Divisibility by 3 of Even Multiperfect Numbers of Abundancy 3 and 4

Kevin A. Broughan and Qizhi Zhou University of Waikato Hamilton, New Zealand

                       kab@waikato.ac.nz

Abstract

We say a number is flat if it can be written as a non-trivial power of 2 times an odd squarefree number. The power is the “exponent” and the number of odd primes the “length”. Let N be flat and 4-perfect with exponent a and length m. If a ≢ 1 mod 12, then a is even. If a is even and 3 ∤ N then m is also even. If a ≡ 1 mod 12 then 3 | N and m is even. If N is flat and 3-perfect and 3 ∤ N, then if a ≢ 1 mod 12, a is even. If a ≡ 1 mod 12 then m is odd. If N is flat and 3 or 4-perfect then it is divisible by at least one Mersenne prime, but not all odd prime divisors are Mersenne. We also give some conditions for the divisibility by 3 of an arbitrary even 4-perfect number.

1 Introduction

We say a natural number N is multiperfect of abundancy k (or k-perfect) if σ(N) = kN, (k ≥ 2), where σ(N) denotes the sum of all of the divisors of N. A perfect number is a multiperfect number of abundancy 2.

A number of writers have taken an interest in the presence or absence of divisibility by 3 for classes of multiperfect numbers. Carmichael [3] showed that an even multiperfect number with exactly 4 primes must be divisible by 3, and that there are only two of these numbers, one with abundancy 3 and the other with abundancy 4. If the abundancy is 3 then the number is c_{3} := 2^{9} · 3 · 11 · 31, and if the abundancy is 4 then the number is d_{1} := 2^{5} · 3^{3} · 5 · 7. Kishore [5, 6] and also Hagis [4] showed that any odd perfect number not divisible by 3 must have at least 11 different prime factors. We also have the immediate relationship that if 3 ∤ N, N is 3-perfect if and only if 3N is 4-perfect.

Now consider the table of all known multiperfect numbers of abundancy 3:

c_{1} = 2^{3} · 3 · 5,
c_{2} = 2^{5} · 3 · 7,
c_{3} = 2^{9} · 3 · 11 · 31,
c_{4} = 2^{8} · 5 · 7 · 19 · 37 · 73,
c_{5} = 2^{13} · 3 · 11 · 43 · 127,
c_{6} = 2^{14} · 5 · 7 · 19 · 31 · 151.

Definition 1. We say a number N is flat if its odd part is square free, i.e. if N can be written in the form N = 2^{a} · p_{1} ⋯ p_{m} where a ≥ 0, m ≥ 0 and p_{1} < p_{2} < ⋯ < p_{m}, where the p_{i} are odd primes. If N is flat then the value of a is called its exponent and the value of m its length.

All the known 3-perfect numbers are flat and divisibility of the numbers in this table by 3 is precisely correlated with the parity of the exponent.

Similarly one could consider 4-perfect numbers with the flat shape, for example d_{7} and d_{10} in the partial listing of 4-perfect numbers below:

d_{1} = 2^{5} · 3^{3} · 5 · 7,
d_{2} = 2^{3} · 3^{2} · 5 · 7 · 13,
d_{3} = 2^{2} · 3^{2} · 5 · 7^{2} · 13 · 19,
d_{4} = 2^{9} · 3^{3} · 5 · 11 · 31,
d_{5} = 2^{7} · 3^{3} · 5^{2} · 17 · 31,
d_{6} = 2^{9} · 3^{2} · 7 · 11 · 13 · 31,
d_{7} = 2^{8} · 3 · 5 · 7 · 19 · 37 · 73,
d_{8} = 2^{10} · 3^{3} · 5^{2} · 23 · 31 · 89,
d_{9} = 2^{13} · 3^{3} · 5 · 11 · 43 · 127,
d_{10} = 2^{14} · 3 · 5 · 7 · 19 · 31 · 151.

There are 36 known examples of 4-perfect numbers, of any shape, and they are all even and all divisible by 3. (None of the other over 2000 known multiperfect numbers other than the six 3-perfect and two 4-perfect numbers given above, as of the year this is being written, are flat.) We are able to demonstrate in part the apparent relationships between flatness, the divisibility of the number by 3, the parity of the exponent, and its length.

Theorem 2. Let N be flat and 4-perfect with exponent a and length m. If a ≢ 1 mod 12, then a is even. If a is even and 3 ∤ N then m is also even. If a ≡ 1 mod 12 then 3 | N and m is even.

Theorem 3. Let N be flat and 3-perfect with exponent a and length m and with 3 ∤ N. If a ≢ 1 mod 12 then a is even. If a ≡ 1 mod 12 then m is odd and every odd prime divisor of N is congruent to 1 modulo 3.

Although the table of examples suggests that all even multiperfect numbers of abundancy 4 are divisible by 3, we are not able to show this completely, but have the following conditions:

Theorem 4. Let N be 4-perfect and even and let N = 2^{a}p_{1}^{α_{1}} ⋯ p_{m}^{α_{m}} be its standard prime factorization. Then in the following three cases N is divisible by 3:

(A) If a is odd,

(B) If there exists an i with α_{i} odd and p_{i} ≡ 2 mod 3,

(C) If there exists an i with α_{i} ≡ 2 mod 3 and p_{i} ≡ 1 mod 3.

If 3 | N with a even then necessarily at least one of (A) or (B) or (C) hold.

It is also of some interest to observe the existence of Mersenne primes in the factorizations of the multiperfect numbers. Of course every 2-perfect number must be divisible by a Mersenne prime. We are able to show this persists for flat multiperfect numbers of multiplicities 3 and 4, but that non-Mersenne primes must always be present:

Theorem 5. Let N be even, flat and multiperfect. (A) If the multiplicity is not greater than 4 then N is divisible by at least one Mersenne prime. (B) If all odd prime divisors of N are Mersenne primes then N is perfect.

2 Proofs of Theorems 2 and 3

In order to prove Theorem 2, we need a number of lemmas. First an elementary divisibility result:

Lemma 6. Let d and n be positive integers and p be a prime. Then d + 1 | n + 1 if and only if σ(p^{d}) | σ(p^{n}).

Lemma 7. If N is a flat 4-perfect number with exponent a, then a ≢ 3 mod 4 and a ≢ 5 mod 6.

Proof. (1) Let a ≡ 3 mod 4 and N = 2^{a} · p_{1} ⋯ p_{m}. Since 4 | a + 1 we have, by Lemma 6, 15 = 2^{4} − 1 | 2^{a+1} − 1 so we can write

(2^{a+1} − 1)(p_{1} + 1) ⋯ (p_{m} + 1) = 2^{a+2}p_{1} ⋯ p_{m},
15 · \frac{2^{a+1} − 1}{15}(p_{1} + 1) ⋯ (p_{m} + 1) = 2^{a+2}3 · 5 · p_{3} ⋯ p_{m},
15 · \frac{2^{a+1} − 1}{15} · 2^{2} · 2 · 3 · (p_{3} + 1) ⋯ (p_{m} + 1) = 2^{a+2}3 · 5 · p_{3} ⋯ p_{m},

and therefore 3^{2} divides the right hand side, a contradiction, showing that a ≢ 3 mod 4.

(2) Let a ≡ 5 mod 6 and N = 2^{a} · p_{1} ⋯ p_{m}. Then

(2^{a+1} − 1)(p_{1} + 1) ⋯ (p_{m} + 1) = 2^{a+2}p_{1} ⋯ p_{m}.

Since a + 1 ≡ 0 mod 6, so 5 + 1 | a + 1, so 63 = 3^{2} · 7 = σ(2^{5}) | σ(2^{a}) = 2^{a+1} − 1. So, 3^{2} | p_{1} ⋯ p_{m}, which is a contradiction. Therefore, a ≢ 5 mod 6.

Lemma 8. If N is a flat 4-perfect number with exponent a then a ≢ 9 mod 12.

Proof. Let σ(N) = 4N and a = 12b + 9 with b ≥ 0, then

σ(N) = σ(2^{a})σ(p_{1}) ⋯ σ(p_{m})
= (2^{a+1} − 1)(p_{1} + 1) ⋯ (p_{m} + 1)
= (2^{6b+5} + 1)(2^{6b+5} − 1)(p_{1} + 1) ⋯ (p_{m} + 1)
= 2^{12b+11}p_{1}p_{2} ⋯ p_{m}.

Note that 3 | 2^{6b+5} + 1. If for any i, (2 ≤ i ≤ m), p_{i} ≡ 2 mod 3, then 3 | p_{i} + 1, which implies N has too many 3’s. So we can say p_{i} ≡ 1 mod 3, for all i, (2 ≤ i ≤ m).

Since

2^{6b+5} + 1 = 3(21x + 11) = 3(3y + 2),

where x = 2^{6b−1} + 2^{6b−7} + ⋯ + 2^{5} and y = 7x + 3, then p_{1} = 3, and 3y + 2 is the product of some odd prime factors of N. So 3y + 2 = ∏_{i∈I} p_{i}, where I ⊆ {2, 3, ⋯, m}. Since 3y + 2 ≡ 2 mod 3, but ∏_{i∈I} p_{i} ≡ 1 mod 3, we reach a contradiction.

Therefore, N is not a 4-perfect number, if a ≡ 9 mod 12.

Lemma 9. Let N be flat, 4-perfect with exponent a, N = 2^{a}p_{1} ⋯ p_{m}. If a ≡ 1 mod 12 then 3 | N, for 2 ≤ i ≤ m, p_{i} ≡ 1 mod 3, and m is even.

Proof. Suppose a = 12b + 1, b ≥ 0. We can assume b ≥ 1. Because σ(N) = 4N we have

σ(N) = (2^{12b+2} − 1)(p_{1} + 1)(p_{2} + 1) ⋯ (p_{m} + 1)
= 2^{12b+3}p_{1}p_{2} ⋯ p_{m}, and
σ(2^{a}) = 2^{12b+2} − 1
= 3(21x + 1)
= 3(3y + 1),

where x = 2^{12b−4} + 2^{12b−10} + ⋯ + 2^{2} and y = 7x.

So, p_{1} = 3, and

\frac{(2^{12b+2} − 1)}{3}(p_{2} + 1) ⋯ (p_{m} + 1) = 2^{12b+1}p_{2} ⋯ p_{m}  (1)

If any p_{i} ≡ 2 mod 3, with 2 ≤ i ≤ m, then p_{i} + 1 ≡ 0 mod 3, implying there would be too many 3’s, so for all i with 2 ≤ i ≤ m, we must have p_{i} ≡ 1 mod 3. Now taking the equation (1) modulo 3, we get

∏_{i=2}^{m}(p_{i} + 1) ≡ 2^{m−1} ≡ 2^{a} ≡ 2 mod 3

and therefore 2^{m} ≡ 1 mod 3 so m must be even.

Lemma 10. Let N be flat and 4-perfect with even exponent and suppose also 3 ∤ N. Then the length of N is even.

Proof. Let N = 2^{a}p_{1} ⋯ p_{m} and a = 2b. Since 3 ∤ N, for 1 ≤ i ≤ m each p_{i} ≡ 1 mod 3, and if 2^{β_{i}}∥p_{i} + 1, since (p_{i} + 1)2^{−β_{i}} is a product of primes congruent to 1 modulo 3, it must also be congruent to 1 modulo 3. Thus each β_{i} is odd. Since β_{1} + ⋯ + β_{m} = 2b + 2, m must be even.

Now, we can provide the proof of Theorem 2 as follows:

Proof. Suppose N = 2^{a}p_{1}p_{2} ⋯ p_{m} is a 4-perfect number. By Lemma 7, we know a ≢ 5 mod 6, which implies a ≢ 5 mod 12 and a ≢ 11 mod 12. By Lemma 8 a ≢ 9 mod 12. By Lemma 7 again, since a ≢ 3 mod 4, we have a ≢ 3 mod 12 and a ≢ 7 mod 12. Therefore, since by hypothesis a ≢ 1 mod 12, a must be even. By Lemma 10 if a is even and 3 ∤ N, then m is even. By Lemma 9 if a ≡ 1 mod 12 then 3 | N and m is even.

Lemma 11. If N is a flat 3-perfect number with odd exponent and 3 ∤ N then every odd prime divisor of N is congruent to 1 modulo 3.

Proof. Let N be flat and 3-perfect with N = 2^{a} · p_{1} ⋯ p_{m}, where the exponent a is odd, and suppose that 3 ∤ N. Then

3 · 2^{a} · p_{1} ⋯ p_{m} = σ(2^{a}) · (p_{1} + 1) ⋯ (p_{m} + 1).  (2)

Since a is odd and 2^{2} ≡ 1 mod 3, then σ(2^{a}) = 2^{a+1} − 1 ≡ 0 mod 3. Therefore

2^{a} · p_{1} ⋯ p_{m} = \frac{σ(2^{a})}{3}(p_{1} + 1) ⋯ (p_{m} + 1).  (3)

Since 3 ∤ N, then for all i = 1, 2, ⋯, m, p_{i} ≠ 3.

If there exists a prime factor p_{i} of N with p_{i} ≡ 2 mod 3, for some i ∈ {1, 2, ⋯, m}, then in the right hand side of equation (3), (p_{i} + 1) ≡ 0 mod 3, giving 3 | N, a contradiction.

Theorem 3 is a corollary of Theorem 2:

Proof. Let M = 3N. Then M is a flat 4-perfect number with the same exponent a as N. By Theorem 2, when a ≢ 1 mod 12, a is even. When a ≡ 1 mod 12, again by Theorem 2 the length of M is even so the length of N is odd and, by Lemma 11, every odd prime divisor of N is congruent to 1 modulo 3.

3 Proof of Theorem 4

Definition 12. Let p and q be distinct primes. The exponent of q modulo p, exp_{p} q, is the minimum natural number l such that p | q^{l} − 1, [1, Chapter 10].

For example exp_{2} q = 1 for all odd primes q. If q ≠ 3, exp_{3} q = q mod 3 = \frac{3−(q|3)}{2}, where we have used the least positive residue and (a|b) is the Legendre symbol. If p > q then exp_{p} q > 1. If exp_{p} q > 1 and α > 1 then exp_{p} q | α if and only if p | q^{α} − 1.

Definition 13. If p is a prime and N a natural number let v_{p}(N) be the exponent of the highest power of p dividing N, or 0 if p does not divide N.

We are able to get the following result from Theorems 94 and 95 in Nagell [7, pp. 164–166] or Pomerance [8, p. 269]:

Lemma 14. If p is an odd prime and x > 1 an integer with p | x − 1 then for every e ≥ 1

v_{p} (\frac{x^{e} − 1}{x − 1}) = v_{p}(e).

Lemma 15. (Prime factorization of σ(q^{e}))

Let i ≥ 1 and p be any odd prime, q a prime with q ≥ 2 such that p ≠ q. Then

(1) if exp_{p} q = 1 then p | σ(q^{e}) if and only if p | e + 1, and

(2) if exp_{p} q > 1 then p | σ(q^{e}) if and only if exp_{p} q | e + 1.

Proof. (1) Now exp_{p} q = 1 if and only if p | q − 1. By Lemma 14

v_{p}(σ(q^{e})) = v_{p}(\frac{q^{e+1} − 1}{q − 1}) = v_{p}(e + 1)

and both implications of this part follow directly.

(2) If exp_{p} q > 1 we have p ∤ q − 1 so q − 1 ≢ 0 mod p. Hence p | σ(q^{e}) ⇔ p | q^{e+1} − 1 ⇔ exp_{p} q | e + 1.

Now we are able to prove Theorem 4.

Proof. (A) If N = 2^{a} · p_{1}^{α_{1}} ⋯ p_{m}^{α_{m}} and a is odd, then since

(2^{a+1} − 1)σ(p_{1}^{α_{1}}) ⋯ σ(p_{m}^{α_{m}}) = 2^{a+2}p_{1}^{α_{1}} ⋯ p_{m}^{α_{m}}

and 3 | 2^{a+1} − 1, one of the p_{i} must be 3, so 3 | N.

(B) If one of the α_{i} is odd and the corresponding p_{i} ≡ 2 mod 3, then, since 2 | α_{i} + 1, by Lemma 6, 1 + p_{i} | σ(p_{i}^{α_{i}}), so again 3 | N.

(C) Let us suppose that 3 does not divide N. Let b = p_{1}^{α_{1}} ⋯ p_{m}^{α_{m}} so 3 ∤ b. And, because of point (A) we may assume that a is even. Then the hypothesis σ(N) = 4N gives

(2^{a+1} − 1)σ(b) = 2^{a+2}b

which implies

σ(b) = 2b + \frac{2b}{2^{a+1} − 1}.

Suppose b ≡ 2 mod 3. Then since each divisor d of b satisfies 3 ∤ d, each sum b/d + d ≡ 0 mod 3. But from the equation above, σ(b) ≡ 0 mod 2, so, since each divisor of b is odd, b has an even number of divisors. Arrange them in pairs {b/d, d} and add to show that 3 | σ(b) leading to 3 | b, a contradiction.

Since 3 ∤ b, by what we have just shown this means b ≡ 1 mod 3. Then by the given hypothesis and definition of b, there is a p_{i} ≡ 1 mod 3 and, by (B) if any of the p_{i} ≡ 2 mod 3, then its corresponding α_{i} is even (otherwise 3 | b).

Now consider the equation σ(N) = 4N:

(2^{a+1} − 1)σ(p_{1}^{α_{1}}) ⋯ σ(p_{m}^{α_{m}}) = 2^{a+2} · b

with a even, and take this equation modulo 3. This leads to

(1 + α_{1}) ⋯ (1 + α_{l}) ≡ 1 mod 3,

where, if needed, we have reordered the α_{i} to place the non-empty set of those with p_{i} ≡ 1 mod 3 first, and l is the number of primes congruent to 1 modulo 3. But given an α_{i} ≡ 2 mod 3 we obtain 0 ≡ 1 mod 3, a contradiction which implies therefore 3 | b, so finally 3 | N.

For the necessary condition assume N = 2^{2a} · p_{1}^{α_{1}} ⋯ p_{m}^{α_{m}} and 3 | N. Because 2^{2a+1} − 1 ≡ 1 mod 3 we must have an i with 3 | σ(p_{i}^{α_{i}}). If exp_{3} p_{i} = 1 ⇔ p_{i} ≡ 1 mod 3, we must have, by Lemma 15, 3 | α_{i} + 1 so α_{i} ≡ 2 mod 3 which is (C). If however exp_{3} p_{i} = 2 then 3 ∤ p_{i} − 1 and 3 | p_{i}^{2} − 1, so we must have 2 | α_{i} + 1 so α_{i} is odd and p_{i} ≡ 2 mod 3, which is (B).

Lemma 16. Let N be a flat 3-perfect integer, not divisible by 3 and whose exponent a is even. Let N = 2^{a}p_{1}p_{2} ⋯ p_{m} be its standard prime factorization.

(A) There exists an unique j ∈ {1, 2, ⋯, m} such that 3 divides p_{j} + 1.

(B) For each i, 1 ≤ i < j, p_{i} + 1 divides N.

Proof. (A) We have (2^{a+1} − 1)(p_{1} + 1)(p_{2} + 1) ⋯ (p_{m} + 1) = 3N. Since a is even, 3 ∤ 2^{a+1} − 1. Since 3 ∤ N, there is an unique j ∈ {1, 2, ⋯, m} such that 3 | p_{j} + 1.

(B) By point (A), p_{i} + 1 is coprime with 3 and p_{i} + 1 divides σ(N) = 3N. Thus p_{i} + 1 | N.

4 Mersenne Prime Divisors

Recall the statement of Theorem 5: Let N be even, flat and multiperfect. (A) If the multiplicity is not greater than 4 then N is divisible by at least one Mersenne prime. (B) If all odd prime divisors of N are Mersenne primes then N is perfect.

Proof. Let N = 2^{a}p_{1} ⋯ p_{m} with m ≥ 1.

(A) We can assume that 3 ∤ N. If the multiplicity k = 2 then N = 2^{p−1}M_{p} where p is prime and M_{p} is a Mersenne prime.

Let k = 4. Write

(2^{a+1} − 1)(p_{1} + 1) ⋯ (p_{m} + 1) = 2^{a+2}p_{1} ⋯ p_{m}.

If p_{1} is not Mersenne, the least odd divisor of p_{1} + 1 is an odd prime q < p_{1} which divides p_{1} ⋯ p_{m} and, therefore, divides N. This contradicts the fact that p_{1} is the least odd divisor of N. Thus p_{1} is Mersenne.

Now let k = 3. If a is odd, write

(\frac{2^{a+1} − 1}{3}) (p_{1} + 1) ⋯ (p_{m} + 1) = 2^{a}p_{1} ⋯ p_{m}.

Like in the case k = 4, we deduce from this equation that p_{1} is Mersenne.

If a is even, by Lemma 16, there exists an unique j such that 3 divides p_{j} + 1. Then either 3 is the unique odd prime divisor of p_{j} + 1, either there is an odd prime q_{1} > 3 which divides p_{j} + 1.

In the latter case let us suppose that no prime factor of N less than p_{j} is Mersenne. Then there exists an odd prime factor q_{2} of q_{1} + 1. By Lemma 16, q_{2} | q_{1} + 1 | N. Thus, q_{2} < q_{1} is an another odd prime factor of N less than p_{j}. Repeating this construction we get a decreasing sequence (q_{n}) of odd prime divisors of N. This is absurd.

In the former case, since p_{1} < p_{2} ⋯ < p_{m} and p_{m} ∤ p_{i} + 1 for 1 ≤ i ≤ m, we must have p_{m} | 2^{a+1} − 1, so, assuming j_{o} < m, we can write

(\frac{2^{a+1} − 1}{p_{m}}) (\frac{p_{1} + 1}{2^{α_{1}}}) ⋯ (\frac{p_{j_{o}} + 1}{3 · 2^{α_{o}}}) ⋯ (\frac{p_{m} + 1}{2^{α_{m}}}) = p_{1} ⋯ p_{m−1},

where each α_{i} = v_{2}(p_{i}+1), so that each term on the left is odd. Each of the terms (p_{i}+1)/2^{α_{i}} for i ≠ j_{o} on the left hand side can be written as a non-empty product of distinct primes from {p_{1}, ⋯, p_{m−1}} and there are exactly m − 1 such terms. Therefore we can cancel each from both sides to derive 2^{a+1} − 1 = p_{m}, so p_{m} is Mersenne. Also, we note that (p_{1} + 1)/2^{α_{1}} must be 1, meaning that p_{1} is a Mersenne prime.

If however j_{o} = m, then since there are at least two odd prime divisors [2] then the smallest odd prime divisor of N is Mersenne: canceling the 3’s in the expression

σ(2^{a})(p_{1} + 1) ⋯ 3 · 2^{α_{o}} = 3 · 2^{a}p_{1} ⋯ p_{m}

shows that all potential odd prime divisors of p_{1} + 1 are too large to actually occur so p_{1} must be Mersenne.

(B) Let σ(N) = kN for some k ≥ 2 and suppose that all of the p_{i} are Mersenne. Then

σ(2^{a})(p_{1} + 1) ⋯ (p_{m} + 1) = k · 2^{a} · p_{1} ⋯ p_{m}.

There exist primes q_{i} such that p_{i} = 2^{q_{i}} − 1. Therefore

(2^{a+1} − 1) · 2^{q_{1}} ⋯ 2^{q_{m}} = k · 2^{a} · (2^{q_{1}} − 1) ⋯ (2^{q_{m}} − 1)

so a ≤ q_{1} + ⋯ + q_{m} and for each i, 2^{q_{i}} − 1 | 2^{a+1} − 1. But then Lemma 6 implies q_{i} | a + 1, and, since necessarily these q_{i}’s are distinct primes, q_{1} ⋯ q_{m} | a + 1 giving

q_{1} ⋯ q_{m} ≤ a + 1 ≤ q_{1} + q_{2} + ⋯ q_{m} + 1.

It follows (say by induction on m) that m = 1, therefore N = 2^{a} · p_{1}. Then

(2^{a+1} − 1)(p_{1} + 1) = k2^{a}p_{1}

implies p_{1} | 2^{a+1} − 1 and 2^{a} | p_{1} + 1, so 2^{a} ≤ p_{1} + 1 ≤ 2^{a+1} or

1 ≤ \frac{p_{1} + 1}{2^{a}} ≤ 2.

If (p_{1} + 1)/2^{a} = 1 then p_{1} + 1 = 2^{a}, so p_{1} = 2^{a} − 1 and 2^{a} − 1 | 2^{a+1} − 1, which implies a = 1. It leads to the perfect number 6. If (p_{1} + 1)/2^{a} = 2, then p_{1} = 2^{a+1} − 1 giving k = 2, so N is perfect.

5 Comments

The six flat 3-perfect numbers given in the introduction have been known for over 100 years. There are no flat multiperfect numbers known of abundancy 5 or more, so in addition to the conjecture that all even 4-perfect numbers, flat or otherwise, are divisible by 3, an additional problem in this area is to find an upper bound for the possible multiplicities of flat multiperfect numbers. We have not been able to do this.

6 Acknowledgements

We gratefully acknowledge the helpful contributions made by two referees to improve an earlier version of this paper. The second author was supported by a grant from the New Zealand Institute for Mathematics and its Applications (NZIMA).

References

[1] T. M. Apostol, Introduction to analytic number theory, Springer, New York, 1976.

[2] R. D. Carmichael, Multiply perfect numbers of three different primes, Ann. Math. 8(1) (1906), 49–56.

[3] R. D. Carmichael, Multiply perfect numbers of four different primes, Ann. Math. 8(4) (1907), 149–158.

[4] P. Hagis, Jr., Sketch of a proof that an odd perfect number relatively prime to 3 has at least eleven prime factors, Math. Comp. 40 (1983), 399–404.

[5] M. Kishore, Odd perfect numbers not divisible by 3 are divisible by at least ten distinct primes, Math. Comp. 31 (1977), 274–279.

[6] M. Kishore, Odd perfect numbers not divisible by 3. II, Math. Comp. 40 (1983), 405– 411.

[7] T. Nagell, Introduction to Number Theory, Chelsea, New York, 1964.

[8] C. Pomerance, Odd perfect numbers are divisible by at least seven distinct primes, Acta Arith. 25 (1974), 265–300.

2000 Mathematics Subject Classification: Primary 11A05; Secondary 11A51. Keywords: multiperfect number, flat number, abundancy.

(Concerned with sequences \underline{A005820}, \underline{A027687}.)

Received June 30 2009; revised versions received October 12 2009; January 7 2010. Published in Journal of Integer Sequences, January 8 2010.

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