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volume 2, issue 1, article 3, 2000.

Received 7 January, 2000;

accepted 18 August 2000.

Communicated by:L. Toth

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Journal of Inequalities in Pure and Applied Mathematics

ON MULTIPLICATIVELY PERFECT NUMBERS

J. SÁNDOR

Department of Mathematics, Babes-Bolyai University, Str. Kogalniceanu,

3400 Cluj-Napoca, ROMANIA EMail:[email protected]

2000c Victoria University ISSN (electronic): 1443-5756 019-99

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On Multiplicatively Perfect Numbers J. Sándor

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J. Ineq. Pure and Appl. Math. 2(1) Art. 3, 2001

http://jipam.vu.edu.au

Abstract

We study multiplicatively perfect, superperfect and analogous numbers . Con- nection to various arithmetic functions is pointed out. New concepts, inequalities and asymptotic evaluations are introduced.

2000 Mathematics Subject Classification:11A25,11N56, 26D15

Key words: Perfect numbers, arithmetic functions, inequalities in Number theory.

1. Introduction

It is well-known that a number n is said to be perfect if the sum of aliquot divisors ofnis equal ton. By introducing the functionσ(sum of divisors), this can be written equivalently as

(1.1) σ(n) = 2n.

The Euclid-Euler theorem gives the form of even perfect numbers: n= 2^kp, wherep= 2^k+1−1is prime (“Mersenne prime”). No odd perfect numbers are known. The numbernis said to be super-perfect if

(1.2) σ(σ(n)) = 2n.

The Suryanarayana-Kanold theorem [16], [4] gives the general form of even super-perfect numbers: n = 2^k, where2^k+1−1 =pis a prime. No odd super- perfect numbers are known. For new proofs of these results, see [10], [11].

Many open problems are stated e.g. in [1], [10].

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2. m-Perfect Numbers

Let T(n)denote the product of all divisors of n. There are many numbers n with the propertyT(n) = n², but none satisfyingT(T(n)) =n². Let us call the numbern >1multiplicatively perfect (or, for short,m-perfect) if

(2.1) T(n) = n²,

and multiplicatively super-perfect (m-super-perfect), if

(2.2) T(T(n)) =n².

To begin with, we prove the following little result:

Theorem 2.1. All m-perfect numbers nhave one of the following forms: n = p₁p₂ or n = p³₁, where p₁, p₂ are arbitrary, distinct primes. There are no m- super-perfect numbers.

Proof. Firstly, we note that ifd₁, d₂, . . . , d_sare all divisors ofn, then {d₁, d₂, . . . , d_s}=

n d₁, n

d₂, . . . , n d_s

, implying that

d1d2. . . ds = n d₁ · n

d₂ . . . n d_s, i.e.

(2.3) T(n) = n^s/2,

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wheres=d(n)denotes the number of (distinct) divisors ofn.

Let n = p^α₁¹. . . p^α_r^r be the prime factorisation of n > 1. It is well-known thatd(n) = (α₁+ 1). . .(α_r+ 1), so equation (2.1) combined with (2.3) gives (2.4) (α₁+ 1). . .(α_r+ 1) = 4.

Sinceα_i + 1 > 1, for r ≥ 2 we can have only α₁ + 1 = 2, α₂ + 1 = 2, implying α₁ = α₂ = 1, i.e. n = p₁p₂. Forr = 1we have α₁ + 1 = 4, i.e.

α₁ = 3, givingn = p³. There are no other solutions n > 1(n = 1is a trivial solution) of equation (2.1).

On the other hand, let us remark that forn≥2one hasd(n)≥2, so

(2.5) T(n)≥n

with equality only for n = prime. If n 6= prime, then it is immediate that d(n)≥3, giving

(2.6) T(n)≥n^3/2 for n6=prime.

Now, relations (2.5) and (2.6) together give

(2.7) T(T(n))≥n^9/4, n 6=prime.

Thus, by 9/4 > 2, there are no non-trivial (i.e. n 6= 1) m-super-perfect numbers.

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In fact, we have found that the equation

(2.8) T(T(n)) =n^a, a∈

1,9

4

has no nontrivial solutions.

Note. According to the referee the notion of “m-perfect numbers”, as well as Theorem2.1appears in [3].

Corollary 2.2. n= 6is the only perfect number, which is alsom-perfect.

Indeed,ncannot be odd, since by a result of Sylvester, an odd perfect number must have at least five prime divisors. If n is even, thenn = 2^kp = p₁p₂ ⇔ k = 1, when2 =p₁ and2²−1 = 3, when3 = p₂. Thusn= 2·3 = 6.

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3. k-m-Perfect Numbers

In a similar manner, one can definek-m-perfect numbers by

(3.1) T(n) =n^k

where k ≥ 2 is given. Since the equation (α₁ + 1). . .(α_r + 1) = 2k has a finite number of solutions, the general form ofk-multiply perfect numbers can be determined. We collect certain particular cases in the following.

Theorem 3.1. 1) All tri-m-perfect numbers have the formsn=p₁p²₂ orn = p⁵₁;

2) All 4-m-perfect numbers have the formsn =p1p³₂ orn =p1p2p3 orn = p⁷₁;

3) All 5-m-perfect numbers have the formsn=p₁p⁴₂orn=p⁹₁;

4) All 6-m-perfect numbers have the formsn=p1p2p²₃,n=p1p⁵₂,n=p¹¹₁ ; 5) All 7-m-perfect numbers have the formsn=p₁p⁶₂, orn =p¹³₁ ;

6) All 8-m-perfect numbers have the formsn = p₁p₂p₃p₄ or n = p₁p₂p³₃ or n=p³₁p³₂,n=p¹⁵₁ ;

7) All 9-m-perfect numbers have the formsn=p₁p²₂p²₃,n=p₁p⁸₂,n=p¹⁷₁ ; 8) All 10-m-perfect numbers have the formsn =p₁p₂p⁴₃,n=p₁p⁹₂,n =p¹⁹₁ ,

etc.

(Herep_idenote certain distinct primes.)

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Proof. We prove only the case 6). By relation (2.3) we must solve the equation (3.2) (α₁+ 1). . .(α_r+ 1) = 16

inα_randr. It is easy to see that the following four cases are possible:

i) α1+ 1 = 2,α2+ 1 = 2,α3+ 1 = 2,α4+ 1 = 2;

ii) α₁+ 1 = 2,α₂+ 1 = 2,α₃+ 1 = 4,α₄+ 1 = 4;

iii) α₁+ 1 = 4,α₂+ 1 = 4;

iv) α₁+ 1 = 16.

This gives the general forms of all 8-m-perfect numbers, namely(α₁ =α₂ = α₃ =α₄ = 1)n = p₁p₂p₃p₄; (α₁ = 1, α₂ = 1, α₃ = 3)n = p₁p₂p³₃; (α₁ = 3, α₂ = 3)n=p³₁p³₂;(α₁ = 15)n =p¹⁵₁ .

Corollary 3.2. 1) n= 28is the single perfect and tri-perfect number.

2) There are no perfect and 4-perfect numbers;

3) n= 496is the only perfect number which is 5-m-perfect;

4) There are no perfect numbers which are 6-m-perfect;

5) n= 8128is the only perfect number which is 7-m-perfect.

In fact, we have:

Theorem 3.3. Letpbe a prime, with2^p−1prime too (i.e. 2^p−1is a Mersenne prime). Then2^p−1(2^p−1)is the only perfect number, which isp-m-perfect.

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Proof. By writing(α₁+ 1). . .(α_r+ 1) = 2p(pprime), the following cases are only possible:

i) α₁+ 1 = 2,α₂+ 1 =p;

ii) α₁+ 1 = 2p.

Thenn = p₁p^p−1₂ orn = p^2p−1₁ are the general forms ofp-m-perfect numbers. By the Euclid-Euler theorem p₁p^p−1₂ = 2^p−1(2^p − 1) iff p₂ = 2 and p₁ = 2^p−1is prime.

Remark 3.1. Forp <10000the following Mersenne primes are known; namely forp= 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941. It is an open problem to show the existence of infinitely many Mersenne primes ([1]).

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4. Some Results for k-m-Perfect Numbers

As we have seen, the equation (2.2), i.e. T(T(n)) = n² has no nontrivial solutions. A similar problem arises for the equation

(4.1) T(T(n)) =n^k; n >1

(k ≥2, fixed). By (2.3) we can see that this is equivalent to

(4.2) d(n)d(T(n))

4 =k.

Let n = p^α₁¹. . . p^α_r^r > 1 be the canonical representation of n. By d(n) = (α₁+ 1). . .(α_r+ 1), and (2.3) we have

T(n) =p^α₁¹^(α¹^+1)...(α^r^+1)/2. . . p^α_r^r^(α¹^+1)...(α^r^+1)/2,

so (4.2) becomes equivalent to (4.3) (α₁+ 1). . .(α_r+ 1)

α₁(α₁+ 1). . .(α_r+ 1)

2 + 1

. . .

α_r(α₁+ 1). . .(α_r+ 1)

2 + 1

= 4k,

and this, clearly has at most a finite number of solutions.

Theorem 4.1. 1) Equation (4.1) is not solvable fork= 4,5,6;

2) Fork = 3the general solutions aren =p²₁;

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3) Fork = 7the solutions aren =p³₁;

4) Fork = 9the solutions are:n =p1p2 (p1 6=p2 primes).

Proof. For k = 4,5,6, from (4.3) we must solve the corresponding equations for 16, 20, 24. It is a simple exercise to verify these impossibilities. Fork = 3 we have the single equality 12 = 3·4, when α₁ = 2, ^α¹^(α₂¹⁺¹⁾ + 1 = 4. For k = 7,α1 = 3by ^3·4₂ + 1 = 7and4·7 = 28. Fork= 9we have2·2·3·3 = 36 andα₁ =α₂ = 1.

Corollary 4.2. n = 6 is the single perfect number which is also 9-super-m- perfect.

Indeed,p₁p₂ = 2·(2²−1) = 2·3 = 6by Theorem4.1and the Euclid-Euler theorem.

Remark 4.1. By relation (2.6), by consecutive iteration we can deduce

T(T(. . . T(n). . .))

| {z }

k

≥n³^k^/2^k

for n 6=prime. Since 3^k > 2^k·k for allk ≥ 1(induction: 3^k+1 = 3·3^k >

3·2^k·k >2·2^k(k+1) = 2^k+1(k+1)) we can obtain the following generalization of equation (2.2):

T(T(. . . T(n). . .))

| {z }

k

=n^k has no nontrivial solutions.

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5. Other Results

By relation (2.3) we have

(5.1) logT(n)

logn = d(n) 2 .

Clearly, this implies

n→∞lim inf logT(n)

logn = 1, lim

n→∞suplogT(n)

logn = +∞

(take e.g. n = p(prime);n = 2^k (k ∈ N)). Since 2 ≤ d(n) ≤ 2√

n(see e.g.

[13]) forn ≥2we get

1≤ logT(n) logn ≤√

n.

By2^ω(n) ≤d(n)≤2^Ω(n) (see e.g. [12]) we can deduce:

2^ω(n)−1 ≤ logT(n)

logn ≤2^Ω(n)−1 (n≥2).

Since it is known by a theorem of Hardy and Ramanujan [2] that the normal order of magnitude ofω(n)andΩ(n)is log logn, the above double inequality implies that:

the normal order of magnitude of

(5.2) log logT(n)−log lognis(log 2)(log logn−1).

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By a theorem of Wiegert ([17]) we have

n→∞lim suplogd(n) log logn

logn = log 2,

so by (5.1) we get:

(5.3) lim

n→∞sup (log logT(n))(log logn)

logn = log 2.

In fact, by a result of Nicolas and Robin ([7]), forn≥3one has logd(n)

log 2 ≤c logn

log logn (c≈1,5379. . .),

we can obtain the following inequality:

(5.4) log logT(n)≤log logn+ klogn

log logn −log 2,

wherek=clog 2andn≥3. This gives

(5.5) lim

n→∞

log logT(n) f(n) = 0 for any positive function with f(n) log log^logⁿ n →0 (n→ ∞).

Byϕ(n)d(n)≥n(see [14]) andϕ(n)d²(n)≤n² forn 6= 4(see [8]) we get n

ϕ(n) ≤d(n)≤ n

pϕ(n) for n >4,

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and this, by (5.1) yields

(5.6) n

2ϕ(n) ≤ logT(n)

logn ≤ n 2p

ϕ(n). Hereϕis the usual Euler totient function.

Hence, the arithmetic function T is connected to the other classical arithmetic functions.

By√

n ≤ ^σ(n)_d(n) ≤ ⁿ⁺¹₂ (see [14], [5], [6]), we get

(5.7) σ(n)

n+ 1 ≤ logT(n)

logn ≤ σ(n) 2√

n. For infinitely many primespwe have

d(p−1)>exp

c logp log logp

(c > 0, constant, see [9]), so we have:

(5.8) log logT(p−1)>log log(p−1) + clogp

log logp −log 2 for infinitely many primesp, implying, e.g.

(5.9) lim

p→∞

p prime

sup log logT(p−1)

log logp = +∞

and

(5.10) lim

n→∞inf (log logT(n))(log logn) logn >0.

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References

[1] R.K. GUY, Unsolved Problems in Number Theory, Springer Verlag, 2^nd Ed., 1994.

[2] G.H. HARDY ANDS. RAMANUJAN, The normal number of prime fac- tors of a numbern, Quart. J. Math., 48 (1920), 76–92.

[3] K. IRELANDANDM. ROSEN, A Classical Introduction to Modern Num- ber Theory, Springer, 1982, Chapter 2.

[4] H.J. HANOLD, Über super-perfect numbers, Elem. Math., 24 (1969), 61–

62.

[5] E.S. LANGFORD, See D.S. Mitrinovi´c and M.S. Popadi´c, Inequalities in Number Theory, Niš, 1978 (p. 44).

[6] D.S. MITRINOVI ´CANDJ. SÁNDOR (in coop. with B. CRSTICI), Hand- book of Number Theory, Kluwer Acad. Publ. 1995.

[7] J.L. NICOLAS AND G. ROBIN, Majorantions explicites pour le nombre de diviseurs den, Canad. Math. Bull., 26 (1983), 485–492.

[8] S. PORUBSKY, Problem E2351, Amer. Math. Monthly, 79 (1972), 394.

[9] K. PRACHER, Primzahlverteilung, Berlin, Göttingen, Heidelberg, 1957 (p. 370).

[10] J. SÁNDOR, On the composition of some arithmetic functions, Studia Univ. Babe¸s-Bolyai, Math., 34(1) (1989), 7–14.

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[11] J. SÁNDOR, On perfect and super-perfect numbers (Romanian), Lucr.

Semin. Didactica Mat., 8 (1992), 167–168.

[12] J. SÁNDOR, On certain inequalities for arithmetic functions, Notes Numb.

Th. Discr. Math., 1 (1995), 27–32.

[13] W. SIERPINSKI, Elementary theory of numbers, Warsawa, 1964.

[14] R. SIVARAMAKRISHNAN AND C.S. VENKATARAMAN, Problem 5326, Amer. Math. Monthly, 72 (1965), 915.

[15] R. SIVARAMAKRISHNAN, Problem E1962, Amer. Math. Monthly, 74 (1967), 198.

[16] D. SURYANARAYANA, Super-perfect numbers, Elem. Math., 24 (1969), 16–17.

[17] S. WIEGERT, Sur l’ordre de grandeur du nombre des diviseurs d’une en- tière, Arkiv. für Math., 3 (1907), 1–9.