DOI: 10.2478/auom-2014-0005

An. St. Univ. Ovidius Constanta Vol. 22(1),2014, 45–50

About k-perfect numbers

Mihaly Bencze

Abstract

ABSTRACT. In this paper we present some results about k-perfectnumbers, and generalize two inequalities due to M. Perisastri (see [6]).

1 Introduction

Definition. A positive integer n is k-perfect if σ (n) = kn, when k > 1,k ∈ Q. The special case k = 2 corresponds to perfect numbers, which areintimately connected with Mersenne primes. We have the followingsmallest k-perfect numbers. For k = 2 (6, 28, 496, 8128, ...) , for k = 3(120, 672, 523776, 459818240, ...) , for k = 4 (30240, 32760, 2178540, ...) , for k =5 (14182439040, 31998395520, ...) , for k = 6 (154345556085770649600, ...) .

For a given prime number p, if n is p-perfect and p does not divide n, thenpn id (p+ 1)− perfect. This imples that an integer n is a 3− perfect numberdivisible by 2 but not by 4, if and only if n

2 is an odd perfect number, ofwhich none are known. If 3n is 4k− perfect and 3 does not divide n, then nis 3k−perfect.

A k−perfect number is a positive integer n such that its harmonic sum ofdivisors is k.

For the perfect numbers we have the followings: 28 = 13 + 33, 496 =13 + 33 + 53 + 73, 8128 = 13 + 33 + 53 + 73 + 93 + 113 + 133 + 153 etc. Weposted the following conjecture:

Conjecture. (Bencze, M., 1978) If n is k-perfect, then exist odd positiveintegers ui (i = 1, 2, ..., r) such that

Key Words: perfect numbers2010 Mathematics Subject Classification: 11A25Received: May, 2013.Revised: September, 2013.Accepted: November, 2013.

45

46 Mihaly Bencze

n =

r∑i=1

uk+1i

MAIN RESULTS

Theorem 1. If f : R → R is convex and increasing, N = pα11 pα2

2 ...pαnnwritten in cannonical form is k-perfect, then:

n∑i=1

f

(1

pi

)≥

nf(n

√32 − 1

)if N is even

nf(

3n√k2 − 1

)if N is odd

Proof. If N is even then it follows

n∏i=1

pi + 1

pi>

3

2

For x ≥ 3 holds x+1x ≥

3

√(xx−1

)2(see [9]), therefore if N is odd then yields

n∏i=1

pi + 1

pi> 3

√√√√ n∏i=1

(pi

pi − 1

)2

>3√k2

because

n∏i=1

pipi − 1

= k

n∏i=1

pαi+1i

pαi+1i − 1

> k

Using the AM-GM inequality we obtain:

n∏i=1

pi + 1

pi≤

(1

n

n∑i=1

pi + 1

pi

)n=

(1 +

1

n

n∑i=1

1

pi

)nFinally

n∑i=1

1

pi>

n(n

√32 − 1

)if N is even

n(

3n√k2 − 1

)if N is odd

Because f is convex and increasing from Jensen’s inequality we get

About k-perfect numbers 47

n∑i=1

f

(1

pi

)≥ nf

(1

n

n∑i=1

1

pi

)≥

nf(n

√32 − 1

)if N is even

nf(

3n√k2 − 1

)if N is odd

(1)

Theorem 2. If g : R → R is convex and increasing, N = pα11 pα2

2 ...pαnnwritten in cannonical form is k-perfect, then:

n∑i=1

g

(1

pi

)≤

ng(

1− n

√6kπ2

)if N is even

ng(

1− n

√8kπ2

)if N is odd

Proof. We have the following:

n∏i=1

pipi − 1

=

n∏i=1

pαi+1i − 1

(pi − 1) pαii

n∏i=1

pαi+1i

pαi+1i − 1

= k

n∏i=1

1

1− 1

pαi+1

i

= k

n∏i=1

∞∑j=0

(1

pi

)j ≤

≤ kn∏i=1

∞∑j=0

1

p2ji

<

k∞∑n=1

1n2 if N is even

k∞∑n=0

1(2n+1)2

if N is odd=

{kπ2

6 if N is evenkπ2

8 if N is odd

From AM-GM inequality yields

n∏i=1

pipi − 1

≥

nn∑i=1

pi−1pi

n

=

n

n−n∑i=1

1pi

n

therefore

n∑i=1

1

pi<

n(

1− n

√6kπ2

)if N is even

n(

1− n

√8kπ2

)if N is odd

According to Jensen’s inequality yields

n∑i=1

g

(1

pi

)≤ ng

(1

n

n∑i=1

1

pi

)≤

ng(

1− n

√6kπ2

)if N is even

ng(

1− n

√8kπ2

)if N is odd

48 Mihaly Bencze

Corolloary 1. If N = pα11 pα2

2 ...pαnn written in cannonical form is k-perfectthen:

n(n

√32 − 1

)if N is even

n(

3n√k2 − 1

)if N is odd

<

n∑i=1

1

pi<

n(

1− n

√6kπ2

)if N is even

n(

1− n

√8kπ2

)if N is odd

Theorem 3. If x, t > 0 then

(x+ 1) t1x+1 − xt 1

x ≤ 1

Proof. For t = 1 we have the equality. Let 0 < t < 1. Since the functionu (x) = xt

1x is continuous and differentiable we can apply the Lagrange’s

theorem and we obtain

(x+ 1) t1x+1 − xt 1

x

(x+ 1)− x=u (x+ 1)− u (x)

(x+ 1)− x= u′ (z)

when x < z < x+ 1 hence we have the inequality

t1z

(1− 1

zln t

)< 1 or 1− 1

zln t < t−

1z .

Developing t−1z into McLauren’s series it results

1− 1

zln t < 1− 1

1!zln t+

1

2!z2ln2 t− 1

3!z3ln3 t+ ...

or

∞∑r=2

(−1)r

lnr t

r!zr> 0 or

∞∑r=2

lnr 1t

r!zt> 0

that is obvious because ln 1t > 0 due to 1

t > 1. Let be t > 1. Then is

enough to show that the function V (x) = x(t

1x − 1

)is decreasing.

Differentiable V we get

V ′ (x) = t1x − t 1

x · 1

xln t− 1 = −

∞∑r=2

lnr t

xr (r − 1)!

(1− 1

r

)< 0

Since V is decreasing and we may say that V (x+ 1) < V (x) hence andfrom it follows the inequality of the ennunciation.

Corollary 2. If N = pα11 pα2

2 ...pαnn is a k-perfect number written in can-nonical form, then:

About k-perfect numbers 49

{ln 3

2 if N is even23 ln k if N is odd

<

n∑i=1

1

pi<

{ln kπ2

6 if N is even

ln kπ2

8 if N is odd

Proof. Using the Theorem 3 it is proved that the series(n

(n

√3

2− 1

))n∈N∗

and(n(

3n√k2 − 1

))n∈N∗

are decreasing, and the series(n

(1− n

√6

kπ2

))n∈N∗

and

(n

(1− n

√8

kπ2

))n∈N∗

are increasing. It means that the minimum and maximum are reached onlythen n→∞.

Since n → ∞ we have 0 · ∞. That is why L’Hospital rule and so we findthe results of the enunciation.

Remark 1. For k = 2 we reobtain the M.Perisastri’s inequality

n∑i=1

1

pi< 2 ln

π

2

(see [6]).Corollary 3. Let N = pα1

1 pα22 ...pαnn be a k−perfect number written in

cannonical form and Pmax = {p1, p2, ..., pn} and Pmin = min {p1, p2, ..., pn} ,then

Pmin <

{ 1n√

32−1

if N is even

13n√k2−1

if N is odd

and

Pmax >

1

1− n√

6kπ2

if N is even

1

1− n√

8kπ2

if N is odd

Proof. Considering that

n

Pmax<

n∑i=1

1

pirespective

n∑i=1

1

pi<

n

Pmin

from the theorem if follows the affirmation.

50 Mihaly Bencze

Remark 2. Let N = pα11 pα2

2 ...pαnn be a k-perfect number written in can-nonical form, then

Pmin <2n

k2 − 1+ 2

(see the method of M. Perisastri’s)

Acknowledgements. The author wishes to express his gratitude to the Or-ganizing Committee of the workshop ”Workshop on Algebraic and AnalyticNumber Theory and their Applications” (PN-II-ID-WE - 2012 - 4 -161). Thepublication of this paper is supported by the grant of CNCS-UEFISCDI (Ro-manian National Authority for Scientific Research): PN-II-ID-WE - 2012 - 4-161.

References

[1] B. Apostol, Extremal orders of some functions connected to regular in-tegers modulo n, An. St. Univ. Ovidius Constanta, Vol. 21(2),2013, 5-19.

[2] M. Bencze, On perfect numbers, Studia Mathematica, Univ. Babes-Bolyai, Nr. 4, 1981, 14-18.

[3] G. Hardy, D.E. Littlewood, G. Polya, Inequalities, Cambridge, UniversityPress, 1964.

[4] H.J. Kanold, Uber mehrfach volkommene Zahlen, II. J. Reine Angew.Math., 1957, 197, 82-96.

[5] Octogon Mathematical Magazine (1993-2013).

[6] M. Perisastri, A note on odd perfect numbers, The Mathematics Student26(1958), 179-181.

[7] J. Sandor, B. Cristici, eds.: Handbook of number theory II, Dordrecht,Kluwer Academic, 2004.

[8] W. Sierpinski, Elementary theory of numbers, Warsawa, 1964.

[9] The American Mathematical Monthly, E.2308(1971), E.2162(1969)(Simeon Reich’s note).

Mihaly Bencze,Str. Harmanului 6,,505600 Sacele-Negyfalu, Jud. Brasov, RomaniaEmail: benczemihaly@gmail.com