 961] [17] (1958) X(log X)k−1(log log X)4

Bredikhin [3, Teopema, p. 778] (1963) X(log X)k−1(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) X1−1

2k+

Drappeau [7, Theorem 1.5, p. 687] (2017) X1−δ/k (δ > 0)

Topacogullari [27, Theorem 1.1, p. 7682] (2018) X1−4

15k−9++X1−1

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

4Qp1−4

p2+4

p3−1

p4Xlog4X(large dash in green),

for X≤106.

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

p1−1

pw1+w2+1 1−1

pw2+1

×

1 + 1−1

ps3

1−1

p1

pw1+w2+1 −1+1

pw2+1 −1+1

(pw1+w2+1 −1)(pw2+1 −1)

,

5

A2(s, w1, w2) = Y

p1−1

pw1+w2+1−s1−1

pw2+1−s

×

1 + 1−1

ps3

1−1

p1

pw1+w2+1−s−1+1

pw2+1−s−1+1

(pw1+w2+1−s−1)(pw2+1−s−1)

,

and

A3(s, w1, w2) = Y

p1−1

pw1+1−s1−1

pw2+1−s

(1.8)

×

1 + 1−1

ps3

1−1

p1

pw1+1−s−1+1

pw2+1−s−1+1

(pw1+1−s−1)(pw2+1−s−1)

,

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

n≤X

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)

(k−1)!(`−1)!Xlogk+`−2X,

where

Ck,` =Y

p 1−1

pk−1

+1−1

p`−1

−1−1

pk+`−2!,

and fk,`(h)is given by equation (5.11)below. In particular, for k=`= 3, we have

(1.11) X

n≤X

τ3(n)τ3(n+ 1) ∼1

4Y

p1−4

p2+4

p3−1

p4f3,3(h)Xlog4X,

6

where

f3,3(h) = Y

p|h−νp(h)2(p−1)2(p+ 1) + pνp(h)+2 + 4pνp(h)+3

+pνp(h)+4 +νp(h)−4p3+ 6p−2−4p3−5p2+ 4p−1

/pνp(h)(p−1)2p2+ 2p−1,

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 1≤h≤X2/3.

Corollary 2is derived from the following unconditional

Theorem 2. We have, as X→ ∞,

X

`1≤X1/k X

`2≤X(k−1)/k

`1

· · · X

`k−1≤X(k−1)/k

`1···`k−2X

`k≤X

`1···`k−1

1

ϕ(q1)Res

s=1

(X/δ)s

sX

(n,q1=1)

τ`(nδ)

ns

(1.12)

∼Ck,`fk,`(h)

k!(`−1)! Xlogk+`−2X,

where q=`1· · · `k−1,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 1≤h≤X1/2, the asymptotic equality

X

n≤X

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) = (4γ−2)fh(1,0) + 2f(0,1)

h(1,0) + f(1,0)

h(1,0),

and

c0(h)=2−f(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(γ−1)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

n≤20,220,000

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:q≤nA

τ`−1(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 n≡h(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

h≤HX

n≤X

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 n≤X, we have

(2.1) τk(n) = kΣk(n) + O(E(n)),

where

Σk(n) = X

`1`2···`k=n

`1`2···`k−1≤X(k−1)/k;`1≤X1/k

1

and

E(n) = X

`1`2···`k=n

`1`2···`k−1≤X(k−1)/k;`k≤X1/k

1.

9

Proof. This follows from the identity

X

`1···`k=n

`1,··· ,`k≤X1/k

=X

`1···`k=n

1−X

1≤i≤kX

`1···`k=n

`i>X1/k

1 + X

1≤i1<i2≤kX

`1···`k=n

`i1,`i2>X1/k

1 + · · ·

+ (−1)jX

1≤i1<···<ij≤kX

`1···`k=n

`i1,··· ,`ij>X1/k

1 + · · · + (−1)kX

`1···`k=n

`1,··· ,`k>X1/k

1.

Lemma 2. For any h≥1, we have

(2.2)

∞

X

n=1

τk(nh)

ns=ζk(s)Ah(s),(σ > 1),

where

(2.3) Ah(s) = Y

p|h1−1

pskk+νp(h)−1

k−12F1(1, k +νp(h); 1 + νp(h); p−s),

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

k−11

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

k−11

pjs =k+νp(h)−1

k−12F1(1, k +νp(h); 1 + νp(h); p−s).

This, together with

Y

p ∞

X

j=0

τk(pj)

pjs !=ζk(s),

give (2.2).

Lemma 3. For any h≥1, we have

X

(n,h)=1

τk(n)

ns=ζk(s)Y

p|h1−1

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|h1−1

psk

.

3. Proof of Theorem 1

We start with Hooley’s identity (2.1) specializing to k= 3.

Lemma 4. For any n≤X, we have

(3.1) τ3(n) = 3Σ1(n)−3Σ2(n)+Σ3(n),

where

Σ1(n) = X

`1`2`3=n

`1`2≤X2/3;`1≤X1/3

1,

Σ2(n) = X

`1`2`3=n

`1`2≤X2/3;`1,`3≤X1/3

1,

Σ3(n) = X

`1`2`3=n

`1,`2,`3≤X1/3

1.

Substituting (3.1) in for τ3(n) in D3,3(X, h), we have

D3,3(X, h)(3.2)

= 3 X

n≤X

τ3(n+h)Σ1(n)−3X

n≤X

τ3(n+h)Σ2(n) + X

n≤X

τ3(n+h)Σ3(n)

= 3Σ11(X)−3Σ21(X)+Σ31(X),

say. Interchanging the order of summations in Σ11(X), we have

Σ11(X) = X

`1≤X1/3X

`2≤X2/3

`1X

`3≤X

`1`2

τ3(`1`2`3+h).

Making a change of variables in the `3sum, we get

(3.3) Σ11(X) = X

`1≤X1/3X

`2≤X2/3

`1X

n≤X+h

n≡h(`1`2)

τ3(n).

Similarly, we obtain

(3.4) Σ21(X) = X

`1≤X1/3X

`2≤X2/3

`1X

n≤`1`2X1/3+h

n≡1(`1`2)

τ3(n)

and

(3.5) Σ31(X) = X

`1≤X1/3X

`3≤X1/3X

n≤`1`3X1/3+h

n≡h(`1`3)

τ3(n).

11

By (1.3), we have

X

n≤Y

n≡h(q)

τ3(n)∼1

ϕ(q)Res

s=1

Ys

sζ3(s)fq(s),

where

(3.6) fq(s) = Y

p|q1−1

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

`1≤X1/3X

`2≤X2/3

`1

f`1`2(s)

ϕ(`1`2)

(3.7)

−3Res

s=1

ζ3(s)

sX

`1≤X1/3X

`2≤X2/3

`1

f`1`2(s)

ϕ(`1`2)(`1`2X1/3+h)s

+ Res

s=1

ζ3(s)

sX

`1≤X1/3X

`2≤X1/3

f`1`2(s)

ϕ(`1`2)(`1`2X1/3+h)s

.

We treat the double sums by Perron’s formula. By Perron’s formula, we have

X

`1≤X1/3X

`2≤X2/3

`1

f`1`2(s)

ϕ(`1`2)=1

(2πi)2Z(2) Z(2)

Xw1/3

w1

X2w2/3

w2

∞

X

`1,`2=1

f`1`2(s)

ϕ(`1`2)

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

f`1`2(s)

ϕ(`1`2)

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 + 1−1

ps3

1−1

pX

j1,j2≥1

1

pj1(w1+w2)

1

pj2w2

(3.9)

= 1 + 1−1

ps3

1−1

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

`1≤X1/3X

`2≤X2/3

`1

f`1`2(s)

ϕ(`1`2)=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

Pn≤Xτ3(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 q≥1, we have

1

ϕ(q)X

n≤X

(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

3γ−7

6+7

3X

p|q

log p

p−1

,(4.3)

a3(q) = ϕ(q)2

q3

3γ2−3γ+ 3γ1+X

p|q

log p

p−1

4γ−3 + X

p|q

log p

p−1

.

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

`1≤X1/3X

`2≤X2/3

`1

a1(`1`2),(4.5)

b2(X) = X

`1≤X1/3X

`2≤X2/3

`1

a2(`1`2),(4.6)

b3(X) = X

`1≤X1/3X

`2≤X2/3

`1

a3(`1`2),

and

(4.7) E(X)=(X+ 1)2/3log(X+ 1) X

`1≤X1/3X

`2≤X2/3

`1

τ(`1`2)

ϕ(`1`2).

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

`1≤X1/3X

`2≤X2/3

`1

ϕ(`1`2)2

(`1`2)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

p1−4

p2+4

p3−1

p4log2(X) + c1log X+c2+OX−2

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|n1−1

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

`1≤X1/3

f(`1)

`1

Σ(X, `1),

where

(4.13) Σ(X, `1) = X

n≤X2/3

`1

g`1(n)

n.

14

By Euler products, we have

∞

X

n=1

g`1(n)

ns+1 =ζ(s+ 1)A(s)B`1(s),(σ > 0),

where

(4.14) A(s) = Y

p1−2

ps+2 +1

ps+3 ,(σ > −1),

(4.15) Bn(s) = Y

p|n1−1

p2

1−2

ps+2 +1

ps+3

,(σ > −1),

and A(s) and B`1(s) are convergent in the larger region σ > −1. Thus, by (4.13) and Perron’s

formula, we have

Σ(X, `1) = A(0)B`1(0) log X2/3

`1+ (AB`1)0(0) + γA(0)B`1(0) + O X2/3

`1−1!.

Hence, by (4.12) and the above, we have

(4.16) b1(X) = b11(X) log X+b12 (X) + b13(X) + OX−1

3+,

where

b11(X) = 2

3A(0) X

n≤X1/3

Bn(0)

n,(4.17)

b12(X) = −A(0) X

n≤X1/3

Bn(0)

nlog n,(4.18)

b13(X) = ((AB`1)0(0) + γA(0)B`1(0)) X

n≤X1/3

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

1−1

ps+1 +1

ps+1 1−1

p2

1−2

p2+1

p3

,(σ > −1).

By (4.14) and (4.21), we have

(4.22) A(0)B(0) = Y

p1−4

p2+4

p3−1

p4.

15

Thus, by (4.17), Perron’s formula, (4.20), and (4.22), we have

b11(X) = 2

9Y

p1−4

p2+4

p3−1

p4log X(4.23)

+2

3A(0) (B0(0) + γB(0)) + OX−1

3+.

Next, by (4.18), partial summation, and the above, we have

b12(X) = −1

18 Y

p1−4

p2+4

p3−1

p4log2X(4.24)

−1

2A(0) (B0(0) + γB(0)) log X+OX−1

3+.

Lastly, we have, from (4.19)

(4.25) b13(X) = ((AB`1)0(0) + γA(0)B`1(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

p−11.

Thus, by (4.6), (4.3), the above, and (4.5), we have

b2(X)X

`1≤X1/3X

`2≤X2/3

`1

ϕ(`1`2)2

(`1`2)3b1(X)log2X,

by (4.9). Similarly, we get

b3(X)log2X.

We now estimate E1(X). We have

X

`1≤X1/3X

`2≤X2/3

`1

τ(`1`2)

ϕ(`1`2)XX

`1≤X1/3

1

`1X

`2≤X2/3

`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

p1−4

p2+4

p3−1

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 1≤N < N 0<2X,N0−N > Xd, and (a, d) = 1.

Then for j, ν ≥1we have

(4.26) X

N≤n≤N0

n≡a(d)

τj(n)νN0−N

ϕ(d)(log X)jν−1.

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(`1`3)(`3X2/3+ 1) log2(`3X2/3+ 1).

Thus, by (3.4) and (4.27),

Σ21(X)X2/3log2XX

`1≤X1/3X

`3≤X1/3

`3

ϕ(`1`3)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

n≤X

τ`(n+h)X

`1`2···`k=n

`1`2···`k−1≤X(k−1)/k;`1≤X1/k

1

=kX

`1≤X1/k X

`2≤X(k−1)/k

`1X

`3≤X(k−1)/k

`1`2

· · · X

`k−1≤X(k−1)/k

`1···`k−2X

`k≤X

`1···`k−1

τ`(`1· · · `k+h).

Making a change of variables n=`1· · · `k+hin the most inner `ksum, the above becomes

kX

`1≤X1/k X

`2≤X(k−1)/k

`1X

`3≤X(k−1)/k

`1`2

· · · X

`k−1≤X(k−1)/k

`1···`k−2X

n≤X+h

n≡h(`1···`k−1)

τ`(n).

By the bound (1.3), the error term is negligible and the above is in turns asymptotic to

kX

`1≤X1/k X

`2≤X(k−1)/k

`1X

`3≤X(k−1)/k

`1`2

· · · X

`k−1≤X(k−1)/k

`1···`k−2

1

ϕ`1···`k−1

(h,`1···`k−1)X

n≤X+h

n, `1···`k−1

(h,`1···`k−1)=1

τ`(n).

Thus, by Perron’s formula, we obtain that

Dk,`(X, h)∼kRes

s=1

w1=···=wk−1=0 Xs+w1

k+k−1

k(w2+···wk−1)

sw1· · · wk−1

Tk,`(s, w1,· · · , wk−1;h)!,(5.1)

where

Tk,`(s, w1,· · · , wk−1;h) =

∞

X

`1,...,`k−1=1

1

(h, `1· · · `k−1)s

1

ϕ`1···`k−1

(h,`1···`k−1)

×X

n, `1···`k−1

(h,`1···`k−1)=1

τ`(n(h, `1· · · `k−1))

ns

k−1

Y

j=1

`−Pk−1

i=jwi

j.

By multiplicativity and Euler products, the above generating function Tk,` can be written as

(5.2) Tk,`(s, w1,· · · , wk−1;h) = Y

p|h

Ap(s;w1,· · · , wk−1;h)

Bp(s;w1,· · · , wk−1)Y

p

Bp(s;w1,· · · , wk−1),

18

where

Ap(s;w1,· · · , wk−1;h)(5.3)

=X

j1,··· ,jk−1

1

pmin(j1+···+jk−1,νp(h))s

1

ϕ(pj1+···+jk−1−min(j1+···+jk−1,νp(h)))

×X

(n,pj1+···+jk−1−min(j1+···+jk−1,νp(h))=1

τ`(npmin(j1+···+jk−1,νp(h))

ns

1

pPk−1

i=1 jiPk−1

κ=iwκ,

and

(5.4) Bp(s;w1,· · · , wk−1) = ζ`(s)

1 + 1−1

p`

1−1

p

k−1

X

j=1 X

σ∈Ξj,k−1

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,· · · , wk−1) = ζ`(s)ζ(w1+w2+· · · wk−1+ 1)ζ(w2+· · · wk−1+ 1) × · · ·(5.5)

×ζ(wk−1+ 1) Y

p

BBp(s;w1,· · · , wk−1),

where

(5.6)

BBp(s;w1,· · · , wk−1) =

k−1

Y

i=1 1−1

pwi+···+wk−1+1

1 + 1−1

p`

1−1

p

k−1

X

j=1 X

σ∈Ξj,k−1

j

Y

i=1

1

pwσ(i)+1 −1

.

The product QpBBp(s;w1,· · · , wk−1) 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,· · · , wk−1;h) = ζ`(s)AAp(s;w1,· · · , wk−1;h)(5.7)

with AAp(s;w1,· · · , wk−1;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,· · · , wk−1;h) (5.2) can thus be written as

Tk,`(s, w1,· · · , wk−1;h) = ζ`(s)ζ(w1+w2+· · · wk−1+ 1)ζ(w2+· · · wk−1+ 1) × · · ·(5.8)

×ζ(wk−1+ 1) Y

p|h

Ap(s;w1,· · · , wk−1;h)

Bp(s;w1,· · · , wk−1)Y

p

BBp(s;w1,· · · , wk−1),

and, hence, we conclude that Tk,`(s, w1,· · · , wk−1;h) has poles at s= 1 and w1=· · · =

wk−1= 0. Therefore, by (5.8) above, we obtain from (5.1) that

(5.9) Dk,`(X, h)∼Ck,`fk,`(h)

(k−1)!(`−1)!X(log X)k+`−2,

19

where

(5.10) Ck,` =Y

p

BBp(0; ~

0) = Y

p1−1

pk−1 1 + 1−1

p`−1k−1

X

j=1 k−1

j

(p−1)j!

and

(5.11) fk,`(h) = Y

p|h

Ap(1;~

0; h)

Bp(1;~

0) ,

where we have abbreviated ~

0 for 0,· · · ,0k−1 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 1−1

pk−1

+1−1

p`−1

−1−1

pk+`−2!.

Proof. We have the identity

k−1

X

j=1 k−1

j

(p−1)j=−1 + p

p−1k

−1

pp

p−1k

.

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

n≤X

τ3(n)τ`(n+h)∼3X

`1≤X1/3X

`2≤X2/3

`1X

n≤X+h

n≡h(`1`2)

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, `1`2)s

1

ϕ(`1`2/(h, `1`2))

×X

n, `1`2

(h,`1`2)=1

τ`(n(h, `1`2))

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+j2,νp(h))s

1

ϕ(pj1+j2−min(j1+j2,νp(h)))

×X

(n,pj1+j2−min(j1+j2,νp(h))=1

τ`(npmin(j1+j2,νp(h))

ns

1

pj1(w1+w2)+j2w2.

Thus, (5.13) predicts that

X

n≤X

τ3(n)τ`(n+h)∼1

4Y

p1−4

p2+4

p3−1

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+ 3h−4

h(h2+ 2h−1) .

In particular, assuming the bound (1.3)for k= 3, we have, for hprime,

D3,3(X, h)∼1

4Y

p1−4

p2+4

p3−1

p4h3+ 6h2+ 3h−4

h(h2+ 2h−1) Xlog4X.(5.16)

Proof. By Perron’s formula and (1.3), we have

X

`1≤X1/3X

`2≤X2/3

`1X

n≤X+h

n≡h(`1`2)

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(nδ)

ns

1

`w1+w2

1

1

`w2

2

with

δ= (h, `1`2)

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+j2−1)

1

hsX

(n,pj1+j2−1)=1

τ3(nh)

ns

1

pj1(w1+w2)+j2w2.

We now evaluate the functions Aand B. We start with B.

We split the jisums in (5.18) into

X

j1,j2

j1j26=0

=X

j1≥1

j2=0

+X

j1=0

j2≥1

+X

j1≥1

j2≥1

.

We have

X

j1≥1

j2=0

1

ϕ(pj1+j2)X

(n,pj1+j2)=1

τ3(n)

ns

1

pj1(w1+w2)+j2w2

=

∞

X

j1=1

1

pj11−1

pX

(n,p)=1

τ3(n)

ns

1

pj1(w1+w2)

=ζ3(s)1−1

ps3

1−1

p

∞

X

j1=1

1

pj1(w1+w2+1)

=ζ3(s)1−1

ps3

1−1

p

1

pw1+w2+1 −1,

X

j1=0

j2≥1

1

ϕ(pj1+j2)X

(n,pj1+j2)=1

τ3(n)

ns

1

pj1(w1+w2)+j2w2

=ζ3(s)1−1

ps3

1−1

p

1

pw2+1 −1,

22

and

X

j1≥1

j2≥1

1

ϕ(pj1+j2)X

(n,pj1+j2)=1

τ3(n)

ns

1

pj1(w1+w2)+j2w2

=ζ3(s)1−1

ps3

1−1

p

1

pw1+w2+1 −1

1

pw2+1 −1.

Thus,

Bp(s;w1, w2) = ζ3(s)

1 + 1−1

ps3

1−1

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

p1−1

pw1+w2+1 1−1

pw2+1

×

1 + 1−1

ps3

1−1

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

p1−1

p2 1 + 1−1

p21

p−1+1

p−1+1

(p−1)2!

(5.21)

=Y

p1−4

p2+4

p3−1

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+j2−1 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

j1≥2

j2=0

+X

j1=0

j2≥2

+X

j1≥1

j2≥1

.

We have

X

j1=1

j2=0

1

ϕ(pj1+j2−1)

1

hsX

(n,pj1+j2−1)=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+j2−1)

1

hsX

(n,pj1+j2−1)=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

j1≥2

j2=0

1

ϕ(pj1+j2−1)

1

hsX

(n,pj1+j2−1)=1

τ3(nh)

ns

1

pj1(w1+w2)+j2w2

=

∞

X

j1=2

1

ϕ(pj1−1)

1

hsX

(n,h)=1

τ3(h)τ3(n)

ns

1

pj1(w1+w2)

=1

hsτ3(h)ζ3(s)Y

p|h1−1

ps3∞

X

j1=2

1

pj1−11−1

p

1

pj1(w1+w2)

=3

ps−1ζ3(s)1−1

ps3

1−1

p

∞

X

j1=2

1

pj1(w1+w2+1)

=3

ps−1ζ3(s)1−1

ps3

1−1

p

1

pw1+w2+1

1

pw1+w2+1 −1,

24

X

j1=0

j2≥2

1

ϕ(pj1+j2−1)

1

hsX

(n,pj1+j2−1)=1

τ3(nh)

ns

1

pj1(w1+w2)+j2w2

=

∞

X

j2=2

1

ϕ(pj2−1)

1

hsX

(n,h)=1

τ3(h)τ3(n)

ns

1

pj2w2

=3

ps−1ζ3(s)1−1

ps3

1−1

p

1

pw2+1

1

pw2+1 −1,

and

X

j1≥1

j2≥1

1

ϕ(pj1+j2−1)

1

hsX

(n,pj1+j2−1)=1

τ3(nh)

ns

1

pj1(w1+w2)+j2w2

=X

j1≥1

j2≥1

1

ϕ(pj1+j2−1)

1

hsX

(n,h)=1

τ3(h)τ3(n)

ns

1

pj1(w1+w2)+j2w2

=3

ps−1ζ3(s)1−1

ps3

1−1

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

ps−11−1

ps3

1−1

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

ps−1

(1−1

ps)3

1−1

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

1 + (1−1

ps)3

1−1

p1

pw1+w2+1−1+1

pw2+1−1+1

pw1+w2+1−1

1

pw2+1−1.

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+ 3p−4

p(p2+ 2p−1) .

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|h−νp(h)2(p−1)2(p+ 1) + pνp(h)+2 + 4pνp(h)+3

(5.23)

+pνp(h)+4 +νp(h)−4p3+ 6p−2−4p3−5p2+ 4p−1

/pνp(h)(p−1)2p2+ 2p−1,

f3,4(h) = Y

p|h

p−νp(h)3(p+ 1)(p−1)3−νp(h)27p2+ 6p−4(p−1)2

(5.24)

+νp(h)−16p4+ 33p2−22p+ 5+ 2 −pνp(h)+2 + 5pνp(h)+3

+5pνp(h)+4 +pνp(h)+5 −6p4−9p3+ 9p2−5p+ 1

/2pνp(h)(p−1)3p3+ 2p2−3p+ 1,

and

f3,5(h) = Y

p|h−νp(h)4(p+ 1)(p−1)4−νp(h)311p2+ 8p−7(p−1)3

(5.25)

−νp(h)244p3+ 31p2−50p+ 17(p−1)2−νp(h)76p5+p4

−200p3+ 200p2−94p+ 17+ 6 6pνp<