LOWER BOUNDS FOR NUMBERS WITH THREE PRIME Paul …math.colgate.edu/~integers/t22/t22.pdf#A22 INTEGERS 19 (2019) LOWER BOUNDS FOR NUMBERS WITH THREE PRIME FACTORS Paul Kinlaw Department

#A22 INTEGERS 19 (2019)

LOWER BOUNDS FOR NUMBERS WITH THREE PRIMEFACTORS

Paul KinlawDepartment of Mathematics, Husson University, Bangor, Maine

[email protected]

Received: 5/2/18, Revised: 11/27/18, Accepted: 2/21/19, Published: 3/15/19

AbstractA k-almost prime number is a product of k prime numbers, some of which may berepeated. By a 1900 theorem of Landau, the number of k-almost prime numbers notexceeding x is asymptotic to x(log log x)k�1/((k�1)! log x). We prove a numericallyexplicit lower bound for 3-almost prime numbers which is asymptotic to Landau’sformula, and hence to the actual count. It exceeds Landau’s formula for all x �500194. We prove an analogous lower bound for products of three distinct primenumbers. This expands on previously known results for k 2.

1. Introduction

For a natural number n, the arithmetic functions !(n) and ⌦(n) denote the numberof prime factors of n, counted without (respectively with) repeated prime factors.Thus for n with prime factorization n = pa1

1 · . . . · pakk , we have !(n) = k and

⌦(n) = a1 + . . . + ak.Numbers n such that ⌦(n) = k are called k-almost primes. Let ⇡k(x) = |{n

x : !(n) = ⌦(n) = k}| denote the counting function of squarefree k-almost primes,and let ⌧k(x) = |{n x : ⌦(n) = k}| denote the counting function of k-almostprimes.

For k = 1 we have ⇡1(x) = ⌧1(x) = ⇡(x), the prime counting function. Theprime number theorem asserts that ⇡(x) is asymptotic to x/ log x. In 1900, Landau[6] proved that for each k 2 N, the estimate

⇡k(x) =x(log log x)k�1

(k � 1)! log x

✓1 + O

✓1

log log x

◆◆

holds, and that the same estimate also holds for ⌧k(x).Selberg [11] proved that for any 0 < � < 1, uniformly for all x � 3 and 1 k

(2� �) log log x, we have

⌧k(x) = G

✓k � 1

log log x

◆x(log log x)k�1

(k � 1)! log x

✓1 + O

✓k

(log log x)2

◆◆,

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where

G(z) = F (1, z)/�(z + 1) and F (s, z) =Y

p

✓1� z

ps

◆�1✓1� 1

ps

◆z

.

Classic references [5, 7] provide details on these and similar results. A 1962 paper[10] of Rosser and Schoenfeld and contemporary work [3, 4] of Dusart give explicitbounds for ⇡(x), see Lemma 1. A 2018 paper [1] of Bayless et al. established explicitupper bounds for ⇡k(x), ⌧2(x), and ⌧3(x), as well as explicit lower bounds for ⇡2(x)and ⌧2(x). In particular, it is shown [1, Thm. 3.5] that for all k � 2 and x � 3, wehave

⇡k(x) 1.028x(log log x + 0.26153)k�1

(k � 1)! log x,

and [1, Thm. 5.2] that for all x � 1012, we have

⌧2(x) � ⇡2(x) � x(log log x + 0.1769)log x

✓1 +

0.4232log x

◆.

Furthermore [1, Thm. 5.3], for all x � 1012 we have

⌧3(x) 1.028x((log log x + 0.26153)2 + 1.055852)2 log x

.

A relatively sharp explicit lower bound for ⇡3(x) requires more work, and asidefrom the explicit results above, the literature consists of implied constants. Weprove explicit lower bounds for ⇡3(x) and ⌧3(x) which are asymptotic to Landau’sformula, and hence to the actual count. In particular, we prove the following threetheorems.

Theorem 1. For all x � 500194,

⌧3(x) >x(log log x)2

2 log x.

The constant 500194 is optimal, since the inequality is violated at x = 500194�✏for all su�ciently small ✏ > 0. We also have an analogue of Theorem 1 for squarefree3-almost primes, also called sphenic numbers.

Theorem 2. For all x � 10203553,

⇡3(x) >x((log log x)2 � 1)

2 log x.

The constant 10203553 is also optimal. We also obtain the following.

Theorem 3. For all su�ciently large x,

⇡3(x) >x(log log x)2

2 log x.

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Theorem 3 follows readily from the proof of Theorem 2 by comparing secondaryterms, however the reader will see that x must be very large. Without more work,it is unclear where the optimal cuto↵ c0 is for Theorem 3 to apply. Lifchitz andRenner [8] computed the values of ⇡3(10k) for 1 k 19, and this data shows thatc0 > 1019.

2. Notation and Preliminary Lemmas

Throughout the paper we let p, q, and r denote prime numbers and we let log xdenote the natural logarithm. We define L = log log x, x0 = 1012, and y0 = x1/3

0 =104. Furthermore, ⇡(t) denotes the prime counting function and T (t) =

Ppt 1/p

denotes the sum of reciprocals of prime numbers p t. We let B = 0.2614972128 . . .denote the Mertens constant and c = 0.26146521 (see Lemmas 2 and 3).

In bounding expressions, we will make use of manipulations such as

log logx

a= log log x + log

✓1� log a

log x

◆,

the Maclaurin series expansion of the righthand term, and the bounds

1log x

✓1 +

log a

log x

◆<

1log x

a

1log x

+1

log2 x· log a log x0

log x0a

for a > 1 and x � x0. We will use the following bounds [10, Thm. 2] of Rosser andSchoenfeld and [4, Cor. 5.2] of Dusart on the prime counting function.

Lemma 1 (Rosser, Schoenfeld, Dusart). We have

x

log x< ⇡(x)

for all x � 17. Additionally, we have

x

log x

✓1 +

1log x

◆ ⇡(x) x

log x

✓1 +

1.2762log x

◆,

where the lower bound holds for all x � 599 and the upper bound holds for all x > 1.Furthermore, for all x > 1 we have

⇡(x) x

log x

✓1 +

1log x

+2.53816log2 x

◆.

We also use the following bounds [10, Theorem 5] of Rosser and Schoenfeld and[3, Thm. 6.10], [4, Thm. 5.6] of Dusart which give numerically explicit versions ofMertens’ second theorem.

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Lemma 2 (Rosser, Schoenfeld, Dusart). Let T (t) =P

pt 1/p denote thereciprocal sum of prime numbers up to t. We have

T (t) = log log t + B + E(t),

where B = 0.2614972128 . . . denotes the Mertens constant, and

� 12 log2 t

< E(t) <1

log2 t

for all t > 1,

|E(t)| <1

2 log2 t

for all t � 286,

|E(t)| 110 log2 t

+4

15 log3 t

for all t � 10372, and

|E(t)| 15 log3 t

for all t � 2278383.

Combining bounds on the prime number reciprocal sum in [10, 4] readily yieldsthe following lower bound.

Lemma 3. For x > 1,

X

px

1p

> log log x + 0.26146521.

Proof. By Lemma 2, for all x � 2278383,

X

px

1p� log log x + B � 0.2

log3 x, (1)

where B is the Mertens constant. By [10, Thm 20],

X

px

1p

> log log x + B

for all x 108. Substituting 108 in (1) gives the lemma.

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3. The Proof of Theorem 2

We have⇡3(x) = |{n = pqr x : p < q < r}|

where p, q, and r denote prime numbers. To determine the count, we note thepossible values for each of p, q, r. We have p < x1/3, and q2 < qr x/p, so thatp < q <

px/p. Since pqr x, we also have q < r x/(pq). Therefore, ⇡3(x) is

equal to the sum of 1 over all possible values of p, q, r, so that

⇡3(x) =X

p<x1/3

X

p<q<p

xp

X

q<r xpq

1 =X

px1/3

X

p<qp

xp

✓⇡

✓x

pq

◆� ⇡(q)

◆.

Here we have used the fact that the strict inequalities p < x1/3 and q <p

x/p can bewritten to include the case of equality, since the quantity vanishes when q =

px/p

or when p = x1/3. We checked the inequality using the computer program Pari/GPfor the interval 10203553 x 1012. Thus we may assume x � x0. By Lemma 1,since x/(pq) � 599 we have

⇡

✓x

pq

◆� x

pq log xpq

1 +1

log xpq

!

� x

pq log xpq

✓1 +

1log x

◆.

We thus have⇡3(x) �

✓1 +

1log x

◆ X

px1/3

S1 �X

px1/3

S2,

whereS1 =

X

p<qp

xp

x

pq log xpq

andS2 =

X

p<qp

xp

⇡(q).

We determine a lower bound for S1 and an upper bound for S2. By Lemma 7 below,we have X

px1/3

S2 x

✓2L + 0.1436

log2 x+

10.9113L + 3.1227log3 x

◆.

We next consider S1. We have

S1 =x

p

X

p<qpy

1q log y

q

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where y = x/p > x2/3 � x2/30 . Recall that the function T (t) =

Ppt 1/p denotes the

prime reciprocal sum up to t. We bound S1 below by applying partial summationto obtain

X

p<qpy

1q log y

q

=T (py)logpy

� T (p)log y

p

�Z p

y

p

T (t) dt

t log2 yt

. (2)

By Lemma 2 this is bounded below by the expression

2(log logpy + B � 12 log2py )

log y�

log log p + B + 1log2 p

log yp

�Z p

y

p

log log t + B + 1log2 t

t log2 yt

dt.

Substituting u = log t and integrating, this is equal to

�4log3 y

� log log p

log yp

� 1log2 p log y

p

+log log y

p

log y+

log p log log p

log y log yp

+1

log2 y log yp

� 1log2 y log p

�2 log log y

p

log3 y+

2 log log p

log3 y.

Thus a lower bound for S1 is

x

p log xp

log logx

p2+

log p log log p

log xp2

+1

log xp log x

p2

+2 log log p

log2 xp

!

� x

p log xp

4

log2 xp

+ log log p ·log x

p

log xp2

+1

log2 p·

log xp

log xp2

+1

log xp log p

+2 log log x

p2

log2 xp

!

.

(3)To simplify this expression, we compare the second (respectively, third) term of thefirst line above to the second (respectively, first) term of the second line. We have

log p log log p

log xp2

� log log p ·log x

p

log xp2

= � log log p.

Also, 1/(log(x/p) log(x/p2)) > 1/ log2(x/p). Therefore, a lower bound for expres-sion (3) is

x

p log xp

log logx

p2+

2 log log p

log2 xp

!

� x

p log xp

log log p +3

log2 xp

+1

log2 p·

log xp

log xp2

+1

log xp log p

+2 log log x

p2

log2 xp

!

.

(4)

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Write the sum of this expression over p x1/3 as

x (S5 + S6 � (S7 + S8 + S9 + S10 + S11)) .

Since 2 log log p� 3 is negative for p < 89 and positive for p � 89,

S6 � S8 =X

px1/3

2 log log p� 3p log3 x

p

� 1log3 x

X

p<89

2 log log p� 3p

log3 x0

log3 x0p

+1

log3 x

X

89py0

2 log log p� 3p

� �3.9322log3 x

.

Note that

S10 =X

px1/3

1p log p log2 x

p

1log2 x

X

py0

1p log p

log2 x0

log2 x0p

+2.25

log2 x

X

y0<p

1p log p

1.7496log2 x

+2.25(0.1085)

log2 x

1.9938log2 x

.

Here we verified the valueP

p 1/(p log p) = 1.63661 . . . found to much higher preci-sion [2, p. 6] by H. Cohen. Similarly,

S9 =X

px1/3

1p log2 p log x

p2

X

py0

1p log2 p

1

log x+

1log2 x

2 log p log x0

log x0p2

!

+3

log x

X

y0<px1/3

1p log2 p

1.5151log x

+3.5585log2 x

+3

log x

X

y0<p

1p log2 p

1.5328log x

+3.5585log2 x

.

Here we computed the sum over y0 < p 109 using Pari/GP and then appliedpartial summation together with Lemmas 2 and 3 to bound the sum over p > 109.By Lemma 8,

S5 � S7 �0.5L2 + 0.121434L� 0.22

log x

+�1.5244L + 1.112

log2 x+�2L2 + 2.9067L + 2.5389

log3 x,

and by Lemma 5,

S11 2L2 + 1.3972L

log3 x.

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Finally, note that the application of the bounds

L + c < T (x) < L + B +1

log2 x,

valid for all x > 1, incur error terms which are particularly large for small values ofx. We may improve our bounds as follows. In equation (2) we replaced the quantity�T (p)/(p log(x/p2)) with

�log log p + B + 1

log2 p

p log xp2

.

Thus we may add the following expression to the lower bound:

X

px1/3

log log p + B + 1log2 p � T (p)

p log xp2

.

A computation gives a lower bound of

1log x

4

X

py0

log log p + B + 1log2 p � T (p)

p� 0.9

log x4

� 0.9log x

✓1 +

log 4log x

+log2 4log2 x

◆.

Again considering equation (2) and the following displayed expression, we may addto the lower bound

X

px1/3

Z pxp

p

log log t + B + 1log2 t � T (t)

t log2 xpt

dt

� 1log2 x

X

k1228

k

Z log pk+1

log pk

✓log u + B +

1u2� T (pk)

◆du

� 20.4395log2 x

.

Here pk denotes the k-th prime number. Combining all bounds, we obtain

⇡3(x) � x(0.5L2 + 0.121434L� 0.8528)log x

+x(0.5L2 � 4L + 15)

log2 x

� x(4L2 + 10.9262L)log3 x

� x(0.5L2 + 0.121434L� 0.8528)log x

+x(0.5L2 � 4L + 12)

log2 x

� x(0.5L2 + 0.121434L� 0.8528)log x

+4x

log2 x

>x((log log x)2 � 1)

2 log x+

4xlog2 x

(5)

for all x � x0. This completes the proof of Theorem 2.

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4. Additional Lemmas

We prove several lemmas used in the proof of Theorem 2.

Lemma 4. For x � x0, we have

X

px1/3

1p log2 x

p

L + 0.071781log2 x

.

Proof. For x0 x 103723, we have

X

px1/3

1p log2 x

p

1log2 x

X

p10372

log2 x0

p log2 x0p

log log x0 � 0.0279log2 x

L

log2 x,

by directly computing the sum over p 10372. Suppose next that x > 103723. ByLemmas 2 and 3 and partial summation,

X

px1/3

1p log2 x

p

=T (x1/3)

log2(x2/3)�Z x1/3

2

2T (t) dt

t log3 xt

94 (log log x1/3 + 0.26300402)

log2 x�Z x1/3

2

2(log log t + c) dt

t log3 xt

,

recalling that c = 0.26146521. The integral is equal to

"log log t + c

log2 xt

+log log x

t � log log t

log2 x� 1

log x log xt

#x1/3

2

.

Subtracting and bounding the resulting expression, we obtain

S3 94 (0.26300402� c)� log 2 + 1

2 + c + log log x2

log2 x L + 0.071781

log2 x

for all x � 103723, and therefore for all x � x0.

Lemma 5. For x � x0, we have

X

px1/3

1p log3 x

p

L + 0.6986log3 x

.

Proof. Note that for x0 x 103723, we have

X

px1/3

1p log3 x

p

X

p10372

1p log3 x

log3 x0

log3 x0p

L + 0.5411log3 x

,

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by directly computing the sum over p 10372. Suppose next that x � 103723. Bypartial summation,

X

px1/3

1p log3 x

p

=T (x1/3)log3 x2/3

�Z x1/3

2

3T (t) dt

t log4 xt

log log x1/3 + 0.26300402log3 x2/3

�Z x1/3

2

3(log log t + c) dt

t log4 xt

,

using Lemmas 2 and 3. The integral is equal to

"log log t + c

log3 xt

+log log x

t � log log t

log3 x� 1

log2 x log xt

� 12 log x log2 x

t

#x1/3

2

.

Subtracting and bounding the resulting expression, we find that

X

px1/3

1p log3 x

p

278 (0.26300402� c)� log 2 + 9

8 + log log x2 + c

log3 x,

from which we obtain the lemma.

Lemma 6. For x � x0 we have

X

px1/3

1p

p log xp

0.4383x1/6

log x.

Proof. Let x � x0. By partial summation,

X

px1/3

1p

p log xp

1.5log x

X

px1/3

1p

p

=1.5

log x

⇡(x1/3)

x1/6+

12

Z x1/3

2

⇡(t) dt

t3/2

!

1.5log x

⇡(x1/3)

x1/6+ 16.85461 +

12

Z x1/3

y0

⇡(t) dt

t3/2

!

.

By Lemma 1, an upper bound for this expression is therefore

1.5log x

3.4154822x1/6

log x+ 16.85461 + 0.06180521

Z x1/3

y0

dt

t1/2

!

,

from which the lemma readily follows.

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Lemma 7. For x � x0 we haveX

px1/3

S2 x

✓2L + 0.1436

log2 x+

10.9113L + 3.1227log3 x

◆.

Proof. We have the general formulaX

qt

⇡(q) = 1 + 2 + . . . + ⇡(t) =12(⇡(t)2 + ⇡(t)).

Therefore,

S2 =12

⇡

✓rx

p

◆2

+ ⇡

✓rx

p

◆� ⇡(p)2 � ⇡(p)

!

.

We will address the third term below and drop the fourth term. Let x � x0. ByLemma 1, we have

12⇡

✓rx

p

◆ 1.1385

px

pp log x

p

and12⇡

✓rx

p

◆2

2xp log2 x

p

1 +2.551155

log xp

!2

2xp log2 x

p

+10.9113xp log3 x

p

.

We thus haveX

px1/3

S2 xX

px1/3

2

p log2 xp

+10.9113p log3 x

p

+1.1385p

x

1p

p log xp

!

.

Combining Lemmas 4, 5, and 6, we obtainX

px1/3

S2 x

✓2L + 0.1436

log2 x+

10.9113L + 7.6227log3 x

+1.1

log4 x

◆.

We now return to consider the remaining term:

12

X

px1/3

⇡(p)2 =12

· ⇡(x1/3)(⇡(x1/3) + 1)(2⇡(x1/3) + 1)6

� ⇡(x1/3)3

6

� 16

✓x1/3

log x1/3

✓1 +

1log x1/3

◆◆3

� 4.5xlog3 x

✓1 +

9log x

◆

=4.5xlog3 x

+40.5xlog4 x

.

Here we have used Lemma 1. Combining these bounds, we obtain the lemma.

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Lemma 8. For all x � x0, we have

S5 � S7 �0.5L2 + 0.121434L� 0.22

log x

+�1.5244L + 1.112

log2 x+�2L2 + 2.9067L + 2.5389

log3 x.

Proof. Recall the definitions

S5 =X

px1/3

log log xp2

p log xp

, S7 =X

px1/3

log log p

p log xp

.

We apply partial summation with f(t) = log logx

t2/ log

x

tto obtain

S5 =T (x1/3) log log x1/3

log x2/3�Z x1/3

2

T (t) log log xt2 dt

t log2 xt

+Z x1/3

2

2T (t) dt

t log xt log x

t2.

We next apply partial summation with g(t) = log log t/ logx

tto obtain

S7 =T (x1/3) log log x1/3

log x2/3�Z x1/3

2

T (t) log log t dt

t log2 xt

�Z x1/3

2

T (t) dt

t log t log xt

.

Subtracting,S5 � S7 = �I1 + I2 + I3 + I4, (6)

where

I1 =Z x1/3

2

T (t) log log xt2 dt

t log2 xt

,

I2 =Z x1/3

2

2T (t) dt

t log xt log x

t2,

I3 =Z x1/3

2

T (t) log log t dt

t log2 xt

,

and

I4 =Z x1/3

2

T (t) dt

t log t log xt

.

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Let x � x0. Splitting the interval at x1/6,

I1 ✓

L� log32

◆Z x1/3

x1/6

T (t)t log2 x

t

dt + L

Z x1/6

2

T (t)t log2 x

t

dt

✓

L� log32

◆Z x1/3

x1/6

log log t + B + 1log2 t

t log2 xt

dt + L

Z x1/6

2

log log t + B + 1log2 t

t log2 xt

dt

=✓

L� log32

◆⇣F (x1/3)� F (x1/6)

⌘+ L

⇣F (x1/6)� F (2)

⌘

= L⇣F (x1/3)� F (2)

⌘�✓

log32

◆⇣F (x1/3)� F (x1/6)

⌘,

where the antiderivative F is given by

log log xt

log x+

log t log log t

log x log xt

+B

log xt

+1

log2 x log xt

� 1log2 x log t

�2 log log x

t

log3 x+

2 log log t

log3 x.

Bounding the resulting expression gives

I1 0.5L2 � 0.94565L + 0.1361

log x+

2.2153Llog2 x

+2L2 � 2.9067L� 2.0810

log3 x,

where we made use of the identity

log logx

2= log log x + log

✓1� log 2

log x

◆

and bounded the Maclaurin series of the right term by comparison to a geometricseries. We next bound

I2 �Z x1/3

2

2(log log t + c) dt

t log xt log x

t2=Z log x1/3

log 2

2(log u + c) du

(log x� u)(log x� 2u).

Integrating, a lower bound for I2 is

2log x

Li2✓

log t

log x

◆� Li2

✓log t2

log x

◆+ (log log t + c)

⇣log log

x

t� log log

x

t2

⌘�x1/3

2

=2

log x

✓Li2✓

13

◆� Li2

✓23

◆+ (log 2)(L� log 3 + c)

◆

+2

log x

✓�Li2

✓log 2log x

◆+ Li2

✓log 4log x

◆+ (log log 2 + c) log

✓1� log 2

log x2

◆◆

� (log 4)L� 2.0947log x

+(log 4)(1� log log 2� c)

log2 x+

log2 22 log3 x

.

INTEGERS: 19 (2019) 14

Here Li2 denotes the dilogarithm, defined by the improper integral

Li2(z) = �Z z

0

log(1� t)t

dt.

For the terms (log 4)/ log2 x + (log2 2)/(2 log3 x) above, we have used the fact thatthe dilogarithm is concave up, together with the expansion

Li2(z) =X

k�1

zk

k2

for |z| < 1, putting z = log 2/ log x, to bound

Li2✓

log 4log x

◆� Li2

✓log 2log x

◆� Li2

✓log 4� log 2

log x

◆� Li2(0)

= Li2✓

log 2log x

◆� log 2

log x+

log2 24 log2 x

.

We have also used the bound � log(1� z) � z, here putting z = log 2/ log(x/2) andnoting that log log 2 + c < 0.

We now turn to I3. In order to bound I3 in the right direction, we write

I3 =Z e

2

T (t) log log t dt

t log2 xt

+Z x1/3

e

T (t) log log t dt

t log2 xt

.

The first integral is equal to

12 log x

✓log✓

1� 1� log 2log x

2

◆� log 2 log log 2

log x2

◆� �0.0304

log2 x.

A lower bound for the second integral is

Z x1/3

e

(log log t + c) log log t dt

t log2 xt

=Z log x1/3

1

(log u + c) log u du

(log x� u)2

=1

log x

c log

✓1� u

log x

◆+

cu log u

log x� u+ 2Li2

✓u

log x

◆�log x1/3

1

+1

log x

2(log u) log

✓1� u

log x

◆+

u log2 u

log x� u

�log x1/3

1

� 0.5L2 � 1.77881L + 1.9771log x

+c� 2log2 x

� 0.5188log3 x

.

Next, note that the use of the inequality T (t) > log log t + c allows us to add to the

INTEGERS: 19 (2019) 15

lower boundZ x1/3

3

(T (t)� (log log t + c)) log log t dt

t log2 xt

� 1log2 x

3

X

2k1228

Z log pk+1

log pk

(T (pk)� c� log u) log u du

� 0.3099log2 x

✓1 +

log 9log x

◆.

We therefore have

I3 �0.5L2 � 1.77881L + 1.9771

log x� 1.4591

log2 x+

0.1621log3 x

.

We now consider

I4 =Z x1/3

2

T (t) dt

t log t log xt

�Z x1/3

2

(log log t + c) dt

t log t log xt

=Z log x1/3

log 2

(log u + c) du

u(log x� u).

Integrating, we obtain the expression

1log x

c log u�(log u) log

✓1� u

log x

◆�c log(log x� u)�Li2

✓u

log x

◆+

log2 u

2

�log x1/3

log 2

.

We bound this expression to obtain

I4 �0.5L2 � 0.4317L� 0.3608

log x+

0.7658log2 x

+0.0556log3 x

.

Note that the inequality T (t) > log log t+c was used to bound I4. We may thereforeadd the following expression to the lower bound:

Z x1/3

2

T (t)� (log log t + c)t log t log x

t

dt � 1log x

2

Z x1/3

2

T (t)� (log log t + c)t log t

dt.

Splitting the interval [2, y0] into subintervals between primes [pk, pk+1], k = 1, 2, . . .1228, and using the numerical value of T (pk) in each subinterval, we find by com-putation in Pari/GP that

1log x

2

Z x1/3

2

T (t)� (log log t + c)t log t

dt � 0.3945log x

2

� 0.3945log x

✓1 +

log 2log x

◆.

A similar argument allows us to improve our upper bound on the integral I1 inequation (6). Here we used the upper bound

L

Z x1/6

2

T (t)t log2 x

t

dt L

Z x1/6

2

log log t + B + 1log2 t

t log2 xt

dt.

INTEGERS: 19 (2019) 16

Bounding the di↵erence in the interval [2, 100], we add 0.6909L/ log2 x to the lowerbound. Combining the bounds for I1, I2, I3, and I4 completes the proof of Lemma8.

5. The Proof of Theorem 1

We now prove Theorem 1. Note that ⌧3(x) = ⇡3(x) + N(x), where N(x) = |{n x : n = p2q}|, and p and q denote (possibly equal) primes. For each such p, thenumber of possibilities for q is ⇡(x/p2), and we thus have

N(x) =X

pp

x2

⇡

✓x

p2

◆.

The range 500194 x 1012 was checked using the computer program Pari/GP.Thus we may assume x � x0. Now, a lower bound for ⇡3(x) is given in Theorem 2.The contribution from N(x) is given by the following lemma, from which Theorem 1follows by an argument analogous to that of inequality (5) in the proof of Theorem2.

Lemma 9. For all x � x0 we have

X

pp

x2

⇡

✓x

p2

◆� x

✓↵

log x+

0.9861log2 x

+1.2861log3 x

◆,

where ↵ =P

p 1/p2 = 0.4522474 . . . denotes the reciprocal sum of squares of primes.

Proof. Let x � x0. We wish to apply the lower bound ⇡(t) > t/ log t, valid fort � 17, in Lemma 1. Thus we write

X

pp

x2

⇡

✓x

p2

◆=

X

pp

x17

⇡

✓x

p2

◆+

X

px17<p

px2

⇡

✓x

p2

◆.

The right sum is

X

px17<p

px2

⇡

✓x

p2

◆= �6 · ⇡

✓rx

17

◆+ ⇡

✓rx

13

◆+ . . . + ⇡

✓rx

2

◆.

Also by Lemma 1, this expression is bounded below by

2p

x

�6 · 1.0972p17 log x

17

+6X

k=1

1p

pk log xpk

!

� 1.8188p

x

log x,

INTEGERS: 19 (2019) 17

where pk denotes the k-th prime number. For the remaining term, let S(t) =Ppt(1/p2). By partial summation,

X

pp

x17

1p2 log x

p2

=S(p

x17 )

log 17�Z p x

17

2

2S(t) dt

t log2 xt2

=S(p

x17 )

log 17�Z p x

17

2

2↵ dt

t log2 xt2

+Z p x

17

2

2(↵� S(t)) dt

t log2 xt2

.

Now,Z p x

17

2

2↵ dt

t log2 xt2

=↵

log 17� ↵

log x4

↵

log 17� ↵

log x

✓1 +

log 4log x

◆.

Also,Z p x

17

2

2(↵� S(t)) dt

t log2 xt2

�X

k21433

(↵� S(pk))

1log x

t2

�pk+1

pk

=X

k21433

(↵� S(pk))log p2

k+1p2

k

log xp2

k+1log x

p2k

� 0.3592log x

9 log x4

� 0.3592log2 x

✓1 +

log 9log x

◆✓1 +

log 4log x

◆

� 0.3592log2 x

+1.2872log3 x

.

Thus,X

pp

x17

1p2 log( x

p2 )� ↵

log x

✓1 +

log 4log x

◆+

0.3592log2 x

+1.2872log3 x

� 1log 17

X

p>p

x17

1p2

.

By [9, Lem. 2.7], we have

1log 17

X

p>p

x17

1p2

<1/ log 17px17 log

px17

=2p

17/ log 17px log x

17

3.2431px log x

.

Combining these bounds, we obtain Lemma 9.

Acknowledgements. The author is very grateful to Jonathan Bayless, DominicKlyve, and Larry Wilson for helping to improve the mathematics and computerprogramming, and to the anonymous referee for helpful suggestions improving thequality of the paper.

INTEGERS: 19 (2019) 18

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[11] A. Selberg, Note on a paper by L. G. Sathe, J. Indian Math. Soc. 18 no. 1 (1954), 83–87.