INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 4 (2004), #A11

A VARIATION ON PERFECT NUMBERS

Roger Woodford^{1}

Department of Mathematics, University of Manitoba, Winnipeg, MB R3M 2N2, Canada

                           rogerw@math.ubc.ca

Received: 7/28/03, Revised: 3/4/04, Accepted: 7/6/04, Published: 7/9/04

Abstract

For k ∈ ℕ we define a new divisor function s_{k} called the k^{th} prime symmetric function. By analogy with the sum of divisors function σ, we use the functions s_{k} to consider variations on perfect numbers, namely k-symmetric-perfect numbers as well as k-cycles. We find all k-symmetric-perfect numbers for k = 1, 2, 3. We also consider the problem of whether a natural number n can be expressed in the form s_{k}(n), and show that for n large enough, it always can be for k = 1, 2.

1. Introduction

Definition 1: Let k be a nonnegative integer. We define s_{k} : ℕ → ℕ ∪ {0} as follows: If k = 0, s_{k}(n) ≡ 1. If k > 0, and n = p_{1} ⋯ p_{r}, where r = Ω(n) is the number of prime factors (with multiplicity) of n, then

s_{k}(n) = ∑ p_{i_{1}} ⋯ p_{i_{k}},

where the sum is taken over all products of k prime factors from the set {p_{1}, …, p_{r}}. We say s_{k} is the k^{th} prime symmetric function.

Note that if Ω(n) < k, we have s_{k}(n) = 0.

There is an alternate way of defining the functions s_{k}. Given n = p_{1} ⋯ p_{r} ∈ ℕ, set

S_{n}(x) = ∏_{i=1}^{r}(x + p_{i}).

Then s_{k}(n) is the coefficient of x^{r−k} in S_{n}(x). The empty product is taken to be 1.

Example 1: s_{0}(12) = 1, s_{1}(12) = 2 + 2 + 3 = 7, s_{2}(12) = 2 · 2 + 2 · 3 + 2 · 3 = 16, s_{3}(12) = 12, and s_{4}(12) = 0.

1Supported by an NSERC undergraduate student research award.

Several good texts detailing the basic theory of perfect numbers exist, see for instance [1], [4], [7], and [13]. In addition, many variations on perfect numbers have been defined and studied. For examples, see the remaining references. We now define a new variation of perfect, defective, and excessive numbers using the divisor functions s_{k}.

Definition 2: Let n ∈ ℕ. If s_{k}(n) < n, we say n is k-symmetric-defective. If s_{k}(n) > n, we say n is k-symmetric-excessive. If s_{k}(n) = n, and Ω(n) = k, we say n is trivially k-symmetric-perfect. If s_{k}(n) = n, and Ω(n) > k, we say n is k-symmetric-perfect. If n is k-symmetric-perfect or k-symmetric-excessive, we say n is k-symmetric-special.

Notation: For the sake of brevity we write k-SD for k-symmetric-defective, k-SP for k-symmetric-perfect, k-SE for k-symmetric-excessive, and k-SS for k-symmetric-special.

Example 2: If p is prime, then p^{p} is a (p − 1)-SP number, since

s_{p−1}(p^{p}) = \binom{p}{p − 1}p^{p−1} = p^{p}.

In fact, this example has a form of converse:

Theorem 1: The prime power p^{α} is k-SP if and only if α = p and k = p − 1.

Proof. We have seen that this is sufficient, now suppose k < α, and s_{k}(p^{α}) = p^{α}. Then

\binom{α}{k} = p^{α−k}.  (1)

For 1 < k < α − 1, \binom{α}{k} is divisible by two distinct prime factors, hence we must have k = 1 or α − 1. Now 4 = 2^{2}, is the only 1-SP number, and corresponds to the case where k = 1 = α − 1. Hence we may assume k = α − 1, which from (1) implies that α = p and k = p − 1. This proves the theorem. □

Definition 3: A finite sequence {n_{0}, …, n_{ℓ}} is called a k-cycle if the following conditions are satisfied:

1. ℓ > 1,

2. n_{0}, …, n_{ℓ−1} are distinct and n_{ℓ} = n_{0}, and

3. s_{k}(n_{i}) = n_{i+1}, for i = 0, 1, …, ℓ − 1.

2. Basic Properties of Prime Symmetric Functions

The following proposition is an immediate consequence of the definition.

Proposition 2: If n = p_{1}^{α_{1}} ⋯ p_{r}^{α_{r}}, then

s_{k}(n) = ∑ ; \begin{matrix}i_{1} + ⋯ + i_{r} = k \\ i_{1}, …, i_{r} ≥ 0 \end{matrix} \binom{α_{1}}{i_{1}} ⋯ \binom{α_{r}}{i_{r}}p_{1}^{i_{1}} ⋯ p_{r}^{i_{r}}.

Proposition 3:

s_{k}(mn) = ∑_{i=0}^{k} s_{k−i}(m)s_{i}(n).

Proof. If m = 1, or n = 1, the result is immediate, as it is if k = 0. If k > 0, m = p_{1} ⋯ p_{r}, and n = q_{1} ⋯ q_{s}, let

S = {p_{1}, …, p_{r}, q_{1}, …, q_{s}}.

Then

s_{k}(mn) = ∑_{{r_{1},... ,r_{k}}⊂S} r_{1} ⋯ r_{k}.

We collect the terms of this sum having k − i factors from m, and i factors from n. The sum of these is equal to s_{k−i}(m)s_{i}(n). Summing as i ranges from 0 to k gives the desired result. □

Corollary 4: Let n, k ∈ ℕ, and let p and q be primes, with p < q, and suppose Ω(pn) > k. If pn − s_{k}(pn) > 0, then pn − s_{k}(pn) < qn − s_{k}(qn).

Proof. The following inequality

qn − s_{k}(qn) = qn − qs_{k−1}(n) − s_{k}(n)
> pn − ps_{k−1}(n) − s_{k}(n)
= pn − s_{k}(pn)

is true if n > s_{k−1}(n). But

pn − s_{k}(pn) = pn − ps_{k−1}(n) − s_{k}(n) > 0

by assumption, so

n > s_{k−1}(n) + s_{k}(n)/p > s_{k−1}(n).
□

In searching for k-cycles and k-SP numbers, it is essential to know when s_{k}(n) ≥ n. We search by fixing Ω(n), and systematically checking all products of Ω(n) primes. The corollary tells us that if s_{k}(pn) < pn, then for any q > p, qn is also k-SD.

Lemma 5: Let k, n ∈ ℕ. Then there exists an r > k such that if Ω(n) ≥ r, then n is k-SD. Let r(k) denote the least such r. Then

r(k) = min\{ r : \binom{r}{k} < 2^{r−k} \} .

Proof. There is an r > k such that the function

f(t) = \binom{t}{k}

satisfies f(t) < g(t) for all t ≥ r, where

g(t) = 2^{t−k},

since f is a polynomial, and g is an exponential function. Now suppose t ≥ r, and let p_{1}, …, p_{t} be t primes. Then

\binom{t}{k} = \binom{t}{t − k} < 2^{t−k}, which implies ∑ \frac{1}{p_{i_{1}} ⋯ p_{i_{t−k}}} ≤ \binom{t}{t − k}\frac{1}{2^{t−k}} < 1,

where the sum is taken over all i_{1}, …, i_{t−k} such that 1 ≤ i_{1} < ⋯ < i_{t−k} ≤ t. This implies

∑ p_{i_{1}} ⋯ p_{i_{k}} < p_{1} ⋯ p_{t},

where the sum is taken over all i_{1}, …, i_{k} such that 1 ≤ i_{1} < ⋯ < i_{k} ≤ t. This in turn implies that

s_{k}(p_{1} ⋯ p_{t}) < p_{1} ⋯ p_{t}.

Now we prove the second statement. The inequality

\binom{2k}{k} ≥ 2^{k}

holds for all k ≥ 1, and so r(k) > 2k. This in mind, let r(k) be as claimed in the statement of the theorem. We argue inductively. Let t > r, and suppose that

\binom{t − 1}{k} < 2^{t−1−k}.

Then

2\binom{t − 1}{k} < 2^{t−k}.

Since t > 2k, we have t < 2(t − k), and so

\binom{t}{k} = \frac{t(t − 1) ⋯ (t − k + 1)}{k!} < \frac{2(t − 1)(t − 2) ⋯ (t − k)}{k!} = 2\binom{t − 1}{k}.

Hence

\binom{t}{k} < 2\binom{t − 1}{k} < 2^{t−k},

and the proof is complete by induction. □

The first few values of r(k) are given in the following table:

\begin{matrix}k & r(k) \\ 1 & 3 \\ 2 & 6 \\ 3 & 10 \\ 4 & 14 \\ 5 & 19 \\ 6 & 23 \\ 7 & 27 \\ 8 & 31 \\ 9 & 36 \\ 10 & 40 \end{matrix}

The properties of 1-symmetric-perfection etc. corresponding to the first prime symmetric function s_{1} are easily characterized. The primes are the trivial 1-SP numbers, 4 is the only 1-SP number, and all other numbers are 1-SD. Clearly there are no 1-cycles. We now investigate these properties in the second prime symmetric function.

3. The Second Prime Symmetric Function

Let n be an integer greater than 1. By a family E_{k}(n, r) of k-SS numbers, we mean a set

E_{k}(n, r) = {np_{1} ⋯ p_{r}|p_{1}, …, p_{r} are primes}

such that if m ∈ E_{k}(n, r), then m is k-SS. The family E_{2}(4, 1) of numbers of the form 4p, where p is prime, is one such set, since the elements satisfy s_{2}(4p) = 4p + 4 > 4p. E_{k}(n, 0) merely denotes the singleton set of a k-SS number. To find all 2-SP numbers and all 2-cycles we need to find all numbers n such that 2 < Ω(n) < 6, with s_{2}(n) ≥ n, since r(2) = 6. To do this we use the algorithm mentioned after Corollary 4.

3.1. Ω(n) = 3

s_{2}(2 · 2 · p) = 4p + 4 > 4p,
s_{2}(2 · 3 · p) = 5p + 6 < 6p, when p > 6.

This shows that there are no other infinite families of 2-SS numbers satisfying Ω(n) = 3. Below we find all 2-SS numbers not belonging to this family.

s_{2}(2 · 3 · 3) = 21 > 18,
s_{2}(2 · 3 · 5) = 31 > 30,
s_{2}(2 · 3 · 7) = 41 < 42,
s_{2}(3 · 3 · 3) = 27,
s_{2}(3 · 3 · 5) = 39 < 45.

Thus 27 is the only 2-SP number satisfying Ω(n) = 3. Iterating on the above 2-SE numbers shows none belong to 2-cycles. For example

18 ⟶^{s_{2}} 21 ⟶^{s_{2}} 10 ⟶^{s_{2}} 10 ⟶^{s_{2}} · · · .

3.2. Ω(n) = 4

s_{2}(2 · 2 · 2 · p) = 6p + 12 < 8p, when p > 6.

Thus there are no infinite families of 2-SS numbers with Ω(n) = 4. Iterating on 8p for p = 2, 3, 5, shows that none belong to a 2-cycle. Checking other cases:

s_{2}(2 · 2 · 3 · 3) = 37 > 36,
s_{2}(2 · 2 · 3 · 5) = 51 < 60,
s_{2}(2 · 3 · 3 · 3) = 45 < 54.

Hence there are no 2-SP numbers satisfying Ω(n) = 4. Iterating on the above 2-SE numbers shows that none belong to a 2-cycle.

3.3. Ω(n) = 5

s_{2}(2 · 2 · 2 · 2 · p) = 8p + 24 < 16p, when p > 3.

Thus there are no infinite families of 2-SS numbers with Ω(n) = 5. Iterating on 16p for p = 2, 3, shows that 48 is in fact 2-SP, and 32, which is 2-SE, does not belong to a 2-cycle. Checking other cases:

s_{2}(2 · 2 · 2 · 3 · 3) = 57 < 72.

Hence 48 is the only 2-SP number satisfying Ω(n) = 5. We have proved the following theorem.

Theorem 6: 27 and 48 are the only 2-SP numbers.

Theorem 7: There are no 2-cycles.

Proof. A 2-cycle must have a least element that is 2-SE. We have shown that any such element must belong to the family of numbers of the form 4p. We will show that in all but a few trivial cases s_{2}(s_{2}(4p)) < 4p, giving a contradiction. Now, s_{2}(4p) = 8((p + 1)/2). We may assume that p is odd, and set m = (p + 1)/2. Thus we will have a contradiction if the following holds:

s_{2}(8m) < 8m − 4.

This is equivalent to

12 + 6s_{1}(m) + s_{2}(m) < 8m − 4,  (2)

which is equivalent to

\frac{16}{p_{1} ⋯ p_{s}} + 6 ∑_{i=1}^{s} \frac{1}{p_{1} ⋯ p̂_{i} ⋯ p_{s}} + ∑_{1≤i<j≤s} \frac{1}{p_{1} ⋯ p̂_{i} ⋯ p̂_{j} ⋯ p_{s}} < 8,

where m = p_{1} ⋯ p_{s}. Here p_{1} ⋯ p̂_{i} ⋯ p_{s} is defined to be p_{1} ⋯ p_{s}/p_{i}, and p_{1} ⋯ p̂_{i} ⋯ p̂_{j} ⋯ p_{s} is defined to be p_{1} ⋯ p_{s}/p_{i}p_{j}.

Since p_{i} ≥ 2, this expression is implied by:

\frac{16}{2^{s}} + \frac{6s}{2^{s−1}} + \frac{s(s − 1)}{2} \frac{1}{2^{s−2}} < 8,

which holds for all s ≥ 4. If s = 1, then m = p is prime, and so condition (2) becomes:

12 + 6p < 8p − 4,

which holds for all p > 8. It is easily verified for p = 2, 3, 5 and 7, that 8p does not belong to a 2-cycle.

For s = 2, we can write m = pq. The only values of m for which (2) fails are determined by the prime pairs (p, q) = (2, 2), (2, 3). In both cases, 8m does not belong to a 2-cycle.

Finally for s = 3, if m = pqr, only for the triple (p, q, r) = (2, 2, 2) does m fail to satisfy (2). Again, in this case, 8m does not belong to a 2-cycle. □

Definition 4: A sequence {n_{i}} (finite or infinite) is called a k-ascending sequence if n_{i} < s_{k}(n_{i}) = n_{i+1}. If {n_{i}} = {n_{i}}_{i=0}^{t}, then {n_{i}} is said to have length t.

Remark: The longest 2-ascending sequence is

8 ⟶^{s_{2}} 12 ⟶^{s_{2}} 16 ⟶^{s_{2}} 24 ⟶^{s_{2}} 30 ⟶^{s_{2}} 31.

Definition 5: Let k ∈ ℕ ∪ {0}. We define r_{k} : ℕ → ℕ ∪ {0} by

r_{k}(n) = |{s_{k}^{−1}[{n}]|

Example 3: r_{1}(1) = 0, but for all n ≥ 2, r_{1}(n) ≥ 1. In fact, lim_{n→∞} r_{1}(n) = ∞. To see this, simply set

n = s_{1}(2^{a}3^{b}) = 2a + 3b,

and observe that the number of pairs (a, b) satisfying this equation can be made arbitrarily large for all n sufficiently large.

We prove a weaker result for r_{2}.

Theorem 8: There exists an N ∈ ℕ such that for all m ≥ N, r_{2}(m) ≥ 1.

Proof. It suffices to show that for m sufficiently large, m = s_{2}(2^{a}3^{b}5^{c}7^{d}), for some a, b, c, and d ≥ 0. In general,

s_{2}(2^{a}3^{b}5^{c}7^{d}) = 4\binom{a}{2} + 9\binom{b}{2} + 25\binom{c}{2} + 49\binom{d}{2}

+ 6\binom{a}{1}\binom{b}{1} + 10\binom{a}{1}\binom{c}{1} + 14\binom{a}{1}\binom{d}{1}

+ 15\binom{b}{1}\binom{c}{1} + 21\binom{b}{1}\binom{d}{1} + 35\binom{c}{1}\binom{d}{1}

= \frac{1}{2} [(2a + 3b + 5c + 7d)^{2} − (4a + 9b + 25c + 49d)]

So, given m, we need only find solutions to the equations:

2a + 3b + 5c + 7d = R,
4a + 9b + 25c + 49d = R^{2} − 2m,

with nonnegative integers a, b, c, d, and R ∈ ℕ. These equations are equivalent to:

2a − 10c − 28d = 3R − R^{2} + 2m,  (3)
3b + 15c + 35d = R^{2} − 2R − 2m,  (4)

Since a and b must be nonnegative integers, we have the following necessary and sufficient conditions for a solution to (3) and (4):

1. 2m ≡ R^{2} + R + d (mod 3), 2. R^{2} − 3R − 10c − 28d ≤ 2m,

3. 2m ≤ R^{2} − 2R − 15c − 35d.

Note that equation (3) is always satisfied modulo 2. Condition 1 results from taking equation (4) modulo 3, and conditions 2 and 3 are derived from the fact that a, b ≥ 0.

Consider the interval

I_{R}(c, d) = [R^{2} − 3R − 10c − 28d, R^{2} − 2R − 15c − 35d].

For fixed d, let c_{R}(d) be the least c such that ℓ(I_{R}(c, d)) < 15, where ℓ(I) denotes the length of an interval I. We use the notation L(I) and R(I) to denote the left and right endpoints of an interval I, respectively. Since R(I_{R}(c, d)) = R(I_{R}(c + 1, d)) + 15, when they exist, we have that

⋃_{c=0}^{c_{R}(d)} I_{R}(c, d) = [R^{2} − 3R − 10c_{R}(d) − 28d, R^{2} − 2R − 35d].

Denote the above interval by I_{R}(d). By definition of c_{R}(d),

ℓ(I_{R}(c_{R}(d), d)) = R − 5c_{R}(d) − 7d < 15, so − 10c_{R}(d) < −2R + 30 + 14d,

and c_{R}(d) is the least such c. Consider the interval ⋂_{d=0}^{2} I_{R}(d). Clearly R(⋂_{d=0}^{2} I_{R}(d)) = R^{2} − 2R − 70. We now wish to find an upper bound for L(⋂_{d=0}^{2} I_{R}(d)). From the above inequality, we have that

L(I_{R}(d)) = R^{2} − 3R − 10c_{R}(d) − 28d < R^{2} − 5R − 14d + 30.

Thus

L(⋂_{d=0}^{2} I_{R}(d)) = max{R^{2} − 3R − 10c_{R}(d) − 28d|d = 0, 1, 2}
< max{R^{2} − 5R − 14d + 30|d = 0, 1, 2}
= R^{2} − 5R + 30.

Let J_{R} = [R^{2} − 5R + 30, R^{2} − 2R − 70]. Then J_{R} ⊂ ⋂_{d=0}^{2} I_{R}(d). Now

L(J_{R+1}) ≤ R(J_{R}), if and only if R^{2} − 3R + 26 ≤ R^{2} − 2R − 70,

which holds for all R ≥ 96. So if 2m ≥ L(J_{96}) = 8766, then there is an R ≥ 96 such that 2m ∈ J_{R} ⊂ ⋂_{d=0}^{2} I_{R}(d). Choose d ∈ {0, 1, 2} such that condition 1 is satisfied. Since 2m ∈ I_{R}(d), there is a c ≥ 0 such that 2m ∈ I_{R}(c, d). For these values of R, c, and d, conditions 2 and 3 are satisfied. In other words, there exists an n such that m = s_{2}(n). This completes the proof. □

We end this section with a conjecture.

Conjecture 1: For every k ∈ ℕ, lim_{n→∞} r_{k}(n) = ∞.

4. Higher Prime Symmetric Functions

Theorem 9: (1) Let n ∈ ℕ. If n is k-SS then pn is (k + 1)-SE for every prime p.

(2) If pn is (k + 1)-SS for every prime p, then n is k-SS, and hence by (1), pn is (k + 1)-SE for every prime p.

Proof. (1) Suppose n is k-SS. Then since Ω(n) > k, we have s_{k+1}(n) > 0. So

s_{k+1}(pn) = ps_{k}(n) + s_{k+1}(n)
≥ pn + s_{k+1}(n)
> pn.

(2) If s_{k+1}(pn) = ps_{k}(n) + s_{k+1}(n) ≥ pn, for every prime p, then s_{k}(n) ≥ n − s_{k+1}(n)/p. Letting p → ∞, we have s_{k}(n) ≥ n. □

Corollary 10: For k ∈ ℕ, there are only finitely many k-SP numbers.

Proof. By the previous theorem, any family E_{k+1}(n, r +1) is of the form pE_{k}(n, r), where p ranges over the primes. Furthermore, this family contains only (k + 1)-SE numbers. There are only finitely many (k + 1)-SS numbers not belonging to any such family. □

Thus the infinite families of 3-SE numbers are: E_{3}(4, 2), E_{3}(16, 1), E_{3}(18, 1), E_{3}(24, 1), E_{3}(27, 1), E_{3}(30, 1), E_{3}(32, 1), E_{3}(36, 1), E_{3}(40, 1), E_{3}(48, 1).

By exhaustive search (as was done with k = 2), all other 3-SS numbers can be found. They constitute the following set:

{42p|p = 7, 11, …, 41} ∪ {56p|p = 7, 11, …, 43} ∪ {64p|p = 2, 3, …, 37} ∪
{726, 858, 250, 350, 225, 315, 968, 1144, 300, 420, 162, 270, 378, 243, 400, 560,
216, 360, 504, 324, 288, 480, 672, 432, 256, 384, 640, 576, 512, 768}.

None of the elements in the above sets are 3-SP, hence there are no 3-SP numbers. The diversity of possible 3-ascending sequences makes it difficult to rule out the existence of 3-cycles as we did 2-cycles. This is illustrated in the following example.

Example 4: If p_{1}, q_{1} are odd primes, then s_{3}(4p_{1}q_{1}) = 4(p_{1}q_{1} + p_{1} + q_{1}). It is possible that p_{1}q_{1} + p_{1} + q_{1} = p_{2}q_{2}, where p_{2}, q_{2} are again odd primes, and so on. Several such sequences exist the longest one with p_{1}q_{1} < 50000, and p_{i}, q_{i} > 3 is:

184892 = 4 · 17 · 2719 ⟶^{s_{3}} 195836= 4 · 173 · 283 ⟶^{s_{3}} 197660 = 4 · 5 · 9883
⟶^{s_{3}} 237212 = 4 · 31 · 1913 ⟶^{s_{3}} 244988 = 4 · 73 · 839 ⟶^{s_{3}} 248636 = 4 · 61 · 1019
⟶^{s_{3}} 252956 = 4 · 11 · 5749 ⟶^{s_{3}} 275996 = 4 · 7 · 9857 ⟶^{s_{3}} 315452 = 4 · 17 · 4639
⟶^{s_{3}} 334076 = 4 · 47 · 1777 ⟶^{s_{3}} 341372= 4 · 31 · 2753 ⟶^{s_{3}} 352508 = 4 · 13 · 6779
⟶^{s_{3}} 379676 = 4 · 11 · 8629 ⟶^{s_{3}} 414236= 4 · 29 · 3571 ⟶^{s_{3}} 428636 = 4 · 13 · 8243
⟶^{s_{3}} 461660 = 4 · 5 · 41 · 563.

It seems highly unlikely, however, that a 3-ascending sequence be infinite. This is part of our final conjecture: Conjecture 2: Any k-ascending sequence is finite.

References

[1] Burton, D. M., Elementary Number Theory, Allan and Bacon, Inc. Boston-London-Sydney, 1976.

[2] Chandran, V. R., On generalized unitary perfect numbers, Math. Student, 61 (1992), 54-56.

[3] Cohen G. L. and te Riele, H. J. J., Iterating the sum of divisors function, Experiment. Math., 5 (1996), 91-100.

[4] Dunham, W., Euler The Master of Us All, The Mathematical Association of America, 1999.

[5] Hardy, B. E. and Subbarao, M. V., On hyperperfect numbers, Congr. Numer., 42 (1984), 183-198.

[6] Hunsucker, J. L. and Pomerance, C., There are no odd super perfect numbers less than 7 × 10^{24}, Indian J. Math. 17 (1975), 107-120.

[7] Loweke, G. P., The Lore of Prime Numbers, Vantage Press, New York-Washington-Atlanta-Los Angeles-Chicago, 1982.

[8] McCranie, J. S., A study of hyperperfect numbers, J. of Integer Seq., 3 (2000), 153-157.

[9] Minoli, D., Issues in nonlinear hyperperfect numbers, Mathematics of Computation, 34 (1980), 639-645.

[10] Pomerance, C., Multiply perfect numbers, Mersenne primes and effective computability, Math. Ann. 226 (1977), 195-206.

[11] Suryanarayana, D., Superperfect Numbers, Elem. Math., 24 (1967), 16-17.

[12] te Riele, H. J. J., Hyperperfect numbers with three different prime factors, Mathematics of Computation, 36 (1981), 297-298.

[13] Tattersall, J. J., Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999.