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.
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