the k-fold divisor function in arithmetic progression to

The k-fold Divisor Function inArithmetic Progression to Large Moduli

David T. Nguyen

UC Santa Barbara

West Coast Number Theory ConferenceChico, CA, December 15-19, 2018

D. T. Nguyen (UCSB) Distribution of Divisor Functions WCNT 2018 (Chico, CA) 1 / 9

Background–Where does the sum come from?Gauss (∼1792) conjectured the Prime Number Theorem:

Gauss (1792):∑p≤X

1 ∼X

∑p≤X

1→∑p≤X

log p→∑pα≤X

log p =∑n≤X

Λ(n)→∑n≤X

n≡a(mod d)

Λ(n)→∑n≤X

n≡a(mod d)

For us, we consider the k-fold divisor function

f(n) = τk(n) =∑

d1d2···dk=ndi>0

1, k ≥ 1.

Thus, τk(n) is the coefficient of n−s in the Dirichlet series

ζ(s)k =

∞∑n=1

τk(n)n−s.

Ultimate goal: For each k ≥ 1, d ≥ 1, (a, d) = 1, estimate the sum∑n≤X

n≡a(mod d)

τk(n)

for d as large as possible with an error as small as possible.

1 ∼X

∑p≤X

1→∑p≤X

log p→∑pα≤X

log p =∑n≤X

Λ(n)→∑n≤X

n≡a(mod d)

Λ(n)→∑n≤X

n≡a(mod d)

f(n) = τk(n) =∑

d1d2···dk=ndi>0

1, k ≥ 1.

ζ(s)k =

∞∑n=1

τk(n)n−s.

n≡a(mod d)

τk(n)

1 ∼X

∑p≤X

1→∑p≤X

log p→∑pα≤X

log p =∑n≤X

Λ(n)→∑n≤X

n≡a(mod d)

Λ(n)→∑n≤X

n≡a(mod d)

f(n) = τk(n) =∑

d1d2···dk=ndi>0

1, k ≥ 1.

ζ(s)k =

∞∑n=1

τk(n)n−s.

n≡a(mod d)

τk(n)

Why τk(n)?

It turns out that τk(n) is closely related to primes, e.g., Linnik identity (1963),Vaughan identity (1977), Heath-Brown identity (1982), etc... Heuristically,

ζ(s)=

1− (1− ζ(s))=∞∑n=1

(1− ζ(s))n =

∞∑n=1

n∑k=0

)(−1)kζ(s)k,

so a hard problem with primes on the left side is converted into infinitely many, thoughmore accessible, problems about τk on the right side.

Concretely,

Friedlander & Iwaniec 1 (1985)τ3(n)

↓ GPY sieve 2 ↓

Y. Zhang 3 (2014)Bounded gaps between primes!

Work on large prime gaps, e.g., K. Ford, S. Konyagin, J. Maynard, C.Pomerance, and T. Tao (2018).

1J. B. Friedlander and H. Iwaniec, Ann. of Math. (2) 121 (1985), no. 2, 319-350.2D. Goldston, J. Pintz, and C. Yıldırım, Ann. of Math. 170 (2009), no. 2, 819-862.3Y. Zhang, Ann. of Math. 179 (2014), no. 3, 1121-1174.

Why τk(n)?

ζ(s)=

1− (1− ζ(s))=∞∑n=1

(1− ζ(s))n =

∞∑n=1

n∑k=0

)(−1)kζ(s)k,

Concretely,

↓ GPY sieve 2 ↓

Why τk(n)?

ζ(s)=

1− (1− ζ(s))=∞∑n=1

(1− ζ(s))n =

∞∑n=1

n∑k=0

)(−1)kζ(s)k,

Concretely,

↓ GPY sieve 2 ↓

What is known about distribution of τk?

Only reasonably well understood for k = 1, 2, 3!

Simplest case k = 1: ∑n≤X

n≡a(mod d)

d+O(1).

This tells us, in particular, that d ≤ X.

More precisely, for (a, d) = 1, let

∆(τk;X, d, a) =∑n≤X

n≡a(mod d)

τk(n)−1

∑n≤X

(n,d)=1

τk(n).

Then, for each k, we seek θk > 0 as large as possible such that, for all ε > 0, thereis δ > 0, such that

∆(τk;X, d, a)�X1−δ

for all d ≤ Xθk−ε. The number θk is called the level of distribution for τk.

Conjecture (c.f. Elliott-Halberstam): θk = 1 for all k. GRH =⇒ θk < 1/2 for allk. Moduli d > X1/2 are called large moduli. This is sometimes refers to as the“square-root barrier”.

n≡a(mod d)

d+O(1).

∆(τk;X, d, a) =∑n≤X

n≡a(mod d)

τk(n)−1

∑n≤X

(n,d)=1

τk(n).

∆(τk;X, d, a)�X1−δ

n≡a(mod d)

d+O(1).

∆(τk;X, d, a) =∑n≤X

n≡a(mod d)

τk(n)−1

∑n≤X

(n,d)=1

τk(n).

∆(τk;X, d, a)�X1−δ

n≡a(mod d)

d+O(1).

∆(τk;X, d, a) =∑n≤X

n≡a(mod d)

τk(n)−1

∑n≤X

(n,d)=1

τk(n).

∆(τk;X, d, a)�X1−δ

Recall that, for each k, we seek θk > 0 as large as possible such that, for each ε > 0,there is δ > 0, such that

∆(τk;X, d, a)�X1−δ

ϕ(d)(1)

for all d ≤ Xθk−ε.

Known results for individual distribution estimate (1) for τk. Only for k = 1, 2, 3 is theexponent of distribution θk for τk known to hold for a value larger than 1/2.

k θk References

k = 1 θ1 = 1 See previous slide.k = 2 θ2 = 2/3 Selberg, Linnik, Hooley (independent, unpublished, 1950’s).k = 3 θ3 = 1/2 + 1/230 Friedlander and Iwaniec (1985).

θ3 = 1/2 + 1/82 Heath-Brown (1986).k = 4 θ4 < 1/2 Linnik (1961).k ≥ 4 θk = 8/(3k + 4) Lavrik (1965).k = 5 θ5 = 9/20 Friedlander and Iwaniec (1985).k = 6 θ6 = 5/12 Friedlander and Iwaniec (1985).k ≥ 7 θk = 8/3k Friedlander and Iwaniec (1985).k ≥ 4 θk ≥ 1/2 Open

This stubborn problem (showing that θk > 1/2) has stood static for any k ≥ 4 ever sincethe 80’s. (Also, the value θ2 = 2/3 for τ(n) has not been improved since the 50’s.) Tomake progress, various averages on ∆(τk;X, d, a) are considered by many authors.

∆(τk;X, d, a)�X1−δ

ϕ(d)(1)

k θk References

∆(τk;X, d, a)�X1−δ

ϕ(d)(1)

k θk References

Y. Zhang (2014): restrict to smooth moduli; in his remarkable work on boundedgaps between primes, a crucial step is to prove, for any a 6= 0,∑

d<X1/2+2$

(d,a)=1d is X$-smooth

µ(d)2|∆(Λ;X, d, a)| �X

(logX)A.

Zhang’s method originally for Λ applies equally to τk:

Theorem 0 (F. Wei, B. Xue, and Y. Zhang a (2016))

aF. Wei, B. Xue, and Y. Zhang, Sci. China Math. 59 (2016), no. 9, 1663-1668.

For a 6= 0, denote

D = {d ≥ 1 : (d, a) = 1, µ(d)2, (d,∏

p<X1/1168

p) > X71/584}.

Then, for any k ≥ 4, we have∑d<X293/584

|∆(τk;X, d, a)| �k,a X exp(− log1/2 X).

We provide a sharpening of this error term.

d<X1/2+2$

µ(d)2|∆(Λ;X, d, a)| �X

(logX)A.

For a 6= 0, denote

D = {d ≥ 1 : (d, a) = 1, µ(d)2, (d,∏

p<X1/1168

p) > X71/584}.

d<X1/2+2$

µ(d)2|∆(Λ;X, d, a)| �X

(logX)A.

For a 6= 0, denote

D = {d ≥ 1 : (d, a) = 1, µ(d)2, (d,∏

p<X1/1168

p) > X71/584}.

Our results

Theorem 1 (Main result)

Let a 6= 0. Denote P(y) =∏p<y p, $ = 1/1168, and

D = {d ≥ 1 : (d, a) = 1, µ(d)2, (d,P(X$2)) < X$, and (d,P(X$)) > X1/8−4$},

Then, for all k ≥ 4, we have ∑d∈D

d<X1/2+1/584

|∆(τk;X, d, a)| � X1−θk , (2)

where θk = min{1/12(k + 2), $2}. The implied constant is effective and depends on a and k.

If we assume the Generalized Lindelof Hypothesis, we can prove a stronger result.

Theorem 2

On the Generalized Lindelof Hypothesis, the estimate (2) holds with the right side replaced by

X1−1/20182,

where the θk power saving is replaced by a positive constant independent of k.

Our results

Theorem 1 (Main result)

Let a 6= 0. Denote P(y) =∏p<y p, $ = 1/1168, and

D = {d ≥ 1 : (d, a) = 1, µ(d)2, (d,P(X$2)) < X$, and (d,P(X$)) > X1/8−4$},

Then, for all k ≥ 4, we have ∑d∈D

d<X1/2+1/584

|∆(τk;X, d, a)| � X1−θk , (2)

where θk = min{1/12(k + 2), $2}. The implied constant is effective and depends on a and k.

If we assume the Generalized Lindelof Hypothesis, we can prove a stronger result.

Theorem 2

On the Generalized Lindelof Hypothesis, the estimate (2) holds with the right side replaced by

X1−1/20182,

where the θk power saving is replaced by a positive constant independent of k.

If, in addition with averaging over moduli d, we also average over primitive residueclasses in each modulus, then we can further extend the range of d.

Theorem 3

For k ≥ 4 we have

∑d≤D

d∑a=1

(a,d)=1

∆(τk;X, d, a)2 �{X2−1/6(k+4), for 1 ≤ D ≤ X1−1/6(k+2),

DX(logX)k2−1, for X1−1/6(k+2) < D ≤ X.

In our last result, we prove a distribution estimate involving Λ where the moduli d canbe as large as X2.

Theorem 4

For k ≥ 4 there holds

∑d≤D

d∑a=1

(a,d)=1

∑m,n≤X

m≡an(mod d)

τk(m)Λ(n)−X

∑n≤X

(n,d)=1

τk(n)

�{X4−1/3(k+4), for 1 ≤ D ≤ X2−1/3(k+2),

DX2(logX)2k−2, for X2−1/3(k+2) < D ≤ X2.

If, in addition with averaging over moduli d, we also average over primitive residueclasses in each modulus, then we can further extend the range of d.

Theorem 3

For k ≥ 4 we have

∑d≤D

d∑a=1

(a,d)=1

∆(τk;X, d, a)2 �{X2−1/6(k+4), for 1 ≤ D ≤ X1−1/6(k+2),

DX(logX)k2−1, for X1−1/6(k+2) < D ≤ X.

In our last result, we prove a distribution estimate involving Λ where the moduli d canbe as large as X2.

Theorem 4

For k ≥ 4 there holds

∑d≤D

d∑a=1

(a,d)=1

∑m,n≤X

m≡an(mod d)

τk(m)Λ(n)−X

∑n≤X

(n,d)=1

τk(n)

�{X4−1/3(k+4), for 1 ≤ D ≤ X2−1/3(k+2),

DX2(logX)2k−2, for X2−1/3(k+2) < D ≤ X2.

Thank You!

Figure 1: Chico State Motto, “Today decides tomorrow”.

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