Must a Primitive Non-Deficient Number Always Have a Small Component?

Joshua Zelinsky Department of Mathematics Hopkins School New Haven, CT 06515 USA

                     jzelinsky@hopkins.edu

Abstract Let n be a primitive non-deficient number where n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} where p_{1}, p_{2}, …, p_{k}

are distinct primes. We prove that there exists an i such that

p_{i}^{a_{i}+1} < 2k(p_{1}p_{2}p_{3} ⋯ p_{k}).

We conjecture that in fact one can always find an i such that p_{i}^{a_{i}+1} < 2p_{1}p_{2}p_{3} ⋯ p_{k}.

1 Introduction

Let σ_{d}(n) be the sum of the dth powers of divisors of n. We write σ(n) for σ_{1}(n). Given a positive integer n, we write h(n) = σ_{−1}(n) = \frac{σ(n)}{n}. We write

H(n) = ∏_{p|n, p prime} \frac{p}{p − 1}.

In what follows, we write n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} where p_{1}, p_{2}, …, p_{k} are distinct primes and the a_{i} are positive integers. We write rad(n) to be the radical of n. That is, rad(n) = p_{1}p_{2} ⋯ p_{k}. We also set R = rad(n).

Recall, a number n is said to be deficient if σ(n) < 2n. Deficient numbers form sequence \underline{A005100} in the OEIS (On-Line Encyclopedia of Integer Sequences). If n is non-deficient, then so is any multiple of n. Thus, there is an interest in non-deficient numbers which are primitive in the sense that such a number is non-deficient and also has no non-deficient divisors. Primitive non-deficient numbers form sequence \underline{A006039}. They have been studied since at least Dickson [8] who proved that for any fixed k there are only finitely many odd primitive non-deficient numbers. An asymptotic for the number of primitive non-deficient numbers which are at most x is still open. The best known bounds are due to Avidon [3]. In a different direction, the non-deficient numbers have a positive density and there are estimates of their natural density. Kobayashi [14], showed that their natural density is at least 0.24760444, while Deléglise [7] obtained an upper bound of 0.248. More recently, a preprint by McNew and Setty [17] obtained that their natural density is at least 0.247619608 and at most 0.247619658.

It is a classical observation dating to the 19th century that the functions h(n) and H(n) are closely linked. In particular, h(n) ≤ H(n) with equality if and only if n = 1. Moreover, H(n) is the best upper bound one can have on h(n) that depends only on the distinct prime factors of n. In particular,

lim_{t→∞} h(n^{t}) = H(n) = H(rad(n)).

We tentatively suspect that the following conjecture holds:

Conjecture 1. If n is a primitive non-deficient number with n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}}, and p_{1} < p_{2} < ⋯ < p_{k}, then there exists an i, 1 ≤ i ≤ k such that p_{i}^{a_{i}+1} < 2R.

Conjecture 1 is false if 2R is replaced by R, although the author knows of only one counterexample, namely, n = 4164647056479 = 3^{7}7^{4}13^{3}19^{2}, where R = 5187, and the smallest of p_{i}^{a_{i}+1} is 3^{8} = 6561. It may be that this is the only counterexample. A slightly weaker version of Conjecture 1 would replace 2R with p_{1}R, and we are more confident in that conjecture.

We prove the following weaker result:

Theorem 2. Let n be a primitive non-deficient number with n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} with p_{1} < p_{2} < ⋯ < p_{k}. Let R = rad(n) = p_{1}p_{2} ⋯ p_{k}. Then there is an i such that

p_{i}^{a_{i}+1} < 2kR.  (1)

It is plausible that the right-hand side of Inequality (1) can be replaced with p_{1}R, which would still be weaker than Conjecture 1.

Conjecture 1 and Theorem 2 are motivated by three interrelated lines of thought.

First, we have the following intuition: if n is a non-deficient number (primitive or not), then one must have 2 ≤ h(n) < H(n), but if all of the a_{i} are “large” then for each i, h(p_{i}^{a_{i}}) will be close to H(p_{i}), and thus h(n) will be very close to H(n). Thus, if the a_{i} are large enough, then for at least one p_{j}, h(\frac{n}{p_{j}}) will still be very close to H(n), and thus \frac{n}{p_{j}} will be non-deficient (since it will be very close to H(n)) and so n will not be primitive. Thus, if the p_{i} are all fixed and n is in fact primitive, then at least one of the p_{i}^{a_{i}} must be small.

Conjecture 1 and Theorem 2 are essentially making this intuition precise. The radical of n is a natural function to make this notion precise since the radical exactly contains the information of what the p_{i} are.

Second, one concern in some recent work concerning odd perfect numbers has been the graph representation of the components of the odd perfect number. In particular, given an odd perfect number n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}}, we can construct a weighted directed graph G with k vertices labeled p_{1}, p_{2}, …, p_{k} with an edge from p_{i} to p_{j} of weight m where p_{i}^{m} || σ(p_{j}^{a_{j}}). Recent papers have used this graph framework or similar frameworks. One example is a paper Bibby, Vyncke, and the author [4] which proved new upper bounds on the size of the third largest prime factor of an odd perfect number. A similar graph approach was used in a paper of Yamada [21] to produce new results about solutions to the equation

σ(n) = rad(n)^{2},  (2)

where it is currently unknown if the only solutions are n = 1 and n = 1782. Other notable work on solutions to Equation (2) also has been done by Broughan, De Koninck, Kátai, and Luca, [6] with related work by Luca [15]. In this context, if n is an odd perfect number, then Conjecture 1 and Theorem 2 can be thought of as bounds on the out-degree of vertices in this graph.

The third line of thought also concerns odd perfect numbers. Over the last few years, multiple papers have developed inequalities relating an odd perfect number n to rad(n). Acquaah and Konyagin [1] proved that R < 2n^{\frac{2}{3}} , and Luca and Pomerance [16] proved that R < 2n^{\frac{17}{26}} . Subsequently, Klurman [13] proved that there is a constant c (independent of n) such that R < cn^{\frac{9}{14}} , and it is not hard to show that one may take c = 2 in Klurman’s argument. Ellia [9] showed that under some reasonable technical assumptions one must have R < n^{\frac{1}{2}} , and Ochem and Rao [10] extended Ellia’s results further.

All of these results are upper bounds for R in terms of n. But one can also prove upper bounds for n in terms of R. In particular, Nielsen [18] proved that if n is an odd perfect number then

n < R^{2^{k}−1}.  (3)

An immediate consequence of Equation (3) is that if n is an odd perfect number then there exists an 1 ≤ i ≤ k such that p_{i}^{a_{i}} || n, and where

p_{i}^{a_{i}+1} < R^{\frac{2^{k}}{k}} .  (4)

Thus Conjecture 1 and Theorem 2 can be thought of as generalizations of Equation (4), in that the inequality is tighter and applies also to all primitive non-deficient numbers.

Now, with all of these motivations, a reader may reasonably wonder why look at an inequality involving p_{i}^{a_{i}+1} and R rather than p_{i}^{a_{i}} and R? There are two answers for this, one somewhat unsatisfying and one hopefully more satisfying. The less satisfying answer is that the techniques used in this paper naturally seem to give inequalities involving p_{i}^{a_{i}+1}. The more satisfying answer is that these inequalities are equivalent to an inequality for p_{i}^{a_{i}} in terms of all the other prime factors of n. That is, if one has an inequality of the form

p_{i}^{a_{i}+1} < f(k)R,

for some function f(k) then this is the same as

p_{i}^{a_{i}} < f(k) ∏_{j=1,j≠i}^{k} p_{i},

so this translates to an inequality for a component in terms of the product of all the other prime factors.

2 Main results

We first recall the following basic fact about non-deficient numbers which has a straightforward proof.

Lemma 3. Let n be a number of the form n = 2^{a}p where p is an odd prime. Then n is a primitive non-deficient number if and only if 2^{a} < p ≤ 2^{a+1} − 1. Also, all odd numbers with exactly two distinct prime divisors are deficient.

One consequence of Lemma 3 we prove is that Theorem 2 holds whenever n is even and i = 1.

Lemma 4. Let n be an odd primitive non-deficient number with n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}}, and p_{1} < p_{2} < ⋯ < p_{k}. Assume further that k = 3 or k = 4 then there exists an i, 1 ≤ i ≤ k such that p_{i}^{a_{i}+1} < 2R.

Proof. There are no odd non-deficient numbers with fewer than 2 distinct prime factors, and the only odd primitive non-deficient numbers with 3 distinct prime factors are with a little computation seen to be just the elements of the set

{945, 1575, 2205, 7425, 78975, 131625, 342225, 570375}

and one can check that all elements of this set satisfy the desired inequality. The set for odd primitive non-deficient numbers with 4 distinct prime factors is longer, but this is still a straightforward computation.

The next lemma is old, with a slightly weaker version due to Servais [19]. Although this lemma is often phrased just for odd perfect numbers, the lemma applies to any non-deficient number.

Lemma 5. Let n be a non-deficient number with n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} with primes p_{1} < p_{2} < ⋯ < p_{k}. Then

\frac{3}{2}p_{1} − 2 < k.  (5)

Additionally, if p_{1} ≥ 5, then

p_{1} + 2 ≤ k.  (6)

Proof. Inequality (5) is essentially just the argument from Grün [11].

The second inequality follows immediately from Inequality (5) for p_{1} ≥ 11, so it only needs to be verified for p_{1} = 5 and p_{1} = 7.

We verify it for p_{1} = 5. The proof for p_{1} = 7 is nearly identical. Assume n is a nondeficient number with n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} with primes p_{1} < p_{2} < ⋯ < p_{k}, and that p_{1} = 5. Assume further that k ≤ 6. Since n is non-deficient, we have

2 ≤ h(n) < H(n) = ∏_{i=1}^{k} \frac{p_{i}}{p_{i} − 1}.  (7)

Now, since \frac{x}{x−1} is a decreasing function and k ≤ 6, and p_{1} = 5, we have from Equation (7),

2 < \frac{5}{4} \frac{7}{6} \frac{11}{10} \frac{13}{12} \frac{17}{16} \frac{19}{18} < 2,  (8)

which is a contradiction. So k ≥ 7.

Note that a stronger version of Inequality (5) can be found in a prior paper by the author [22], and similarly tighter bounds have been found by others [2, 20] but we do not need it here. We have then the following result for even primitive non-deficient numbers.

Proposition 6. Let n be an even primitive non-deficient number, with R = rad(n). Let a be the integer such that 2^{a} || n. Then 2^{a+1} < R. Thus every even primitive non-deficient number satisfies the radical inequality.

Proof. Let n be an even primitive non-deficient number. Let a be the positive integer such that 2^{a} || n. If n were a power of 2, it would be deficient, so n must have at least one odd prime divisor. Call that divisor p. Consider then m = 2^{a}p. Since n is a primitive non-deficient number, it must be the case that m is deficient. Then, from Lemma 3, we must have 2^{a} < p, so 2^{a+1} < 2p ≤ R.

Even primitive non-deficient numbers are much more common than odd primitive nondeficient numbers, so we can say that Conjecture 1 is at least true for most primitive nondeficient numbers. We prove a weaker version of the conjecture but we first need a few lemmas.

Lemma 7. If k and m are real numbers such that k ≥ 2 and 0 < m < \frac{1}{k(k+1)}, then

\frac{1}{1 − km} < 1 + (k + 1)m.  (9)

Proof. Assume that k and m are real numbers such that k ≥ 2 and 0 < m < \frac{1}{k(k+1)}. Then, since km < 1, we have that 1 − km is positive, and so our desired inequality is equivalent to showing that

1 < (1 − km)(1 + (k + 1)m) = 1 + m(1 − k(k + 1)m),

and thus our inequality is equivalent to the claim that m(1 − k(k + 1)m) is positive, but this is true since m > 0 and 1 − k(k + 1)m > 0, since m < \frac{1}{k(k+1)}.

Lemma 8. Let n be a positive integer, and let p be a prime such that p^{a} || n. Then

\frac{h(n)}{h(\frac{n}{p})} ≤ 1 + \frac{1}{p^{a}} .

Proof. We break this into two cases, depending on whether a = 1 or a ≥ 2. If a = 1, since h(n) is multiplicative we have,

\frac{h(n)}{h(\frac{n}{p})} = h(p) = 1 + \frac{1}{p}.

Now assume that a ≥ 2. Then we have

\frac{h(n)}{h(\frac{n}{p})} = \frac{1 + \frac{1}{p} ⋯ + \frac{1}{p^{a−1}} + \frac{1}{p^{a}}}{1 + \frac{1}{p} ⋯ \frac{1}{p^{a−1}}} = 1 + \frac{\frac{1}{p^{a}}}{1 + \frac{1}{p} ⋯ \frac{1}{p^{a−1}}} < 1 + \frac{1}{p^{a}} .

Lemma 9. Let n be a primitive non-deficient number with n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} with p_{1} < p_{2} < ⋯ < p_{k}. Assume further that (p_{1}, k) ≠ (3, 3). Let R = rad(n) = p_{1}p_{2} ⋯ p_{k}. Assume further that

H(n) ≥ 2 + α

for some positive α < 1. Then there exists an i, 1 ≤ i ≤ k, such that

p_{i}^{a_{i}+1} < max (\frac{2(k + 2 + p_{1})}{α}, k(k + 1)) .

Proof. This proof is essentially an adaptation of part of the argument used in prior work by Heath-Brown [12]. Assume n is a primitive non-deficient number with n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} with p_{1} < p_{2} < ⋯ < p_{k} and let R = rad(n) = p_{1}p_{2} ⋯ p_{k}. Assume that

H(n) ≥ 2 + α

for some positive α < 1. So we have

H(n) = ∏_{i=1}^{k} \frac{p_{i}}{p_{i} − 1} ≥ 2 + α.

We also have that

h(n) = H(n) ∏_{i=1}^{k} (1 − \frac{1}{p_{i}^{a_{i}+1}} ) = ∏_{i=1}^{k} (\frac{p_{i}}{p_{i} − 1}) (1 − \frac{1}{p_{i}^{a_{i}+1}} ) .

But we also have

∏_{i=1}^{k} (\frac{p_{i}}{p_{i} − 1}) (1 − \frac{1}{p_{i}^{a_{i}+1}} ) ≥ (∏_{i=1}^{k} \frac{p_{i}}{p_{i} − 1}) (1 − ∑ \frac{1}{p_{i}^{a_{i}+1}} ) ,

and

(∏_{i=1}^{k} \frac{p_{i}}{p_{i} − 1}) (1 − ∑ \frac{1}{p_{i}^{a_{i}+1}} ) ≥ (2 + α) (1 − k max \frac{1}{p_{i}^{a_{i}+1}} ) .

Reducing any of the a_{i} by 1 or more to get a new number n′ yields a deficient number since n is primitive non-deficient. Thus for any prime p_{i} we have

h(n/p_{i}) < 2,

which we combine with Lemma 8 to conclude that for any p_{i},

h(n) < 2 (1 + \frac{1}{p_{i}^{a_{i}}} ) .

We have then,

2 (1 + min \frac{1}{p_{i}^{a_{i}}} ) > (2 + α) (1 − k max \frac{1}{p_{i}^{a_{i}+1}} ) .

Set m = max _{\frac{1}{p_{i}^{a_{i}+1}}} . Note that m = _{\frac{1}{min p_{i}^{a_{i}+1}}} and that min \frac{1}{p_{i}^{a_{i}}} ≤ p_{1}m. Thus

\frac{1 + p_{1}m}{1 − km} > 1 + \frac{α}{2}.

If one has a concern about sign issues in the above, note that if km ≥ 1, then one has that min p_{i}^{a_{i}+1} < k, which is stronger than our desired inequality, so we can assume that km < 1. Note that k ≥ p_{1} by Lemma 5, and the fact that no non-deficient number is a power of 2.

Now, we split into two cases, where m < \frac{1}{k(k+1)} and where m ≥ \frac{1}{k(k+1)}. Notice that since k(k + 1) cannot be a power of a prime, these two quantities cannot be equal.

Consider the case where m > \frac{1}{k(k+1)}. This means that we have

min p_{i}^{a_{i}+1} < k(k + 1),

which is one of the terms in our maximum needed. If

m < \frac{1}{k(k + 1)},

then by Lemma 7, we have

\frac{1}{1 − km} < 1 + (k + 1)m.

Hence,

\frac{1 + p_{1}m}{1 − km} < (1 + mp_{1})(1 + (k + 1)m) = 1 + (k + 1)m + mp_{1} + m^{2}p_{1}(k + 1). (10)

Now we use the fact that p_{1} < k with Inequality (10) to obtain the following:

1 + (k + 1)m + mp_{1} + m^{2}p_{1}(k + 1) < 1 + m(k + 1 + p_{1}) + m^{2}k^{2} < 1 + m(k + 2 + p_{1}),

and so 1 + m(k + 2 + p_{1}) > 1 + \frac{α}{2}, which yields

m(k + 2 + p_{1}) > \frac{α}{2}.

Thus

min p_{i}^{a_{i}+1} < 2\frac{(k + 2 + p_{1})}{α}.  (11)

We are now ready to prove Theorem 2.

Proof. We may, based on the earlier results, assume that p_{1} ≥ 3. By Lemma 4 we may also assume that k ≥ 5.

Since n is non-deficient, we have

2 ≤ h(n) < H(n).

We have

H(n) = ∏_{i=1}^{k} \frac{p_{i}}{p_{i} − 1} ≥ 2 + \frac{1}{(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k} − 1)} ≥ 2 + \frac{2}{∏_{i=1}^{k} p_{i} − 1} > 2 + \frac{2}{R}.

Note that in the last step in the above chain of inequalities, we are using that ∏_{i=1}^{k} \frac{p_{i}}{p_{i}−1} > 2 so (p_{1} − 1)(p_{2} − 1) ⋯ (p_{k} − 1) < \frac{R}{2}.

We may then apply Lemma 9 with α = \frac{2}{R}. Thus there exists an i such that

p_{i}^{a_{i}+1} < max(R(k + 2 + p_{1}), k(k + 1)).  (12)

The first term in the above maximum is clearly larger than the second term so we have

p_{i}^{a_{i}+1} < R(k + 2 + p_{1}).  (13)

We now apply Lemma 5 along with the fact that k ≥ 5 to get p_{1} + 2 ≤ k and thus

R(k + 2 + p_{1}) ≤ 2Rk.

We thus have

p_{i}^{a_{i}+1} ≤ 2Rk.

But since the left-hand side is odd and the right-hand side is even, the inequality must be strict and so we are done.

Note that one can get a slightly tighter, but substantially less nice looking bound by replacing Lemma 5 in the above argument with the sort of inequality used in Section 5 of the author’s aforementioned paper [22].

3 Related results and further conjectures

How tight is Conjecture 1? Some of our odd primitive non-deficient numbers with 3 prime factors just barely manage to satisfy this inequality. Some of worst case situations occur when n = 2^{a}p and p is a Fermat prime. That is, we have p = 2^{a} + 1. In this scenario, 2^{a+1} = R − 2 and so these are close to our inequality being violated. However, the standard conjecture is that there are only finitely many Fermat primes.

The main theorem of this paper and Conjecture 1 motivate the following definitions and additional conjectures.

Let A be the set of numbers satisfying p_{i}^{a_{i}+1} < 2R for some i. It is not hard to show that such numbers have density 1. Let B be the set of numbers satisfying p_{i}^{a_{i}+1} < 2R for all i. An example of such a number is 105. Note that there are some primitive nondeficient numbers in the set B. For example, 945 is in B. What can we say about density of numbers from B? Note that B has natural density zero due to classical results about the distribution of distinct prime factors such as those by Billingsley [5]. Given a set S, we write S(x) = |{n ∈ S, n ≤ x}|. It seems natural to ask for an asymptotic for B(x).

Although 945 is in B, such examples seem to be rare. Let C be the set of primitive non-deficient numbers. Let C_{O} be the set of odd primitive non-deficient numbers. Then we suspect that Conjecture 10.

lim_{x→∞} \frac{(B ∩ C)(x)}{B(x)} = 0.

Similarly, it seems likely that

Conjecture 11.

lim_{x→∞} \frac{(B ∩ C)(x)}{C(x)} = 0,

We have similar conjectures for C_{O}. In particular, we have the following:

Conjecture 12.

lim_{x→∞} \frac{(B ∩ C_{O})(x)}{B(x)} = 0.

Conjecture 13.

lim_{x→∞} \frac{(B ∩ C_{O})(x)}{C_{O}(x)} = 0.

We have the following generalization: Let M be a set of primes, and let C_{M} be the set of primitive non-deficient numbers which do not have any element of M as a prime divisor. So, C_{O} = C_{{2}}. Then the following seem plausible.

Conjecture 14. Let M be a non-empty finite set of primes. Then

lim_{x→∞} \frac{(B ∩ C_{M})(x)}{B(x)} = 0.

Conjecture 15. Let M be a non-empty finite set of primes. Then

lim_{x→∞} \frac{(B ∩ C_{M})(x)}{C_{M}(x)} = 0.

We can also apply Lemma 9 to obtain the following result, which essentially says that if n is an odd primitive non-deficient number, then either its largest prime factor is small compared to the radical or it has a component which is small compared to the radical.

Proposition 16. Let n be an odd primitive non-deficient number with n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} with p_{1} < p_{2} < ⋯ < p_{k}. Either

p_{k} < \sqrt{R}  (14)

or there exists an i such that

p_{i}^{a_{i}+1} < (k + 2 + p_{1})(p_{k} − 1).  (15)

Proof. If p_{1} = 3 and k = 3, it is easy to check the short finite list that all of these satisfy at least one of the two bounds. We thus may assume that (p_{1}, k) ≠ (3, 3).

As before, if n is non-deficient we must have

H(n) = ∏_{i=1}^{k} \frac{p_{i}}{p_{i} − 1} > 2,

and thus

\frac{p_{1}p_{2}p_{3} ⋯ p_{k}}{(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k} − 1)} − 2 > 0

which means that

p_{1}p_{2} ⋯ p_{k} = 2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k} − 1) + x  (16)

for some x. Note that x is itself a positive integer. Now Equation (16) modulo p_{k} becomes

0 ≡ 2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k−1} − 1)(−1) + x (mod p_{k}).

So p_{k} | −2(p_{1} −1)(p_{2} −1) ⋯ (p_{k−1} −1)+x. If −2(p_{1} −1)(p_{2} −1) ⋯ (p_{k−1} −1)+x is negative, then p_{k} | 2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k−1} − 1) − x. Thus we get

p_{k} ≤ 2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k−1} − 1) − x < 2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k−1} − 1)  (17)

We have then

2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k−1} − 1) = 2R\frac{p_{1} − 1}{p_{1}} \frac{p_{2} − 1}{p_{2}} ⋯ \frac{p_{k−1} − 1}{p_{k−1}} \frac{1}{p_{k}}  (18)

Since n is non-deficient we have

\frac{p_{1} − 1}{p_{1}} \frac{p_{2} − 1}{p_{2}} ⋯ \frac{p_{k−1} − 1}{p_{k−1}} \frac{p_{k−1}}{p_{k}} < 2,

and so

\frac{p_{1} − 1}{p_{1}} \frac{p_{2} − 1}{p_{2}} ⋯ \frac{p_{k−1} − 1}{p_{k−1}} < \frac{1}{2} \frac{p_{k}}{p_{k} − 1}.  (19)

We then combine inequalities (17), (18) and (19) to get that

p_{k} ≤ \frac{2R}{p_{k}} \frac{1}{2} \frac{p_{k}}{p_{k} − 1}

which implies that

p_{k}(p_{k} − 1) < R.  (20)

Since R is a multiple of p_{k}, we must have

p_{k}(p_{k} − 1) ≤ R − p_{k}

which implies that p_{k} < \sqrt{R}.

Now assume that

p_{k} ≥ \sqrt{R}.

By our above reasoning we must have that −2(p_{1}−1)(p_{2}−1) ⋯ (p_{k−1}−1)+x is non-negative. Then we have that x ≥ 2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k−1} − 1).

Thus, we have that

p_{1}p_{2} ⋯ p_{k} ≥ 2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k} − 1) + 2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k−1} − 1), (21)

and so

∏_{i=1}^{k} \frac{p_{i}}{p_{i} − 1} ≥ 2 + \frac{2(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k−1} − 1)}{(p_{1} − 1)(p_{2} − 1) ⋯ (p_{k} − 1)} = 2 + \frac{2}{p_{k} − 1}.

The rest of the proof then continues from invoking Lemma 9 and proceeding as before.

One can use the same technique as in the proof of Proposition 16 to get a similar bound for even primitive non-deficient numbers. However, the result is weaker than the straightforward bound one gets just from analyzing how the power of 2 must behave per Lemma 3.

Note that 945 satisfies Inequality (14) but not Inequality (15). In contrast, many primitive non-deficient numbers satisfy both inequalities. For example, 4095 satisfies both inequalities. Empirically it seems that most common behavior for odd non-deficient numbers is to satisfy both inequalities. An example of an odd primitive non-deficient number which does not satisfy Inequality (14) is

n = (3^{2})(5^{4})(11^{3})(137^{2})(31873),

for which we have \frac{R^{\frac{1}{2}}}{p_{k}} ≈ 0.84. This raises the following question.

Question 17. For all ϵ > 0, does there exist an odd, primitive non-deficient number n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}} with prime factors p_{1} < p_{2} < ⋯ < p_{k} such that

(rad(n))^{\frac{1}{2}} < ϵp_{k}?

We suspect that the answer to Question 17 is yes, but we are uncertain. At this point, we do not even see any way of proving that there are infinitely many odd primitive non-deficient numbers which do not satisfy Inequality (14).

It seems that most odd primitive non-deficient numbers satisfy both Inequality (14) and Inequality (15). It is not too hard to show that there are infinitely many which satisfy both. In particular, there are infinitely many square free odd primitive non-deficient numbers made from taking a prime and then multiplying by each successive prime until the number is not deficient. All numbers generated this way satisfy both inequalities. Let D be the set of odd primitive non-deficient numbers which satisfy both Inequality (14) and Inequality (15). Then we suspect the following.

Conjecture 18.

lim_{x→∞} \frac{D(x)}{C_{O}(x)} = 1.

4 Acknowledgments

The results in this paper were substantially tightened due to suggestions by Jan-Christoph Schlage-Puchta. The referee also made substantial improvements which greatly improved the presentation and helped fix gaps in the proofs of various lemmas as well as in Theorem 2, fixed the statement and proof of Proposition 16, and pointed out McNew and Setty’s recent work.

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2020 Mathematics Subject Classification: Primary 11A25; Secondary 11N64. Keywords: primitive non-deficient number, odd perfect number, abundant number.

(Concerned with sequences \underline{A000396}, \underline{A005100}, and \underline{A006039}.)

Received December 10 2024; revised versions received March 13 2026; March 14 2026; June 19 2026. Published in Journal of Integer Sequences, July 3 2026.

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