Those Fascinating Numbers 7

21

• the smallest integer 1 whose sum of divisors is a fifth power: here σ(21) = 25;

• the smallest 2-hyperperfect number: a number n is said to be 2-hyperperfect

if it can be written as n = 1 + 2

d|n

1dn

d, which is equivalent to the condition

2σ(n) = 3n + 1; the sequence of numbers satisfying this property begins as

follows: 21, 2 133, 19 521, 176 661, 129 127 041, . . . ; more generally, a number

n is said to be hyperperfect if there exists a positive integer k such that

n = 1 + k

d|n

1dn

d, (1)

in which case we also say that n is k-hyperperfect

9;

the following table contains

some k-hyperperfect numbers along with their factorization:

9A

1-hyperperfect number is simply a perfect number. lt is easy to show that relation (1) is

equivalent to

kσ(n) = (k + 1)n + (k − 1). (2)

Also, it is clear that a prime power

pα,

with α ≥ 1, cannot be hyperperfect. Furthermore, it follows

immediately from (1) that if n is k-hyperperfect, then n ≡ 1 (mod k) and moreover that

σ(n) = n + 1 +

n − 1

k

. (3)

This last relation proves to be an excellent tool to determine if a given integer n is a hyperperfect

number and also to construct, using a computer, a list of hyperperfect numbers. Indeed, it follows

from (3) that

n is a hyperperfect number ⇐⇒

n − 1

σ(n) − n − 1

is an integer.

It also follows from (3) that the smallest prime factor of such an integer n is larger than k. Indeed,

assume that p|n with p ≤ k. We would then have that n/p is a proper divisor of n, in which case

σ(n) n + 1 +

n

p

≥ n + 1 +

n

k

n + 1 +

n − 1

k

= σ(n),

a contradiction. It follows from this that a hyperperfect number which is not perfect is odd.

On the other hand, if n is a square-free k-hyperperfect number, then k must be even. Assume the

contrary, that is that k is odd. As we just saw, n must be odd, unless k = 1, in which case n would

be perfect and even. But then we would have that n =

2p−1(2p

− 1) for a certain prime number

p ≥ 3, in which case n would not be square-free. We therefore have that n is odd. Now, because of

(2), we have

kσ(n) = 2

k + 1

2

n +

k − 1

2

. (4)

If k ≡ 1 (mod 4), then it follows from (4) that

kσ(n) = 2 (odd + even) = 2 × odd,

while if k ≡ 3 (mod 4), then

kσ(n) = 2 (even + odd) = 2 × odd,

which means that 2 σ(n), in which case n is prime, since n is square-free.