An inequality for the function π(n)

Periodica Mathematica Hungarica Vol. 67 (2 ), 2013, pp. 243–249DOI: 10.1007/s10998-013-4674-5

AN INEQUALITYFOR THE FUNCTION π(n)

Horst Alzer

Morsbacher Str. 10, 51545 Waldbrol, Germany

E-mail: [email protected]

(Received January 16, 2012; Accepted March 26, 2012)

[Communicated by A. Sarkozy]

Abstract

We prove that the inequality π2(m) + π2(n) ≤ 54 π2(m + n) holds for

all integers m,n ≥ 2. The constant factor 5/4 is sharp. This complements a

result of Panaitopol, who showed in 2001 that 12 π2(m+ n) ≤ π2(m) + π2(n)

is valid for all m,n ≥ 2. Here, as usual, π(n) denotes the number of primes

not exceeding n.

1. Introduction

In 1923, Hardy and Littlewood [4] published the conjecture that the inequality

π(m+ n) ≤ π(m) + π(n) (1.1)

holds for all integers m,n ≥ 2. This is still an open problem. However, in [5] the

authors give “strong evidence against this assertion” [5, p. 375]. Inequality (1.1) has

attracted the attention of numerous mathematicians, who presented various related

results. We mention a few of them. (Here, m and n are natural numbers.)

π(m) + π(n) ≤ π(mn) (Ishikawa [6], 1934),

π(m+ n) ≤ π(m) + π(n) (2 ≤ min(m,n) ≤ 1731,

Gordon and Rodemich [3], 1998),

π(m+ n) ≤ π(m) + π(n) + π(m− n) (2 ≤ n ≤ m, Panaitopol [8], 2001),

π(m+ n) ≤ 1.11π( m

1.11

)

+ π(n) (7 ≤ n ≤ m, Garunkstis [2], 2002).

For more information on this subject we refer to [1] and [7, Chapter VII].

Mathematics subject classification numbers: 11A25, 11A41, 11N05, 26D07.

Key words and phrases: prime numbers, inequalities.

0031-5303/2013/$20.00 Akademiai Kiado, Budapestc© Akademiai Kiado, Budapest Springer, Dordrecht

244 H. ALZER

The work on this paper has been inspired by an interesting result published

by Panaitopol [8] in 2001. He proved that the following companion of (1.1) is valid

for all m,n ≥ 2:1

2π2(m+ n) ≤ π2(m) + π2(n). (1.2)

If m = n = 2, then the sign of equality holds in (1.2). This implies that the constant

factor 1/2 is sharp. It is natural to look for a converse of (1.2). More precisely we

ask: what is the smallest real number α such that we have for all m,n ≥ 2:

π2(m) + π2(n) ≤ απ2(m+ n)?

It is our aim to answer this question. In the next section we collect some lemmas.

They play an important role in the proof of our main result, which is given in

Section 3.

The numerical values have been calculated via the computer system MAPLE

V Release 5.1.

2. Lemmas

First, we present upper and lower bounds for π(x). This result is due to

Rosser and Schoenfeld [9].

Lemma 1. If x > exp(1.5), then

π(x) <x

log x− 1.5.

If x ≥ 67, thenx

log x− 0.5< π(x).

The following lemmas provide some properties of the functions

λ(x) =( x

log x− 0.5

)2

and µ(x) =( x

log x− 1.5

)2

. (2.1)

Lemma 2.

(i) λ is strictly increasing on (exp(1.5),∞).

(ii) λ′ is strictly increasing on (exp(0.5),∞).

Proof. We have

λ′(x) =8x(2 log x− 3)

(2 logx− 1)3> 0 for x > exp(1.5)

and

λ′′(x) =32

(2 log x− 1)4((log x− 2)2 + 0.75) > 0 for x > exp(0.5).

AN INEQUALITY FOR THE FUNCTION π(n) 245

Lemma 3. If x ≥ 83, then

µ(x) <5

8λ(2x).

Proof. We define

z(x) =√5(log x− 1.5)−

√2(log(2x)− 0.5).

Then we get for x ≥ 83:

z′(x) =

√5−

√2

x> 0 and z(x) ≥ z(83) = 0.004 . . . .

This leads to

5λ(2x) =(

√5 · 2x

log(2x)− 0.5

)2

>(

√2 · 2x

log x− 1.5

)2

= 8µ(x).

Lemma 4. If x ≥ 21 561, then

µ(x) + 2116 <5

4λ(x + 2). (2.2)

Proof. Let x ≥ c0 = 21 561. We obtain

5

4λ(x + 2)− µ(x) = (η(x) + τ(x))(η(x) − τ(x)) (2.3)

with

η(x) =√5

x+ 2

2 log(x+ 2)− 1and τ(x) =

2x

2 logx− 3.

Since η(x) > 0, τ(x) > 0 and

η′(x) =√5

2 log(x+ 2)− 3

(2 log(x+ 2)− 1)2> 0, τ ′(x) =

4 logx− 10

(2 logx− 3)2> 0,

we conclude that η + τ is positive and increasing on I = [c0,∞). Next, we show

that η − τ is positive and increasing on I. Let

σ1(x) =(

8√5

log x

log(x+ 2)− 16

)

log x,

σ2(x) = −(24√5− 16)

log x

log(x + 2)− 12

√5( log x

log(x+ 2)

)2

+ 40,

σ3(x) =c1

log(x+ 2)+

c2 log x− c3

log2(x+ 2)

246 H. ALZER

with

c1 = 18√5− 40 = 0.24 . . . , c2 = 36

√5− 4 = 76.49 . . . , c3 = 27

√5− 10 = 50.37 . . . .

We find( (2 log(x + 2)− 1)(2 logx− 3)

log(x+ 2)

)2

(η′(x)− τ ′(x)) = σ1(x) + σ2(x) + σ3(x). (2.4)

Since

θ(x) =log x

log(x+ 2)

is increasing on (0,∞), we get

8√5θ(x)− 16 ≥ 8

√5θ(c0)− 16 = 1.88 . . . .

This implies that σ1 is increasing on I. Moreover, σ2 is decreasing on I. Let

φ(x) =c2 log x− c3

log2(x+ 2).

The function x 7→ (log(x + 2))/x is decreasing on (0,∞), so that we obtain

(x + 2) log3(x+ 2)φ′(x) = 2c3 + 2c2log(x+ 2)

x+ c2 log

(1

x+

2

x2

)

≤ 2c3 + 2c2log(c0 + 2)

c0+ c2 log

( 1

c0+

2

c20

)

= −662.52 . . . .

This implies that σ3 is decreasing on I.

Let c0 ≤ r ≤ x ≤ s ≤ 450 000. The monotonicity of σ1, σ2, σ3 leads to

σ1(x) + σ2(x) + σ3(x) ≥ σ1(r) + σ2(s) + σ3(s) = L(r, s), say.

We have

L(c0, 100 000) = 0.63 . . . and L(100 000, 450 000) = 2.84 . . . .

This reveals that σ1(x) + σ2(x) + σ3(x) > 0 for x ∈ [c0, 450 000]. Let x ≥ 450 000.

Using

limx→∞

σ2(x) = 56− 36√5 and lim

x→∞

σ3(x) = 0

yields

σ1(x) + σ2(x) + σ3(x) > σ1(450 000) + 56− 36√5 = 0.08 . . . .

Thus, σ1 + σ2 + σ3 is positive on I. From (2.4) we conclude that η′(x)− τ ′(x) > 0

for x ≥ c0. Therefore, η − τ is increasing on I with η(c0) − τ(c0) = 0.41 . . .. It

follows that (η + τ)(η − τ) is increasing on I. Applying (2.3) gives for x ≥ c0:

5

4λ(x + 2)− µ(x) ≥ 5

4λ(c0 + 2)− µ(c0) = 2121.13 . . . .

The proof of (2.2) is complete.


Lemma 5. If x ≥ 200, then

µ′(x) <5

4λ′(x+ 200). (2.5)

Proof. We have

λ′(x+ 200)

µ′(x)= u(x)v(x)w(x) (2.6)

with

u(x) = 1 +200

x,

v(x) = 1− 2

2 log(x + 200)− 1,

w(x) =(2 log x− 3)3

(2 log x− 5)(2 log (x+ 200)− 1)2.

Let x ∈ J = [200,∞). The functions u, v, and w are positive on J . Moreover, u is

decreasing and v is increasing. We obtain

x(x + 200)(2 logx− 5)2(2 log(x+ 200)− 1)3

4(2 logx− 3)2w′(x) = w1(x) +w2(x) +w3(x) (2.7)

with

w1(x) = 4x(log x) log(

1 +200

x

)

, w2(x) = 400(logx− 3)(2 log(x+ 200)− 1),

w3(x) = x(14 logx− 12 log(x + 200)− 9).

Obviously, w1 is positive and w2 is increasing on J . Since

w′′

3 (x) =2(x2 + 400x+ 280 000)

x(x+ 200)2> 0 and w′

3(200) = 1.27 . . . ,

we conclude that also w3 is increasing on J . Thus,

w1(x) + w2(x) + w3(x) ≥ w2(200) + w3(200) = 8752.67 . . . . (2.8)

It follows from (2.7) and (2.8) that w is increasing on J .

Let 200 ≤ r ≤ x ≤ s ≤ 13 500. Then we obtain

u(x)v(x)w(x) ≥ u(s)v(r)w(r) = M(r, s), say.

Since

M(200 + 100k, 200+ 100(k + 1)) > 0.8 for k = 0, 1, 2, . . . , 132,

248 H. ALZER

we get

u(x)v(x)w(x) > 0.8 for x ∈ [200, 13 500]. (2.9)

Let x ≥ 13 500. Then we have

u(x)v(x)w(x) ≥ v(13 500)w(13 500) = 0.80041 . . . .

Thus,

u(x)v(x)w(x) > 0.8 for x ≥ 13 500. (2.10)

From (2.6), (2.9), and (2.10) we conclude that (2.5) is valid.

3. Main result

With the help of the lemmas given in the previous section we are now in a

position to prove the following counterpart of (1.2).

Theorem. For all integers m,n ≥ 2 we have

π2(m) + π2(n) ≤ 5

4π2(m+ n). (3.1)

The constant factor 5/4 is the best possible.

Proof. In order to establish (3.1) we assume thatm ≥ n ≥ 2. We distinguish

three cases.

Case 1: 200 ≤ n ≤ m. Let x be a real number with x ≥ n and

F (x, n) =5

4λ(x + n)− µ(x)− µ(n),

where λ and µ are given in (2.1). We differentiate with respect to x and apply

Lemma 2(ii) and Lemma 5. This leads to

∂

∂xF (x, n) =

5

4λ′(x+ n)− µ′(x) ≥ 5

4λ′(x+ 200)− µ′(x) > 0.

Hence,

F (x, n) ≥ F (n, n) =5

4λ(2n)− 2µ(n). (3.2)

Using (3.2) and Lemma 3 yields

F (m,n) > 0. (3.3)


From Lemma 1 and (3.3) we obtain

π2(m) + π2(n) < µ(m) + µ(n) <5

4λ(m + n) <

5

4π2(m+ n).

Case 2: 2 ≤ n ≤ 200 ≤ m. We consider two subcases.

Case 2.1: m ≤ 21 561. Then, 2 ≤ n ≤ 200 ≤ m ≤ 21 561. Thus, we have a

finite number of pairs (m,n). By direct computation we find that (3.1) is valid for

these pairs.

Case 2.2: 21 561 ≤ m. Applying Lemma 1, Lemma 2(i), and Lemma 4 gives

π2(m) + π2(n) < µ(m) + π2(200) = µ(m) + 2116 <5

4λ(m+ 2) ≤ 5

4λ(m+ n)

<5

4π2(m+ n).

Case 3: 2 ≤ n ≤ m ≤ 200. Computer calculations reveal that (3.1) also holds

for these values.

This completes the proof of inequality (3.1). Moreover, since the sign of

equality is valid for n = 3,m = 7, we conclude that in (3.1) the constant factor 5/4

is sharp.

References

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thesis, University of Limoges, 1998.

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[3] D. M. GORDON and G. RODEMICH, Dense admissible sets, Algorithmic number

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Comput. Sci. 1423 (ed. J. P. Buhler), Springer, Berlin, 1998, 216–225.

[4] G. H. HARDY and J. E. LITTLEWOOD, Some problems of “partitio numerorum” III:

On the expression of a number as a sum of primes, Acta Math., 44 (1923), 1–70.

[5] D. HENSLEY and I. RICHARDS, Primes in intervals, Acta Arith., 25 (1974), 375–391.

[6] H. ISHIKAWA, Uber die Verteilung der Primzahlen, Sci. Rep. Tokyo Bunrika Daigaku

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[7] D. S. MITRINOVIC, J. SANDOR and B. CRSTICI, Handbook of Number Theory, Kluwer,

Dordrecht, 1996.

[8] L. PANAITOPOL, Some generalizations for a theorem by Landau, Math. Inequal.

Appl., 4 (2001), 327–330.

[9] J. B. ROSSER and L. SCHOENFELD, Approximate formulas for some functions of prime

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An inequality for the function π(n)