Determining all Biunitary Triperfect Numbers of a Certain Form

Tomohiro Yamada Center for Japanese Language and Culture Osaka University 562-8678, 3-5-10, Sembahigashi Minoo, Osaka Japan

                     tyamada1093@gmail.com

Abstract

A divisor d of an integer N is called a unitary divisor of N if d and N/d are relatively prime and a biunitary divisor of N if d and N/d have no common unitary divisor except 1. An integer N is called a biunitary triperfect number if the sum σ^{∗∗}(N) of biunitary divisors of N is equal to 3N. We show that 2160 is the only biunitary triperfect number divisible by 27 = 3^{3}.

1 Introduction

As usual, we let σ(N) and ω(N) denote, respectively, the sum of divisors and the number of distinct prime factors of a positive integer N (throughout this paper, we consider only positive divisors). We call a positive integer N perfect if σ(N) = 2N. It is a well-known unsolved problem whether or not an odd perfect number exists. Interest to this problem has produced many analogous concepts and problems concerning divisors of an integer. For example, it is also unknown whether or not there exists an odd multiperfect number, an integer dividing the sum of its divisors. We call a positive integer N k-perfect if σ(N) = kN. Ordinary perfect numbers are 2-perfect numbers and multiperfect numbers are k-perfect numbers for some positive integer k. A 3-perfect number such as 120, 672, and 523776 is also called triperfect, given in sequence \underline{A005820} in the On-Line Encyclopedia of Integer Sequences (OEIS).

On the other hand, some special classes of divisors have also been studied in several papers. One of them is the class of unitary divisors, a divisor d of a positive integer N relatively prime to N/d. As far as we know, Vaidyanathaswamy first mentioned this class of divisors with the term “block-factor” in [10, p. 601, Example 4]. Later, Cohen [2] called such a divisor d of N a unitary divisor. Wall [11] introduced the notion of biunitary divisors, a divisor d of a positive integer N satisfying gcd_{1}(d, N/d) = 1, where gcd_{1}(a, b) is the greatest common unitary divisor of two positive integers a and b.

According to Cohen [2] and Wall [11], respectively, we let σ^{∗}(N) and σ^{∗∗}(N) denote the sum of unitary and biunitary divisors of N. Moreover, we write d || N if d is a unitary divisor of N. Hence, for a prime p, we have p^{e} || N if p divides N exactly e times. Replacing σ by σ^{∗}, Subbarao and Warren [9] introduced the notion of unitary perfect numbers by calling N unitary perfect if σ^{∗}(N) = 2N. They proved that there are no odd unitary perfect numbers, and 6, 60, 90, and 87360 are the first four unitary perfect numbers. Nine years later, the fifth unitary perfect number was found by Wall [12], but no further instance has been found. Subbarao [8] conjectured that there are only finitely many unitary perfect numbers.

Similarly, a positive integers N is called biunitary perfect if σ^{∗∗}(N) = 2N. Wall [11] showed that 6, 60, and 90, the first three unitary perfect numbers, are the only biunitary perfect numbers.

Combining the notion of multiperfect numbers and biunitary perfect numbers, Hagis [4] introduced the notion of biunitary multiperfect numbers, integers N such that σ^{∗∗}(N) = kN for some positive integer k and proved that there is no odd biunitary multiperfect number. Smallest instances for σ^{∗∗}(N) = kN with k ≥ 3 are N = 120, 672, 2160, … with k = 3 and N = 30240, 1028160, 6168960, … with k = 4. All biunitary multiperfect numbers below 4.66 × 10^{12} as well as many larger ones are given in \underline{A189000}.

Now we can call an integer N satisfying σ^{∗∗}(N) = kN biunitary k-perfect and biunitary 3-perfect numbers biunitary triperfect. Haukkanen and Sitaramaiah [5] determined all biunitary triperfect numbers N such that 2^{a} || N with 1 ≤ a ≤ 6 or a = 8, and such ones with a = 7 under several conditions. In this paper, we determine all biunitary triperfect numbers divisible by 27 = 3^{3}.

Theorem 1. There exists the only one biunitary triperfect number N = 2160 divisible by 3^{3}.

Comparison of results on ordinary, unitary, and biunitary divisors such as Wall’s result mentioned above leads us to expect that certain biunitary problems are more accessible than corresponding ordinary and unitary problems. Indeed, we proved [14] that M = 2 and 9 are the only biunitary superperfect numbers, integers satisfying σ^{∗∗}(σ^{∗∗}(N)) = 2N, while the corresponding equation σ^{∗}(σ^{∗}(N)) = 2N has many solutions 2, 9, 165, 238, … as given in \underline{A038843} and we only proved [13] that 9 and 165 are the only odd solutions.

Indeed, our proof is completely elementary, based on the idea used in [14]. If σ^{∗∗}(N) = 3N with a factorization N = ∏_{i} p_{i}^{e_{i}}, then a prime p ≠ 3 dividing σ^{∗∗}(p_{i}^{e_{i}}) must divide N since σ^{∗∗}(p_{i}^{e_{i}}) | σ^{∗∗}(N) = 3N. We see that σ^{∗∗}(2^{e}3^{f})/(2^{e}3^{f}) tends to 3 as e and f grows and for e and f large, σ^{∗∗}(2^{e}) and σ^{∗∗}(3^{f}) produce new prime factors of N. In many cases, this causes σ^{∗∗}(N)/N > 3, which is a contradiction (in [14], we focused on only one prime 2 to arrive at the contradiction σ^{∗∗}(σ^{∗∗}(N))/N > 2 in many cases). In other cases, we are led to a contradiction that p^{f+1} | N under the assumption that p divides N exactly f times or σ^{∗∗}(N)/N < 3.

2 Preliminary lemmas

In this section, we give several preliminary lemmas concerning the sum of biunitary divisors used to prove our main theorems.

Before all, we recall two basic facts from [11]. The sum of biunitary divisors function σ^{∗∗} is multiplicative. Moreover, if p is a prime and e is a positive integer, then

σ^{∗∗}(p^{e}) = \begin{cases}p^{e} + p^{e−1} + ⋯ + 1 = \frac{p^{e+1}−1}{p−1}, & if e is odd; \\ \frac{p^{e+1}−1}{p−1} − p^{e/2} = \frac{(p^{e/2}−1)(p^{e/2+1}+1)}{p−1}, & if e is even. \end{cases}  (1)

We note that, putting e = 2s − 1 − δ with δ ∈ {0, 1}, this can be represented by the single formula

σ^{∗∗}(p^{e}) = \frac{(p^{s−δ} − 1)(p^{s} + 1)}{p − 1},  (2)

provided that e > 0.

From these facts, we can deduce the following lemmas almost immediately. We note that these lemmas were already proved in our previous paper [14]. However, we would like to give proofs for self-containedness.

Lemma 2. Let n be a positive integer. Then, σ^{∗∗}(n) is odd if and only if n is a power of 2 (including 1). More exactly, σ^{∗∗}(n) is divisible by 2 at least ω(n) times if n is odd and at least ω(n) − 1 times if n is even.

Proof. Whether e is even or odd, σ^{∗∗}(p^{e}) is odd if and only if p = 2 by (2). Factoring n = 2^{e} ∏_{i=1}^{r} p_{i}^{e_{i}} into distinct odd primes p_{1}, p_{2}, …, p_{r} with e ≥ 0 and e_{1}, e_{2}, …, e_{r} > 0, each σ^{∗∗}(p_{i}^{e_{i}}) is even. Hence, σ^{∗∗}(n) = σ^{∗∗}(2^{e}) ∏_{i=1}^{r} σ^{∗∗}(p_{i}^{e_{i}}) is divisible by 2 at least r times, where r = ω(n) if n is odd and ω(n) − 1 if n is even.

Lemma 3. For any prime p and any positive integer e, σ^{∗∗}(p^{e})/p^{e} ≥ 1 + 1/p^{2}. Moreover, σ^{∗∗}(p^{e})/p^{e} ≥ 1 + 1/p unless e = 2 and σ^{∗∗}(p^{e})/p^{e} ≥ (1 + 1/p)(1 + 1/p^{3}) if e ≥ 3. More generally, for any positive integers m and e ≥ 2m − 1, we have σ^{∗∗}(p^{e})/p^{e} ≥ σ^{∗∗}(p^{2m})/p^{2m} and, unless e = 2m, σ^{∗∗}(p^{e})/p^{e} ≥ 1 + 1/p + ⋯ + 1/p^{m}.

Proof. If e ≥ 2m − 1 and e is odd, then p^{e}, p^{e−1}, …, p, 1 are biunitary divisors of p^{e}. If e > 2m and e is even, then p^{e}, p^{e−1}, …, p^{e−m} are biunitary divisors of p^{e} since e − m > e/2. Hence, if e ≥ 2m − 1 and e ≠ 2m, then σ^{∗∗}(p^{e}) = p^{e} + p^{e−1} + ⋯ + 1 > p^{e} + ⋯ + p^{e−m} = p^{e}(1+1/p+⋯+1/p^{m}). Since σ^{∗∗}(p^{2m})/p^{2m} < 1+1/p+⋯+1/p^{m}, σ^{∗∗}(p^{e})/p^{e} with e ≥ 2m−1 takes its minimum value at e = 2m.

Now we quote the following lemma of Bang [1], which has been rediscovered (and extended to numbers of the form a^{n} − b^{n}) by many authors such as Zsigmondy [15], Dickson [3] and Kanold [6]. Also see Shapiro [7, Theorem 6.4A.1].

Lemma 4. If a ≥ 2 and n ≥ 2 are integers, then a^{n} − 1 has a prime factor which does not divide a^{m} − 1 for any m < n, unless (a, n) = (2, 1), (2, 6) or n = 2 and a + 1 is a power of 2. Furthermore, such a prime factor must be congruent to 1 modulo n.

As a corollary, we obtain the following lemma:

Lemma 5. If a ≥ 2 and n ≥ 1 are integers, then a^{n} + 1 has a prime factor which does not divide a^{m} + 1 for any m < n unless (a, n) = (2, 3). Furthermore, such a prime factor must be congruent to 1 modulo 2n.

Proof. By Lemma 4, a^{2n} − 1 = (a^{n} − 1)(a^{n} + 1) has a prime factor p which does not divide a^{m} − 1 for any m < 2n. Since p does not divide a^{n} − 1, the prime p must divide a^{n} + 1. On the other hand, for m < n, p cannot divide a^{m} + 1 since p does not divide a^{2m} − 1. Finally, such a prime factor p must be congruent to 1 modulo 2n.

Now we prove that σ^{∗∗}(2^{e}) and σ^{∗∗}(3^{f}) must produce a new prime factor which is not very large.

Lemma 6. Let f be a positive integer and write f = 2t − 1 − η with η ∈ {0, 1} and t a positive integer. If f ≥ 5, then at least one of the following statements holds.

(A) η = 0 and σ^{∗∗}(3^{f}) has a prime factor p_{1} with 5 < p_{1} ≤ (3^{t} − 1)/2,

(B) p_{1} = 5 | σ^{∗∗}(3^{f}) and f ≡ 2 (mod 4) or f = 7, 8,

(C) σ^{∗∗}(3^{f}) has a prime factor p_{1} with 5 < p_{1} ≤ \sqrt{(3^{t−η} − 1)/2}, or

(D) 4 divides t, η = 1, and p_{1} = (3^{t−1} − 1)/2 is prime.

Proof. We begin by observing that σ^{∗∗}(3^{f}) = (3^{t} + 1)(3^{t−η} − 1)/2 from (2). We put t = 2^{h}v with v odd.

If η = 0 and f ≠ 7, then, we see that t = 3 or t ≥ 5. Hence, Lemma 4 implies that (3^{t} −1)/2 has a prime factor greater than 5. Defining p_{1} to be the smallest among such prime factors, we have 5 < p_{1} ≤ (3^{t} − 1)/2 and (A) holds. If f = 7, then σ^{∗∗}(3^{f}) = (3^{8} − 1)/2 = 2^{4} × 5 × 41 and (B) holds.

Henceforth we assume that η = 1. If t is odd and t ≠ 5, 9, then, (t − 1)/2 = 3 or (t − 1)/2 ≥ 5 and, like above, (3^{(t−1)/2} − 1)/2 has a prime factor greater than 5. Take the smallest p_{1} among such prime factors. Then, we have p_{1} ≤ (3^{(t−1)/2} − 1)/2 < \sqrt{(3^{t−1} − 1)/2} and p_{1} | (3^{(t−1)/2} − 1)/2 | (3^{t−1} − 1)/2 | σ^{∗∗}(3^{f}). Hence, (C) holds.

If t ≡ 2 (mod 4), then 5 | (3^{t} + 1) | σ^{∗∗}(3^{f}) and (B) holds.

If 4 | t, then, defining p_{1} to be the smallest prime factor of (3^{t−1} − 1)/2, clearly p_{1} = (3^{t−1} − 1)/2 is prime or p_{1} ≤ \sqrt{(3^{t−1} − 1)/2}. Since t − 1 is odd, we can never have p_{1} = 2 or p_{1} = 5. Thus we see that p_{1} = (3^{t−1} − 1)/2 is prime or 5 < p_{1} ≤ \sqrt{(3^{t−1} − 1)/2}, implying (D) and (C), respectively.

If t = 9, then 7 | (3^{3} + 1) | (3^{9} + 1) | σ^{∗∗}(3^{f}) and (C) holds. Finally, if t = 5, then σ^{∗∗}(3^{f}) = σ^{∗∗}(3^{8}) = 2^{5} × 5 × 61 and (B) holds.

Lemma 7. Let e ≥ 6 be an integer and write e = 2s − 1 − δ with δ ∈ {0, 1} and s a positive integer. If e ≠ 8, 12, then at least one of the following statements holds.

(a) δ = 0 and σ^{∗∗}(2^{e}) has at least two prime factors q_{1} and q_{2} with 5 < q_{1}, q_{2} ≤ 2^{s} − 1,

(b) p_{2} = 5 | σ^{∗∗}(2^{e}).

(c) σ^{∗∗}(2^{e}) has at least two prime factors q_{1} and q_{2} each of which satisfies either q_{i} = 2^{s−1} − 1, q_{i} = 2^{s} + 1, or 5 < q_{i} ≤ \sqrt{2^{s} − 3}. Moreover, if q_{i} = 2^{s−1} − 1 or q_{i} = 2^{s} + 1 for i = 1 or 2, then 4 divides s and δ = 1.

Proof. We begin by observing that σ^{∗∗}(2^{e}) = (2^{s} + 1)(2^{s−η} − 1) from (2). We put s = 2^{g}m with m odd.

If δ = 0, m > 1, and e ≠ 11, 23, then s ≠ 6, 12 and Lemma 4 yields that 2^{s}−1 has a prime factor q_{1} ≡ 1 (mod s) and 2^{2s} − 1 has a prime factor q_{2} ≡ 1 (mod 2s) not dividing 2^{s} − 1 or 2^{2^{g+1}} −1. Clearly, we see that q_{1} ≠ q_{2} and q_{2} divides (2^{s} +1)/(2^{2^{g}} +1). Moreover, since s ≥ 3 is odd, we see that q_{1} ≡ q_{2} ≡ 1 (mod 2s) and q_{2} divides (2^{s} + 1)/(2^{2^{g}} + 1) < 2^{s} − 1. Then, we have 5 < q_{1}, q_{2} ≤ 2^{s} −1 and (a) holds. If e = 11 or e = 23, then 7×13 | (2^{12} −1) | σ^{∗∗}(2^{f}) and therefore we can take (q_{1}, q_{2}) = (7, 13), which yields (a) again.

If δ = 0, m = 1 and s = 2^{g} ≥ 16, then we work as above but with 2^{s/2} − 1 and 2^{s} − 1 instead of 2^{s} − 1 and 2^{2s} − 1, respectively. Now we can take two prime factors q_{1}, q_{2} of 2^{s} − 1 with 5 < q_{1}, q_{2} ≤ 2^{s} − 1, which yields (a). If e = 7 or e = 15, then 5 | (2^{8} − 1) | σ^{∗∗}(2^{e}) and (b) holds.

Henceforth, assume that δ = 1. If s is odd and e ≠ 8, 12, 16, 24, then s ≠ 5, 7, 9, 13 and we see from Lemma 4 that we can take a prime factor q_{1} of 2^{(s−1)/2} − 1 such that q_{1} ≡ 1 (mod (s − 1)/2) and a prime factor q_{2} of 2^{(s−1)/2}+1 such that q_{2} ≡ 1 (mod (s − 1)/2). Since 2^{(s−1)/2} − 1 and 2^{(s−1)/2} + 1 are relatively prime, we have q_{1} ≠ q_{2}. Now 5 < q_{1}, q_{2} ≤ 2^{(s−1)/2} + 1 < \sqrt{2^{s} − 3} and (c) holds.

If g = 1, then 5 = (2^{2} + 1) | (2^{s} + 1) | σ(2^{e}).

Assume that g ≥ 2. Let q_{1} and q_{2} be the smallest prime factors of 2^{s−1} − 1 and 2^{s} + 1, respectively. Since s − 1 is odd, we cannot have q_{1} = 3 or q_{1} = 5. Similarly, since 4 divides s, we cannot have q_{2} = 3 or q_{2} = 5.

Since 2^{s} is square and 2^{s−1} − 1 ≡ 3 (mod 4), neither of 2^{s} + 1, 2^{s} − 1, nor 2^{s−1} − 1 can be square. Hence, we see that either q_{1} = 2^{s−1} − 1 is prime or q_{1} ≤ \sqrt{2^{s} − 3}. Similarly, q_{2} = 2^{s} + 1 is prime or q_{2} ≤ \sqrt{2^{s} − 3}. Hence, we see that (c) holds.

If e = 24, then we can take (q_{1}, q_{2}) = (7, 13) since 7×13 | (2^{12}−1) | σ^{∗∗}(2^{24}) to obtain (c). Finally, if e = 16, then we can take (q_{1}, q_{2}) = (17, 19) since 17×19 | (2^{8}−1)(2^{9}+1) = σ^{∗∗}(2^{16}) to obtain (c) observing that p_{2} = 19 < \sqrt{2^{9} − 3}.

We also use the following miscellaneous divisibility results.

Lemma 8.

(I) For any prime p and g ≥ 2, σ^{∗∗}(p^{g}) has a prime factor ≥ 5.

(II) For any odd prime p and g ≥ 4, σ^{∗∗}(p^{g}) has a prime factor ≥ 7.

(III) If f ≥ 18 or f = 13, 14, then σ^{∗∗}(3^{f}) has at least two distinct prime factors ≥ 127.

(IV) If g ≥ 2, then σ^{∗∗}(13^{g}) has at least two distinct prime factors in addition to 2, 3, 41, and 547.

(V) For any g ≥ 1, σ^{∗∗}(41^{g}) has a prime factor in addition to 2, 3, 5, and 13.

Proof.

(I) We write g = 2u − 1 − γ with γ ∈ {0, 1}. For g ≥ 2, we have u ≥ 2 and, by Lemma 5, p^{u} + 1 has a one prime factor ≥ 5.

(II) We write g = 2u − 1 − γ with γ ∈ {0, 1}. For g ≥ 4, we have u ≥ 3 and, by Lemma 5, p^{u} + 1 has a one prime factor ≥ 7.

(III) By Lemma 4, (3^{s} − 1)/2 has at least one prime factor ≥ 127 for s ≥ 126 and, by Lemma 5, 3^{t} + 1 has at least one prime factor ≥ 127 for t ≥ 126. Now, we can confirm that (3^{s} − 1)/2 has at least one prime factor ≥ 127 for s ≥ 7 and 3^{t} + 1 has at least one prime factor ≥ 127 for t ≥ 10 (see, for example, https://homes.cerias. purdue.edu/~ssw/cun/pmain125.txt). Hence, putting f = 2t−1−η with η ∈ {0, 1}, σ^{∗∗}(3^{f}) = (3^{t} + 1)(3^{t−η} − 1)/2 has at least two distinct prime factors ≥ 127 for f ≥ 18. For f = 13, 14, (I) can be confirmed by noting that σ^{∗∗}(3^{13}) = 2^{2} × 547 × 1093 and σ^{∗∗}(3^{14}) = 2 × 17 × 193 × 547 .

(IV) With the aid of Lemmas 4 and 5, we see that (13^{s} − 1)/12 has at least one prime factor ≥ 11 for s ≥ 3 and 13^{t} + 1 has at least one prime factor ≥ 11 for t ≥ 2 (these can be confirmed by direct calculation for 3 ≤ s ≤ 6 and 2 ≤ t ≤ 5). If 41 divides 13^{t} + 1, then t ≡ 20 (mod 40) and 14281 also divides 13^{t} + 1. Similarly, if 41 divides (13^{s} − 1)/12, then 14281 also divides (13^{s} − 1)/12. If 547 divides (13^{u} − 1)/12, then 21 | s and 61 also divides (13^{s} −1)/12. Since 13^{21} ≡ 1 (mod 547), 547 can never divide 13^{t} + 1. Hence, for g ≥ 5, putting g = 2u − 1 − γ with γ ∈ {0, 1}, each of 13^{u} + 1 and (13^{u−γ} − 1)/12 has at least one prime factor p_{1} and p_{2}, respectively, both other than 2, 3, 7, 41, or 547. If p_{1} = p_{2}, then p_{1} must divide 13^{γ} + 1 and therefore p_{1} = 2 or p_{1} = 7, which is a contradiction. Hence, σ^{∗∗}(13^{g}) = (13^{u} + 1)(13^{u−γ−1} − 1)/12 has at least two prime factors besides 2, 3, 41, and 547 for g ≥ 5. For 2 ≤ g ≤ 4, (IV) can be easily confirmed.

(V) For p = 41, from Lemma 5, we see that 41^{u} + 1 has a prime factor greater than 5 for u ≥ 3. If 13 divides 41^{u} + 1, then u ≡ 6 (mod 12) and 29 also divides 41^{u} + 1. Hence, for g ≥ 2, putting g = 2u − 1 − γ with γ ∈ {0, 1}, 41^{u} + 1 has a prime factor in addition to 2, 3, 5, and 13 and so does σ^{∗∗}(41^{g}). For g = 1, (V) is clear since σ^{∗∗}(41) = 41 + 1 = 2 × 3 × 7.

3 Proof of Theorem 1

Assume that N is a biunitary triperfect number with 3^{3} | N. We put integers e and f by 2^{e} || N and 3^{f} || N and write e = 2s − 1 − δ with δ ∈ {0, 1} and f = 2t − 1 − η with η ∈ {0, 1}.

In this section, once we write p_{i} for a prime factor of N with an index i, e_{i} denotes the exponent of p_{i} dividing N. Clearly, σ^{∗∗}(p_{i}^{e_{i}}) divides 3N for each i. Thus, if p is a prime other than 3 and p^{g} divides σ^{∗∗}(p_{i}^{e_{i}}) for some i, then p^{g} divides N.

We begin with small e and f. Although Haukkanen and Sitaramaiah [5] proved that e ≥ 7 whether 3^{3} | N or not, we give a proof for small e in the view of self-containedness.

Lemma 9. We must have e ≥ 4.

Proof. We cannot have e = 0 as Hagis [4] has shown. Indeed, if N > 1 is odd, then, by Lemma 2, σ^{∗∗}(N) must be even and σ^{∗∗}(N) ≠ 3N. If e = 1, then N can have no odd prime factor other than 3 and therefore we must have σ^{∗∗}(3^{f}) = 2^{e}. By Lemma 4, we must have f = 1, contrary to the assumption that f ≥ 3.

If e = 2, then p_{1} = 5 = σ^{∗∗}(2^{2}) | σ^{∗∗}(N) = 3N and therefore 5 must divide N. From Lemma 2, we see that N has no further prime factor. Hence, σ^{∗∗}(3^{f}) can have no prime factor other than 2 and 5. By Lemma 8 (II), we must have f = 3. Now e_{1} ≥ 2 since 5^{2} | σ^{∗∗}(2^{2}3^{3}) | 3N. However, from Lemma 8 (I), we see that p_{2} | σ^{∗∗}(5^{e_{1}}) | σ^{∗∗}(N) = 3N for some p_{2} > 5 and 2^{3} | N by Lemma 2, which is a contradiction.

If e = 3, then p_{1} = 5 must divide N. If e_{1} ≠ 2, then, with the aid of Lemma 3, we have

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{3}3^{4}5)}{2^{3}3^{4}5} = \frac{28}{9} > 3,  (3)

contrary to the assumption that σ^{∗∗}(N) = 3N. Hence, we must have e_{1} = 2 and p_{2} = 13 must divide N since 13 | σ^{∗∗}(5^{2}) | σ^{∗∗}(N) = 3N. With the aid of Lemma 8 (II), we see that σ^{∗∗}(13^{e_{2}}) has an odd prime factor p_{3} in addition to 3, 5, and 13. By Lemma 2, we must have 2^{4} | σ^{∗∗}(3^{f}5^{2}13^{e_{2}}p_{3}^{e_{3}}) | 3N and 2^{4} | N, which is a contradiction. Thus, we cannot have e = 3.

Lemma 10. We cannot have e = 5.

Proof. Assume that e = 5. Then, σ^{∗∗}(2^{5}) = 3^{2} × 7 and p_{1} = 7 must divide N. If e_{1} ≠ 2, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{5}3^{4}7)}{2^{5}3^{4}7} = \frac{28}{9} > 3  (4)

and if f ≠ 4 and e_{1} = 2, then 5 | σ^{∗∗}(p_{1}^{e_{1}}) = 7^{2} + 1 and

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{5}3^{6}7^{2}5^{2})}{2^{5}3^{6}7^{2}5^{2}} = \frac{6929}{2268} > 3,  (5)

which are both contradictions. If f = 4 and e_{1} = 2, then p_{2} = 5 divides N and 2^{6} | σ^{∗∗}(3^{4}7^{2}5^{e_{2}}) | 3N, which is a contradiction.

Lemma 11. We cannot have e = 6.

Proof. Assume that e = 6. We observe that p_{1} = 7 and p_{2} = 17 must divide N since 7 × 17 = σ^{∗∗}(2^{6}) | σ^{∗∗}(N) = 3N. If e_{1} ≠ 2 and e_{2} ≠ 2, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{6}3^{4}7 × 17)}{2^{6}3^{4}7 × 17} = \frac{28}{9} > 3,  (6)

which is a contradiction. If e_{1} ≠ 2 and e_{2} = 2, then, since 5 divides 17^{2} + 1, p_{3} = 5 must divide N and

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{6}3^{4}5^{2}7)}{2^{6}3^{4}5^{2}7} = \frac{6188}{2025} > 3,  (7)

which is a contradiction again.

Now we must have e_{1} = 2. We see that p_{3} = 5 and e_{3} ≥ 2 since 5^{2} | (7^{2} + 1) = σ^{∗∗}(7^{2}). We observe that we cannot have f = 4 since 2^{7} | σ^{∗∗}(3^{4}7^{2}17^{e_{2}}5^{e_{3}}). If e_{3} > 2, then

\frac{σ^{∗∗}(N)}{N} > \frac{σ^{∗∗}(2^{6}3^{6}7^{2}5)}{2^{6}3^{6}7^{2}5} = \frac{45305}{13608} > 3,  (8)

which is a contradiction. Hence, e_{3} = 2 and p_{4} = 13 must divide N. However, this is impossible. Indeed, if e_{4} = 2, then 5^{3} | σ^{∗∗}(7^{2}13^{2}) | 3N, contrary to e_{3} = 2, and if e_{4} ≠ 2, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{6}3^{6}7^{2}5^{2}13)}{2^{6}3^{6}7^{2}5^{2}13} = \frac{9061}{2916} > 3,  (9)

which is a contradiction.

Lemma 12. We cannot have e ≥ 7 and f = 3.

Proof. Assume that e ≥ 7 and f = 3. Clearly p_{1} = 5 must divide N since 5 | σ^{∗∗}(3^{3}) | 3N. If e_{1} ≠ 2, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{3}5)}{2^{8}3^{3}5} = \frac{55}{16} > 3,  (10)

which is a contradiction.

If e_{1} = 2, then, since σ^{∗∗}(5^{e_{1}}) = 2 × 13, p_{2} = 13 must divide N. We must have e_{2} = 2, p_{3} = 17 (we note that 17 | σ^{∗∗}(13^{e_{2}}) = σ^{∗∗}(13^{2})), and e_{3} = 2 since otherwise

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{3}5^{2}13)}{2^{8}3^{3}5^{2}13} = \frac{77}{24} > 3  (11)

or

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{3}5^{2}13^{2}17)}{2^{8}3^{3}5^{2}13^{2}17} = \frac{165}{52} > 3,  (12)

which is a contradiction. But then, we must have 5^{3} | σ^{∗∗}(3^{3}13^{2}17^{2}) | 3N, contrary to e_{1} = 2.

Lemma 13. We cannot have e ≥ 7 and f = 4.

Proof. Assume that e ≥ 7 and f = 4. Clearly p_{1} = 7 must divide N since 7 | σ^{∗∗}(3^{4}) | 3N. If e_{1} ≠ 2, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{4}7)}{2^{8}3^{4}7} = \frac{55}{18} > 3,  (13)

which is a contradiction.

If e_{1} = 2, then, since σ^{∗∗}(7^{e_{1}}) = 2 × 5^{2}, p_{2} = 5 must divide N. We must have e_{2} = 2, p_{3} = 13, and e_{3} = 2, since otherwise

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{4}7^{2}5)}{2^{8}3^{4}7^{2}5} = \frac{275}{84} > 3  (14)

or

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{4}7^{2}5^{2}13)}{2^{8}3^{4}7^{2}5^{2}13} = \frac{55}{18} > 3,  (15)

which is a contradiction. Hence, we must have 5^{3} | N since 5^{3} | σ^{∗∗}(7^{2}13^{2}), which is also a contradiction.

Now we prove remaining cases. Assume that f ≥ 5, e ≥ 7 and e ≠ 8, 12. Then, a prime p_{1} taken from Lemma 6 must divide N since p_{1} | σ^{∗∗}(3^{f}) | 3N and p_{1} ≠ 3. If (b) in Lemma 7 holds, then p_{2} = 5 must divide N and we have p_{1} > 5 = p_{2}. If (a) or (c) in Lemma 7 holds, then we can choose a prime p_{2} ∈ {q_{1}, q_{2}} not equal to p_{1}. Like above, p_{2} must divide N. We see that if (c) in Lemma 7 holds, then we have either (c′) 5 < p_{2} ≤ \sqrt{2^{s} − 3} or (d) 4 | s, δ = 1, and either of p_{2} = 2^{s−1} − 1 or p_{2} = 2^{s} + 1 is prime. We put N_{1} = 2^{e}p_{2}^{e_{2}} if (a), (b), or (c′) holds or (d) holds with e_{2} ≠ 2. It is clear that N_{1} || N. If (d) holds with e_{2} = 2, then, we observe that p_{i} ≡ 2 (mod 5) and 5 | (p_{i}^{2} + 1) | σ^{∗∗}(N) = 3N and thus we see that N_{1} = 2^{e}5^{e_{3}} || N. Similarly, we put N_{2} = 3^{f}p_{1}^{e_{1}} if (A), (B), or (C) holds or (D) holds with e_{2} ≠ 2 and N_{2} = 3^{f}5^{e_{3}} if (D) holds with e_{2} = 2 to see that N_{2} || N.

We observe that σ^{∗∗}(N_{1})/N_{1} > 2 and σ^{∗∗}(N_{2})/N_{2} > 3/2. Indeed, if (a) holds, then

\frac{σ^{∗∗}(N_{1})}{N_{1}} ≥ \frac{(2^{e+1} − 1)(p_{2}^{2} + 1)}{2^{e}p_{2}^{2}} ≥ \frac{(2^{2s} − 1)((2^{s} − 1)^{2} + 1)}{2^{2s−1}(2^{s} − 1)^{2}} > 2.  (16)

If (c′) holds, then

\frac{σ^{∗∗}(N_{1})}{N_{1}} ≥ \frac{(2^{s−1} − 1)(2^{s} + 1)(p_{2}^{2} + 1)}{2^{2s−2}p_{2}^{2}} ≥ \frac{(2^{s−1} − 1)^{2}(2^{s} − 2)}{2^{2s−3}(2^{s} − 3)} > 2.  (17)

If (d) holds with e_{2} ≠ 2, then

\frac{σ^{∗∗}(N_{1})}{N_{1}} ≥ \frac{(2^{s−1} − 1)(2^{s} + 1)(p_{2} + 1)}{2^{2s−2}p_{2}} ≥ \frac{(2^{s−1} − 1)(2^{s} + 2)}{2^{2s−2}} > 2.  (18)

If (b) holds, then

\frac{σ^{∗∗}(N_{1})}{N_{1}} ≥ \frac{(2^{s−η} − 1)(2^{s} + 1)(5^{e_{2}} + 1)}{2^{2s−1−η}5^{e_{2}}} ≥ \frac{26(2^{s−η} − 1)(2^{s} + 1)}{25 × 2^{2s−1−η}} > 2,  (19)

which also holds with e_{2} replaced by e_{3} in the case (d) with e_{2} = 2. Thus, we see that σ^{∗∗}(N_{1})/N_{1} > 2 in any case.

Similarly, if (A) holds, then

\frac{σ^{∗∗}(N_{2})}{N_{2}} ≥ \frac{(3^{f+1} − 1)(p_{1}^{2} + 1)}{2 × 3^{f}p_{1}^{2}} ≥ \frac{(3^{2t} − 1)((3^{t} − 1)^{2} + 4)}{2 × 3^{2t−1}(3^{t} − 1)^{2}} > \frac{3}{2}.  (20)

If (C) holds, then

\frac{σ^{∗∗}(N_{2})}{N_{2}} ≥ \frac{(3^{t−η} − 1)(3^{t} + 1)(p_{1}^{2} + 1)}{2 × 3^{2t−1−η}p_{1}^{2}} ≥ \frac{(3^{t−η} − 1)(3^{t} + 1)(3^{t−η} + 1)}{2 × 3^{2t−1−η}(3^{t−η} − 1)} > \frac{3}{2}.  (21)

If (D) holds with e_{1} ≠ 2, then

\frac{σ^{∗∗}(N_{2})}{N_{2}} ≥ \frac{(3^{t−1} − 1)(3^{t} + 1)(p_{1} + 1)}{2 × 3^{2t−2}p_{1}} ≥ \frac{(3^{t−1} − 1)(3^{t} + 1)(3^{t−1} + 1)}{2 × 3^{2t−2}(3^{t−1} − 1)} > \frac{3}{2}.  (22)

If (B) holds, then

\frac{σ^{∗∗}(N_{2})}{N_{2}} ≥ \frac{(3^{t−η} − 1)(3^{t} + 1)(5^{e_{1}} + 1)}{2 × 3^{2t−1−η}5^{e_{1}}} ≥ \frac{26(3^{t−1} − 1)(3^{t} + 1)}{2 × 25 × 3^{2t−1−η}} > \frac{3}{2},  (23)

which also holds with e_{1} replaced by e_{3} in the case (D) with e_{1} = 2. Thus, we see that σ^{∗∗}(N_{2})/N_{2} > 3/2 in any case.

If (a) or (c′) holds, then gcd(N_{1}, N_{2}) = 1 and

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(N_{1})}{N_{1}} × \frac{σ^{∗∗}(N_{2})}{N_{2}} > 3,  (24)

which is a contradiction. If (b) or (d) holds and (A) or (C) also holds, then, we have gcd(N_{1}, N_{2}) = 1 and a contradiction (24) again.

Now we settle two cases (i) (b) or (d) holds and (B) or (D) also holds and (ii) e = 8. In both cases, 5 must divide N since 5 divides σ^{∗∗}(2^{8}) = 3^{2} × 5 × 11. We rewrite p_{1} = 5. If e_{1} ≠ 2, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{6}5)}{2^{8}3^{6}5} = \frac{5863}{1728} > 3,  (25)

which is a contradiction. Hence, e_{1} = 2.

Now, rewriting p_{2} = 13, p_{2} must divide N. Hence, if e ≠ 8, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{10}3^{6}5^{2}13^{2})}{2^{10}3^{6}5^{2}13^{2}} > 3  (26)

and, if e = 8 and f ≠ 6, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{8}5^{2}13^{2})}{2^{8}3^{8}5^{2}13^{2}} > 3  (27)

and we have a contradiction. Thus, we must have e = 8 and f = 6. Since f = 6, p_{3} = 7 must divide N. If e_{3} = 2, then 5^{3} | σ^{∗∗}(2^{8}7^{2}) | 3N, which is incompatible with e_{1} = 2. If e_{3} ≠ 2, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{8}3^{6}7)}{2^{8}3^{6}7} = \frac{29315}{9072} > 3,  (28)

a contradiction again.

Now we assume that e = 12. Then, p_{1} = 7 must divide N since 7 | (2^{6} − 1) | σ^{∗∗}(2^{12}). If e_{1} ≠ 2, then

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{12}3^{6}7)}{2^{12}3^{6}7} = \frac{22919}{6912} > 3,  (29)

and, if e_{1} = 2, then 5 divides N and

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{12}3^{6}7^{2}5^{2})}{2^{12}3^{6}7^{2}5^{2}} = \frac{297947}{96768} > 3.  (30)

Thus, we have a contradiction.

Now the only remaining case is the case e = 4. We prove that if e = 4, then N = 2160 to complete the proof of Theorem 1. We begin by showing that if N ≠ 2160, then f ≥ 5 and 5 cannot divide N.

Lemma 14. If e = 4 and f ≤ 4, then N = 2160.

Proof. If f = 4, then σ^{∗∗}(3^{f}) = 2^{4}7 and therefore p_{1} = 7 must divide N and 2^{5} | σ^{∗∗}(3^{f}7^{e_{1}}) | 3N, contrary to the assumption that e = 4.

If f = 3, then σ^{∗∗}(3^{f}) = 2^{3}5 and p_{1} = 5 must divide N. If e_{1} = 1, then σ^{∗∗}(2^{4}3^{3}5) = 2^{4}3^{4}5 and therefore N = 2^{4}3^{3}5. If e_{1} = 2, then p_{1}^{2} + 1 = 2 × 13 and p_{2} = 13 must divide N, which is impossible since 2^{5} | σ^{∗∗}(3^{3}5^{2}13^{e_{2}}). If e_{1} ≥ 3, then

\frac{σ^{∗∗}(N)}{N} > \frac{σ^{∗∗}(2^{4}3^{3}5)}{2^{4}3^{3}5} = 3,  (31)

which is a contradiction. Hence, if f = 3, then we have N = 2^{4}3^{3}5.

Now we must have f ≥ 5 if N ≠ 2160.

Lemma 15. If e = 4 and f ≥ 5, then 5 cannot divide N.

Proof. Assume that p_{1} = 5 divides N. If e_{1} ≠ 2, then we must have f = 6 since otherwise

\frac{σ^{∗∗}(N)}{N} > \frac{σ^{∗∗}(2^{4}3^{3}5)}{2^{4}3^{3}5} = 3,  (32)

which is a contradiction. Noting that 13 | σ^{∗∗}(3^{6}) | 3N, p_{2} = 13 must divide N, e_{2} = 2, p_{3} = 17 must divide N and e_{3} = 2 since otherwise

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{4}3^{6}5 × 13)}{2^{4}3^{6}5 × 13} = \frac{287}{90} > 3  (33)

or

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{4}3^{6}5 × 13^{2}17)}{2^{4}3^{6}5 × 13^{2}17} = \frac{41}{13} > 3,  (34)

a contradiction again. Since 5^{2} | σ^{∗∗}(13^{2}17^{2}) and we assumed that e_{1} ≠ 2, we must have e_{1} ≥ 3 and

\frac{σ^{∗∗}(N)}{N} ≥ \frac{σ^{∗∗}(2^{4}3^{6}5^{4}13^{2})}{2^{4}3^{6}5^{4}13^{2}} = \frac{4879}{1625} > 3,  (35)

which is impossible. Thus, we must have e_{1} = 2 and p_{2} = 13 must divide N since 13 | σ^{∗∗}(5^{2}) | 3N.

If f ≥ 9, then, with the aid of Lemma 8 (III), we see that σ^{∗∗}(3^{f}) must have two odd prime factors p_{3} and p_{4} in addition to 5 and 13, which is incompatible with e = 4 by Lemma 2.

If f = 8, then 61 | σ^{∗∗}(3^{f}) and p_{3} = 61 must divide N. Noting that 2^{2} | (3^{5} + 1) | σ^{∗∗}(3^{8}), we must have 2^{5} | σ^{∗∗}(3^{8}5^{2}13^{e_{2}}61^{e_{3}}), which is a contradiction. If f = 7, then we have a similar contradiction from p_{3} = 41 | (3^{4} + 1) | σ^{∗∗}(3^{f}). If f = 6, then p_{3} = 41 must divide N again. By Lemma 8 (V), σ^{∗∗}(41^{e_{3}}) has an odd prime factor p_{4} other than 3, 5, p_{3} and, we must have 2^{5} | σ^{∗∗}(3^{6}5^{2}13^{e_{2}}41^{e_{3}}p_{4}^{e_{4}}), which is a contradiction.

If f = 5, then 7 | σ^{∗∗}(3^{5}) and p_{3} = 7 must divide N. However, we must have 2^{5} | σ^{∗∗}(3^{5}5^{2}13^{e_{2}}5^{e_{3}}) | 3N from Lemma 2, contrary to e = 4. Thus we conclude that 5 cannot divide N.

Now we prove that we can never have σ^{∗∗}(N) = 3N if N ≠ 2160 and e = 4. By Lemmas 14 and 15, we must have f ≥ 5 and 5 can never divide N.

If f ≥ 18 or f = 13, 14, then, By Lemma 8 (IV), σ^{∗∗}(3^{f}) has at least two odd prime factors p_{1}, p_{2} ≥ 127. By Lemma 2, we see that N can have at most one more prime factor p_{3}. Thus, we must have

\frac{σ^{∗∗}(N)}{N} < \frac{27 × 3 × 7 × 127 × 131}{16 × 2 × 6 × 126 × 130} < 3,  (36)

which is a contradiction.

If 15 ≤ f ≤ 17, then σ^{∗∗}(3^{f}) has at least three odd prime factors not less than 31 and

\frac{σ^{∗∗}(N)}{N} < \frac{27 × 3 × 31 × 37 × 41}{16 × 2 × 30 × 36 × 40} < 3,  (37)

which is a contradiction. If f = 7, 8, 10, 11, then 5 | σ^{∗∗}(3^{f}) and 5 must divide N, which is impossible by Lemma 15.

If f = 6, 12, then p_{1} = 13 and p_{2} = 41, 547, respectively, must divide N. By Lemma 4, we see that σ^{∗∗}(13^{e_{1}}) can have at most one prime factor other than 2, 3, or p_{2}, which is impossible for f = 6 and f = 12 by Lemma 8 (IV). If f = 9, then, 11 × 61 must divide N and, noting that 2^{2} | σ^{∗∗}(3^{9}), N can have no prime factor except 2, 3, 11, or 61. Hence, σ^{∗∗}(61^{e_{2}}) has no prime factor except 2, 3, or 11, which is impossible since 61^{5} + 1 = 2 × 11 × 31 × 1238411.

Finally, if f = 5, then σ^{∗∗}(3^{5}) = 2^{2} × 7 × 13 and therefore p_{2} = 7 and p_{3} = 13 must divide N. If e_{3} ≥ 2, then σ^{∗∗}(13^{e_{3}}) has another prime factor p_{4} in addition to 2, 3, and 7 from Lemma 8 (IV) and 2^{5} | σ^{∗∗}(3^{5}7^{e_{2}}13^{e_{3}}p_{4}^{e_{4}}) | 3N, which is a contradiction. If e_{3} = 1, then we have a similar contradiction that 2^{5} | σ^{∗∗}(3^{5}7) | 3N.

Hence, we can never have σ^{∗∗}(N) = 3N if N ≠ 2160 and e = 4. This completes the proof of Theorem 1.

4 Concluding remarks

Our arguments in this paper and [14] lead us to strongly believe that the equations σ^{∗∗}(σ^{∗∗}(N)) = 3N and σ^{∗∗}(σ^{∗∗}(N)) = 4N have no solutions other than the known twelve solutions given in \underline{A318175}. Moreover, based on our theorem and known biunitary multiperfect numbers, we can pose the following problems, which are so far unsolved:

• For each integer k ≥ 3, are there infinitely or only finitely many integers N for which σ^{∗∗}(N) = kN?

• For a prime power p^{g} ≠ 1, does there exist at least one N for which σ^{∗∗}(N) = kN and p^{g} || N?

• For an integer d > 2, does there exist at least one integer N for which σ^{∗∗}(N) = kN and gcd(d, N) = 1?

Problems involving σ^{∗∗}(σ^{∗∗}(N)) are posed in [14].

5 Acknowledgment

We are thankful for referee’s helpful comments, in particular on the work of Vaidyanathaswamy [10].

References

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[8] M. V. Subbarao, Are there an infinity of unitary perfect numbers?, Amer. Math. Monthly 77 (1970), 389–390.

[9] M. V. Subbarao and L. J. Warren, Unitary perfect numbers, Canad. Math. Bull. 9 (1966), 147–153.

[10] R. Vaidyanathaswamy, The theory of multiplicative arithmetic functions, Trans. Amer. Math. Soc. 33 (1931), 579–662.

[11] Charles R. Wall, Bi-unitary perfect numbers, Proc. Amer. Math. Soc. 33 (1972), 39–42.

[12] Charles R. Wall, The fifth unitary perfect number, Canad. Math. Bull. 18 (1975), 115– 122.

[13] Tomohiro Yamada, Unitary super perfect numbers, Math. Pannon. 19 (2008), 37–47.

[14] Tomohiro Yamada, 2 and 9 are the only biunitary superperfect numbers, Ann. Univ. Sci. Budapest. Sect. Comput. 48 (2018), 247–256.

[15] K. Zsigmondy, Zur Theorie der Potenzreste, Monatsh. Math. 3 (1892), 265–284.

2020 Mathematics Subject Classification: Primary 11A25; Secondary 11A05. Keywords: odd perfect number, biunitary superperfect number, unitary divisor, biunitary divisor, sum of divisors.

(Concerned with sequences \underline{A000396}, \underline{A005820}, \underline{A038843}, \underline{A188999}, \underline{A189000}, \underline{A318175}, \underline{A318781}.)

Received June 27 2025; revised version received August 6 2025. Published in Journal of Integer Sequences, August 6 2025.

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