An inequality for the number of divisors of n

OCTOGON MATHEMATICAL MAGAZINE Vol. 17, No.2, October 2009, pp 746-749 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

746

An inequality for the number of divisors of n

J´ozsef S´andor and Lehel Kov´acs³⁰

ABSTRACT. M.I. Isravilov and I. Allikov proved in 1980 thatd(n)< n^2/3for n >12. The authors proved in [2] thatd(n)<√

nforn≥1262. We shall prove here thatd(n)≤4n^1/3 for alln≥1.

MAIN RESULTS

Letd(n) denote the number of distinct positive divisors ofn.

Then d(1) = 1 and ifn= Q^r

i=1

p^a_iⁱ is the prime factorization ofn (pi distinct primes;ai ≥1), then it is well-known that

d(n) = (a1+ 1)...(ar+ 1) (1) The classical inequality (see e.g. [1], [2])

d(n)<2√

n (2)

is valid for all n≥1. Though the inequality

d(n)< n^2/3 for n >12 (3)

by Isravilar and Allikov holds true, remark that for n≥64 (2) is stronger than (3), as 2n^1/2≤n^2/3 is true forn≥2⁶= 64.

A stronger inequality, namely

d(n)<√

n, n≥1262 (4)

30Received: 12.09.2009

2000Mathematics Subject Classification. 11A25.

Key words and phrases. Arithmetic functions; inequalities.

An inequality for the number of divisors ofn 747 has been proved in our note [2].

For greater values ofn, however we can deduce the following stronger relation:

d(n)≤4n^1/3 forn≥1 (5)

Indeed, as 4n^1/3≤n^1/2 forn≥4⁶ = 4096,we get that (5) is stronger than (4) forn≥4096.

Theorem. For alln≥1 one has

d(n)<4√³

n (6)

Proof. As 1<4 , (6) is true forn= 1. Let n≥2, and use relation (1).

First we need the following.

Lemma. For allp≥11 one has

p^a/3 > a+ 1; (7)

For allp≥2 one has

(a+ 1)p^−a/3 ≤M(p), (8)

where

M(p) = 3

logp ·p^{(logp−3)/3 logp} (9)

Proof. The inequality (7), written equivalentlyp^a>(a+ 1)³ follows by p^a≥11^a and from

11^a>(a+ 1)³, a≥1 (10) inductively. Indeed, (10) is valid fora= 1, as 11>2³ = 8.Now, assuing (10) fora, one has 11^a+1 = 11·11^a>11 (a+ 1)³ >(a+ 2)³,as

(a+ 2)³ (a+ 1)³ =

µ

1 + 1 a+ 1

¶₃

≤ µ

1 +1 2

¶₃

= 27 8 <11;

748 J´ozsef S´andor and Lehel Kov´acs finishing the proof of (10).

For the proof of (8), consider the application f(a) = (a+ 1)p^−a/3 Asf⁰(a) =p^−a/3

h

1−^{(a+1) log}₃ ^p i

,we get that the functionf has a maximum ata=a₀ = _log³_p −1,and the maximum value off(a₀) is

M(p) = 3

logp·p^{(logp−3)/3 logp}, i.e. (8) is valid, with (9).

This finishes the proof of Lemma.

Now, for the proof of the theorem writenas n=Y

p≤7

p^a· Y

p≥11

p^a

As for each term of the second product one has d(p^a) =a+ 1< p^a/3

while for each term of the first product we have d(p^a)≤p^a/3M(p),we get that for alln≥2 one has

d(n)≤M(2)·M(3)·M(5)·M(7)Y

p≥2

p^a/3 =K·n^1/3 By using a computer, one can find that M(2) = 2,006063;M(3) = 1,448847..., M(5) = 1,172580..., M(7) = 1,084934...,soK ≈3,697... <4, finishing the proof of the theorem.

Remark.As in d(n)< K√³

n, one has K≈3, ..., we get even a stronger relation than inequality (6). For example, K <3,7.

An inequality for the number of divisors ofn 749 REFERENCES

[1] S´andor, J., Geometric theorems, diophantine equations and arithmetic functions, New Mexico, 2002.

[2] S´andor, J. and Kov´acs, L., A note on the arithmetical functions d(n) and σ(n), Octogon Mathematical Magazine, Vol. 16, No. 1A, April 2008, pp.

270-274.

Babe¸s-Bolyai University, Cluj and Miercurea Ciuc, Romania

Sapientia Univ. of Targu-Mures, Romania