PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 64 (78), 1998, 9–20

ON A SUM OF DIVISORS PROBLEM

Jean-Marie De Koninck and Aleksandar Ivić

Communicated by Žarko Mijajlović

Abstract. Several results concerning the set S are proved. This set was constructed by A. Granville as follows: Let 1 ∈ S, and for n ∈ ℕ and n > 1 let n ∈ S if ∑_{d|n,d<n,d∈S} d ≤ n. If equality holds then n is S-perfect, and properties of S-perfect are also discussed. Numerical data are given and some conjectures are formulated.

In December 1996 Andrew Granville proposed to us the following construction of the set S involving the sum of divisors of natural numbers. Let 1 ∈ S and for n ∈ ℕ and n > 1 let n ∈ S if

(1)  ∑_{d|n,d<n,d∈S} d ≤ n.

Moreover if equality holds in (1) we shall say that n is S–perfect.

Granville’s problems in connection with S are to determine which numbers n belong to S, and which are S-perfect. Let henceforth p, q denote primes. If α ∈ ℕ, then clearly p^{α} ∈ S and pq ∈ S. There are 17 S-perfect numbers < 200000: 6, 24, 28, 96, 126, 224, 384, 496, 1536, 1792, 6144, 8128, 14336, 15872, 24576, 98304, 114688. Among these only 4 are perfect, namely 6, 28, 496 and 8128, which means that there are 13 S-perfect numbers < 200000 which are not perfect: 24, 126, 224, 384, 496, 1536, 1792, 6144, 14336, 15872, 24576, 98304, 114688. Recall that a number n is perfect if σ(n) = 2n, where σ(n) = ∑_{d|n} d is the sum of all positive divisors of n. It is, in general, difficult to determine which n belong to S and especially which ones are S-perfect. This is not surprising, since the corresponding classical problem of determining the structure of perfect numbers is unsolved. In particular, it is not yet known whether there exists an odd perfect number, but it is known that any such number must be ≥ 10^{300} (see R.P. Brent [2]) and must have at least 8 distinct prime factors (see P. Hagis [6]). However in dealing with perfect numbers one can exploit the fact that σ(n) is multiplicative (σ(mn) = σ(m)σ(n) if (m, n) = 1), whereas the problems involving S-perfect numbers cannot be linked to the distribution of values of a multiplicative function in any obvious way. Traditionally numbers n are called deficient if σ(n) < 2n and abundant if σ(n) > 2n. Trivially deficient and perfect numbers are in S. Also it is an easy matter to check that the smallest odd number not in S is 2835, and that the smallest two consecutive integers not in S are 5984 and 5985. Moreover, as we shall see in Theorem 5 below, the smallest three consecutive integers not in S are 171078830, 171078831 and 171078832.

AMS Subject Classification (1991): Primary 11N25, 11N37 Key Words: sum of divisors, perfect numbers, Mersenne primes Supported by Ministry of Science and Technology of Serbia, grant number 04M03/C

We can prove

Theorem 1. If p ≥ 3 is a given prime and r ≥ 2 is the unique integer such that

(2)  2^{r} − 1 ≤ p < 2^{r+1} − 1,

then

2^{a} · p ∈ S ⟺ a ≢ 0 (mod r).

Proof 1. Since p ≥ 3 is a prime, the condition (2) may be written as

(3)  2^{r} − 1 ≤ p < 2(2^{r} − 1).

Thus 2^{a}p ∉ S is equivalent to r | a, a > 0. The proof is by induction on a ≥ 0. For a = 0, trivially p ∈ S. So assume the result for 0, 1,…, a − 1 and write

a = αr + β, 1 ≤ β ≤ r, α ≥ 0.

By the induction hypothesis we have

∑_{{d|2^{a}p \atop d∈S,d<2^{a}p}} d = 1 + 2 + … + 2^{a} + ∑_{{j=0 \atop j=0 or r |̸ j}}^{a−1} 2^{j}p

= 2^{a+1} − 1 + p(2^{a} − ∑_{j=0,r|j}^{a−1} 2^{j}) = 2^{a}p + 2^{a+1} − 1 − p · \frac{2^{(α+1)r} − 1}{2^{r} − 1}.

Thus

(4)  2^{a}p − ∑_{{d|2^{a}p \atop d∈S,d<2^{a}p}} d = p · \frac{2^{(α+1)r} − 1}{2^{r} − 1} − (2^{a+1} − 1).

If β < r then the left-hand side of (4) is (since p ≥ 2^{r} − 1 by (3))

≥ (2^{(α+1)r} − 1) − (2^{a+1} − 1) = 2^{αr+r} − 2^{αr+β+1} ≥ 0.

If β = r then the left-hand side of (4) is (since p < 2(2^{r} − 1) by (3))

< 2(2^{(α+1)r} − 1) − (2^{a+1} − 1) = (2^{(α+1)r+1} − 2) − (2^{(α+1)r+1} − 1) = −1 < 0.

This proves Theorem 1.

In what concerns the structure of S-perfect numbers we can prove

Theorem 2. An ordinary perfect number is also an S-perfect number. The number n = 2^{m}p (p > 2) is an S-perfect number if and only if p = 2^{r} − 1 (a Mersenne prime) and m = kr − 1 (k ∈ ℕ). Moreover if n = 2 · 3^{2} · p is an S-perfect number, then p = 7, i.e. n = 126.

Proof . Note that d | n implies (by multiplicativity of σ(n)) that σ(d)/d ≤ σ(n)/n, and the inequality is strict if d < n. Hence σ(d) − d ≤ d for all divisors d of n if n is perfect (i.e. σ(n) = 2n) and

∑_{δ|d,δ<d,δ∈S} δ ≤ σ(d) − d ≤ d,

so that d ∈ S. Therefore if n is perfect

∑_{d|n,d<n,d∈S} d = ∑_{d|n,d<n} d = σ(n) − n = n,

hence n is S-perfect.

Now let r be as in (2) and suppose that m ≤ r−1, where n = 2^{m}p is S-perfect. Then by Theorem 1 we have 2^{a}p ∈ S for a = 1,…, m. Thus

\begin{aligned}2^{m}p & = ∑_{d|n,d<n,d∈S} d = 1 + 2 + … + 2^{m} + p(1 + 2 + … + 2^{m−1}) \\ & = 2^{m+1} − 1 + p(2^{m} − 1). \end{aligned}

Hence p = 2^{m+1} − 1, which in view of (2) gives

2^{r} − 1 ≤ 2^{m+1} − 1, m ≥ r − 1,

which combined with m ≤ r − 1 gives m = r − 1. Therefore n = 2^{r−1}p, where p = 2^{r} − 1 is a so-called Mersenne prime. If m = kr + ℓ, k ≥ 1, 1 ≤ ℓ ≤ r − 1 and 2^{m}p is S-perfect, then by using Theorem 1 again we shall obtain

\begin{aligned}2^{m}p & = 2^{m+1} − 1 + p(2^{m} − 1 − 2^{r} −… − 2^{kr}), \\ 2^{kr+ℓ}p & = 2^{kr+ℓ+1} − 1 + p (2^{kr+ℓ} − \frac{2^{(k+1)r} − 1}{2^{r} − 1}) . \end{aligned}

Using this and (2) we have

2^{(k+1)r} − 1 ≤ p\frac{2^{(k+1)r} − 1}{2^{r} − 1} = 2^{kr+ℓ+1} − 1,

giving r ≤ ℓ + 1, which combined with 1 ≤ ℓ ≤ r − 1 gives ℓ = r − 1, as asserted. Hence combining the two above cases we conclude that m = kr − 1, k ∈ ℕ. Conversely if n = 2^{kr−1}p, p = 2^{r} − 1, k ∈ ℕ, then by Theorem 1 we have

∑_{d|n,d<n,d∈S} d = 1 + 2 + ⋯ + 2^{kr−1}
+ p(1 + 2 + ⋯ + 2^{kr−2} − 2^{r} − ⋯ − 2^{(k−1)r})
= 2^{kr} − 1 + p(2^{kr−1} − 1 + 1 − \frac{2^{kr} − 1}{2^{r} − 1}) = 2^{kr} − 1 + n − (2^{kr} − 1) = n,

so that n is S-perfect, as asserted.

For the last part of Theorem 2 note first that 2 · 3^{2} is not in S and that for p > 2 the number 3^{2}p is in S, while 2 · 3 · p is not. Let n = 2 · 3^{2} · p. For p = 2, 3 it is easily checked that n is not S-perfect. If p > 3 then n is S-perfect if and only if

σ(n) − 2 · 3^{2} − 2 · 3 · p = 2n,

giving 39(p + 1) − 18 − 6p = 36p, p = 7, n = 126.

Perhaps 126 is the only even S-perfect number that is not of the form 2^{m}p, although this seems difficult to prove. It is well-known that all even perfect numbers are of the form n = 2^{r−1}p, p = 2^{r} − 1 (see e.g. [8]). Thus Theorem 2 tells us that there exist many more even S-perfect numbers than ordinary even perfect numbers. We have made a numerical check for numbers up to 10^{6}, and have not been able to find any S-perfect number which is not of the form given by Theorem 2. Very likely there are very few S-perfect numbers not given by Theorem 2 (if any). In particular, we have not found any odd S-perfect number, but similarly to the case of ordinary odd perfect numbers, we cannot prove that they do not exist. We can, however, use the characterization of even S-perfect numbers given by Theorem 2 to prove a quantitative result. At present it is not known whether there are infinitely many Mersenne primes, but it is known that there are at least 37 primes q such that 2^{q} − 1 is prime (the 35th, 36th and 37th Mersenne primes were all found using a program written by G.F. Woltman and offered on the Internet – Web site www.mersenne.org – to those PC owners willing to use their own computers to run Woltman’s program and thus find new Mersenne primes; the largest one was actually found on January 29, 1998, by R. Clarkson, a 19-year old student). This is the set

M : = {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281,
3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497,
86243, 110503, 132049, 216091, 756839, 859433, 1257787,
1398269, 2976221, 3021377}.

Then we have

Theorem 3. If P (x) denotes the number of S-perfect numbers not exceeding x, then

(5)  P (x) ≥ \frac{c}{log 2} log x + O(1),  c = ∑_{q∈M} \frac{1}{q} = 1.448 ….

Proof . From Theorem 2 we obtain

P (x) ≥ ∑_{q∈M} ∑_{k≥1} ∑_{2^{kq−1}(2^{q}−1)≤x} 1.

The condition 2^{kq−1}(2^{q} − 1) ≤ x is equivalent to

k ≤ \frac{1}{q} \frac{log x}{log 2} + \frac{1}{q} − \frac{log(2^{q} − 1)}{q log 2} = \frac{1}{q} \frac{log x}{log 2} + O(1).

Therefore it follows that

P (x) ≥ ∑_{q∈M} ∑_{k≤ \frac{1}{q} \frac{log x}{log 2} +O(1)} 1 = \frac{c}{log 2} log x + O(1),

where c = ∑_{q∈M} \frac{1}{q}. Thus \frac{c}{log 2} = 2.089 …, so that in particular from (5) it follows that

P (x) > 2 log x (x ≥ x_{0} > 0).

Let P denote the set of all primes and let

M_{0} := \{q ∈ P : 2^{q} − 1 ∈ P\} , c_{1} := ∑_{q∈M_{0}} \frac{1}{q}.

It is plausible to surmise that the primes in M_{0} are scarce enough so that the series defining c_{1} is convergent. In fact, according to a heuristic argument developed by S.S. Wagstaff [9] (who improved an earlier argument of Gillies), the number of Mersenne primes not exceeding x is about \frac{e^{γ}}{log 2} log log x. It follows that the number of primes p ≤ x, p ∈ M, is certainly < c_{0} log x for some positive constant c_{0}, in which case the series defining c_{1} is indeed convergent. If this is so then it seems to us plausible to conjecture that

P (x) = (\frac{c_{1}}{log 2} + o(1)) log x (x → ∞).

Another type of problem is to investigate the size of S. Here the “size of S” can be understood in terms of the corresponding densities

\underline{δ} = lim inf_{x→∞} \frac{1}{x} ∑_{n≤x,n∈S} 1, \overline{δ} = lim sup_{x→∞} \frac{1}{x} ∑_{n≤x,n∈S} 1, δ = lim_{x→∞} \frac{1}{x} ∑_{n≤x,n∈S} 1,

which represent the lower density, upper density and the density of S, respectively.

Observe that \overline{δ} ≤ 17/18. To prove this note that, for m > 1, at least one of the numbers m, 2m, 3m, 6m is not in S. Namely if all four are in S, then 1, m, 2m, 3m are allowable summands for 6m, but as 1+m+2m+3m > 6m we get that 6m ∉ S, which is a contradiction. More generally, if a is any non-deficient number (meaning that σ(a) ≥ 2a), then for any m > 1 at least one member of the set { dm : d | a } is not in S. Now consider numbers m = 6k ± 1 from [1, x/6] . Then the numbers m, 2m, 3m, 6m are all distinct, lie in [1, x] and at least one of them is not in S. Hence there are ≥ \frac{x}{18} + o(x) numbers in [1, x] which are not in S, and this gives \overline{δ} ≤ 17/18.

Suppose that the density of S

(6)  δ = lim_{x→∞} \frac{1}{x} ∑_{n≤x,n∈S} 1

exists. By elaborating the above argument one can in fact give an explicit upper bound for δ. This is contained in the following

Theorem 4. If the density δ of S exists, then δ ≤ 11/12.

Proof . By the definition (6) of δ there are δx + o(x) numbers m ≤ x from S in [1, x] as x → ∞. Thus there are at least δx + o(x) integers (of the form 2m, 3m or 6m) not in S lying in [1, 2x], [1, 3x] or in [1, 6x]. Let α, β, γ ≥ 0 satisfy α + β + γ = 1. Then for any fixed ε > 0 and x ≥ x_{0}(ε) the interval [1, 2x] contains at least (αδ − ε)x integers not in S, [1, 3x] contains at least (βδ − ε)x integers not in S, and [1, 6x] contains at least (γδ − ε)x integers not in S. In the first case we have

2x + O(1) = ∑_{n≤2x} 1 = ∑_{n≤2x,n∈S} 1 + ∑_{n≤2x,n∉S} 1 ≥ 2δx + o(x) + (αδ − ε)x,

hence if we divide by x and let x → ∞ we shall obtain, since ε may be arbitrarily small,

2 ≥ 2δ + αδ,

and from the remaining cases we likewise have

3 ≥ 3δ + βδ, 6 ≥ 6δ + γδ.

Adding the above inequalities it follows that

11 ≥ 11δ + (α + β + γ)δ = 12δ, δ ≤ 11/12.

Note that the same upper bound follows unconditionally for the lower density of S (since 6 is the smallest non-deficient number > 1). We also remark that S trivially contains deficient numbers which are known to have asymptotic density > 3/4 (see p. 189 of [3]), hence it follows that \underline{δ} > 3/4. Numerical data suggest that the density δ of S exists and is near δ = 0.846.... It is nevertheless unconditionally true that there exists a sequence of numbers of positive density and belonging to S, none of which are deficient numbers. Namely it follows that 40m ∈ S if σ(m)/m < 10/9 − 40/m, and then one has only to apply (8). Thus suppose that σ(m)/m < 10/9 − 40/m. We have that σ(d)/d ≤ σ(n)/n if d|n, so that d ∈ S if d|n and n is deficient. If p(m) is the smallest prime factor of m then σ(m)/m < 10/9 implies that p(m) > 5. Thus (20, m) = 1 and if d|m, then we have 2d, 4d, 5d ∈ S. We want to show that

∑(m) ≤ 40m,

∑(m) := ∑_{d|40m,d<40m,d∈S} d = ∑_{k=0}^{3} 2^{k} ∑_{ℓ=0}^{1} 5^{ℓ} ∑_{δ|m,2^{k}5^{ℓ}δ∈S} δ − 40m.

But 40d < 20d + 10d + 5d + 4d + 2d, so at least one of the numbers 10d, 20d, 40d is not in S, since 2d, 4d, 5d ∈ S. Suppose that 10d ∉ S. Then we use 40d = (20 + 8 + 5 + 4 + 2 + 1)d (d > 1), and it follows that at least one of the numbers 8d, 20d, 40d ∉ S. The conclusion is, after analysing all the cases, that

\begin{aligned}∑(m) & ≤ σ(40)σ(m) − (10 + 8) ∑_{d|m,d>1} d − 40m \\ & = 90σ(m) − 18σ(m) − 40m + 40 ≤ 40m \end{aligned}

for σ(m)/m < 10/9 − 40/m. In a similar way one can prove that, if 1 < σ(m)/m < 4/3, then 6m ∉ S. A consequence of this is that both 12m and 18m are in S if 1 < σ(m)/m ≤ 15/14. Since 12m and 18m are abundant, this is then another way to obtain that there is a set of abundant numbers with positive density which are all in S. One can find more inequalities of the above type which show that certain numbers are in S or not in S. By elaborating the above arguments the upper bound for \overline{δ} and the lower bound for \underline{δ} can be improved, but it seems rather difficult to obtain in this way that \overline{δ} = \underline{δ}, namely that δ exists.

A concept related somewhat to the set S is that of weird numbers. An integer n ≥ 1 is called weird by S.J. Benkoski and P. Erdős [1] if n is abundant, but n is not the distinct sum of some of its proper divisors. There are 24 weird numbers less than one million, the smallest of which is 70. Benkoski and Erdős [1] proved that the density of weird numbers is positive, and obtained also several related results. However it appears that their methods of proof cannot be of help in proving that the density of S exists.

Although we could not prove that the density of S exists, it seems that in the range that we have investigated (up to 300000) the elements of S are quite evenly distributed. This is suggested by the following table, in which S(x) stands for the number of elements of S not exceeding x, and S(x)/x is calculated to the fourth decimal place.

i  x_{i} = 10000i  S(x_{i})  S(x_{i}) − S(x_{i−1}) S(x_{i})/x_{i}
1  10000  8433  8433  0.8433
2  20000  16962  8529  0.8481
3  30000  25421  8459  0.8474
4  40000  33857  8436  0.8464
5  50000  42319  8462  0.8464
6  60000  50759  8440  0.8460
7  70000  59248  8489  0.8464
8  80000  67712  8464  0.8464
9  90000  76175  8463  0.8464
10 100000  84664  8489  0.8466
11 110000  93151  8487  0.8468
12 120000  101630  8479  0.8469
13 130000  110111  8481  0.8470
14 140000  118583  8472  0.8470
15 150000  127063  8480  0.8471
16 160000  135557  8494  0.8472
17 170000  144042  8485  0.8473
18 180000  152538  8496  0.8474
19 190000  160992  8454  0.8473
20 200000  169444  8452  0.8472
21 210000  177901  8457  0.8471
22 220000  186366  8465  0.8471
23 230000  194824  8458  0.8470
24 240000  203283  8459  0.8470
25 250000  211722  8439  0.8469
26 260000  220196  8474  0.8469
27 270000  228654  8458  0.8469
28 280000  237096  8442  0.8468
29 290000  245530  8434  0.8467
30 300000  253975  8445  0.8466.

We can generalize Granville’s construction of the set S in the following way. Fix α > 0 and define the set S_{α} recursively: Let 1 ∈ S_{α} and for n > 1 let n ∈ S_{α} if

s_{α}(n) := ∑_{d|n,d<n,d∈S_{α}} d ≤ αn.

Further we shall say that n is S_{α}-perfect if s_{α}(n) = n. Let

S_{α}(x) := ∑_{n≤x,n∈S_{α}} 1.

We can show that

(7)  S_{α}(x) ≫_{α} x.

Namely by Theorem 5.6 of P. Elliott [3] we have, uniformly for z > 0,

(8)  ∑_{n≤x,σ(n)≤zn} 1 = h(z)x + O (\frac{x log log x}{log x log log log x})

for a suitable distribution function h(z) (> 0). But if σ(n) ≤ (α + 1)n, then

∑_{d|n,d<n,d∈S_{α}} d ≤ ∑_{d|n,d<n} d = σ(n) − n ≤ αn,

hence n is counted by the sum in (8) with z = α + 1 and (7) follows from (8). We state now the following

Conjecture. There is a constant C_{α} > 0 such that

(9)  S_{α}(x) ∼ C_{α}x  (x → ∞).

In other words the density of the set S_{α} should exist for any given α > 0.

Problem 1. If (9) holds, for which α > 0 is C_{α} > h(α + 1)?

It is obvious that S_{α} = S if α = 1, that is, Granville’s construction is the case α = 1 of the general one. Since we have shown that (in the case of S) \underline{δ} > h(2), the answer to the above problem is affirmative in case α = 1 if the density of S exists. The distribution of S_{α}-perfect numbers seems hard, and the set of S_{α}- perfect numbers must be very “thin”, although we have no ideas how to obtain any quantitative result in this direction, even when α = 1. Even proving the assertion that their density is zero, which seems very plausible, seems difficult. Namely it is well-known that the density of ordinary perfect numbers is zero (see e.g. P. Erdős [4], [5]). Actually in [4] Erdős proves that, for any α > 1, one has uniformly

∑_{σ(n)/n=α} \frac{1}{n} ≪ 1.

This result (for α = 2) is obviously stronger than the statement that perfect numbers have density zero. In analogy with the above bound of Erdős we propose

Problem 2. Is it true that for any given α > 0 we have uniformly

∑_{s(n)/n=α} \frac{1}{n} ≪ 1, s(n) := ∑_{d|n,d<n,d∈S} d?

In fact it is known that the number of n ≤ x for which σ(n)/n = α is ≪_{ε} x^{ε} uniformly for all α (see [7], [10]), and very likely this bound holds also for the number of n ≤ x for which s(n)/n = α.

Let S_{α}^{∗} denote the set of S_{α}-perfect numbers. S_{α}^{∗} can be non-empty only if α is a rational number. Determining for which given rational α the set S_{α}^{∗} is non-empty is difficult, and even more difficult is to determine whether S_{α}^{∗} is infinite for a given α, if it is known that it is non-empty. Since S_{1}^{∗} coincides with the set of S-perfect numbers, Theorem 3 shows that S_{1}^{∗} is infinite, and this fact is given in a quantitative form by (5). We shall prove now the infinitude of S∗_{\frac{1}{2}} and S∗_{\frac{3}{2}} , moreover the proof will show that bounds analogous to (5) hold for the counting functions of S_{α}-perfect numbers when α = \frac{1}{2} or α = \frac{3}{2}. The result is

Theorem 5. For k = 0, 1, 2,… we have

(10)  4 · 3^{2k+1} ∈ S∗_{\frac{1}{2}} , 2^{3} · 3 · 5^{k} ∈ S∗_{\frac{3}{2}} .

Proof . The first inclusion in (10) will follow from

(11)  4 · 3^{2k} ∉ S∗_{\frac{1}{2}} , 4 · 3^{2k+1} ∈ S∗_{\frac{1}{2}} (k = 0, 1, 2,…);

note that also 2 ∈ S∗_{\frac{1}{2}} . We prove (11) by induction. Clearly 4 ∉ S_{\frac{1}{2}} , so that the first assertion in (11) is true for k = 0. Obviously 4 · 3 ∈ S∗_{\frac{1}{2}} , 3^{α} ∈ S_{\frac{1}{2}} and 2 · 3^{α} ∉ S_{\frac{1}{2}} for α = 1, 2,…. Suppose now that (11) is true for all non-negative integers less than k. Let s_{α}(n) = ∑_{d|n,d<n,d∈S_{α}} d as before. Then we obtain

\begin{aligned}s_{\frac{1}{2}} (4 · 3^{2k+1}) & = 1 + 3 + 3^{2} + … + 3^{2k+1} + 2 + 4 · 3 + 4 · 3^{3} + … + 4 · 3^{2k−1} \\ & = \frac{3^{2k+2} − 1}{2} + 2 + 4 · 3\frac{9^{k} − 1}{9 − 1} = 2 · 3 · 9^{k} = 2 · 3^{2k+1}, \end{aligned}

yielding 4 · 3^{2k+1} ∈ S∗_{\frac{1}{2}} . On the other hand

\begin{aligned}s_{\frac{1}{2}} (4 · 3^{2k}) & = 1 + 3 + 3^{2} + … + 3^{2k} + 2 + 4 · 3 + 4 · 3^{3} + … + 4 · 3^{2k−1} \\ & = \frac{3^{2k+1} − 1}{2} + 2 + 4 · 3 · \frac{9^{k} − 1}{9 − 1} = 3 · 9^{k} > \frac{4 · 3^{2k}}{2} , \end{aligned}

which implies that 4 · 3^{2k} ∉ S∗_{\frac{1}{2}} .

To prove the second inclusion in (10) note that σ(n) ≤ \frac{5}{2}n implies that n ∈ S_{\frac{3}{2}} . Thus we can readily check that for α = 0, 1, 2,… the following numbers are in S_{\frac{3}{2}} :

5^{α}, 2 · 5^{α}, 3 · 5^{α}, 2^{2} · 5^{α}, 2 · 3 · 5^{α}, 2^{3} · 5^{α}, 2^{2} · 3.

However, 2^{2} · 3 · 5^{α} ∉ S_{\frac{3}{2}} for α ∈ ℕ. This can be seen by induction, the case α = 1 being easy. If the assertion is true for natural numbers less than α, then

s_{\frac{3}{2}} (2^{2} · 3 · 5^{α}) = σ(2^{2} · 3 · 5^{α}) − 2^{2} · 3(5 + … + 5^{α})

= 7 · 4 · \frac{5^{α+1} − 1}{4} − 2^{2} · 3 · 5 · \frac{5^{α} − 1}{4}

= 20 · 5^{α} + 8 > \frac{3}{2} · 2^{2} · 3 · 5^{α},

hence 2^{2} · 3 · 5^{α} ∉ S_{\frac{3}{2}} . On the other hand 2^{3} · 3 · 5 ∈ S∗_{\frac{3}{2}} . Therefore by induction we have, supposing that 2^{3} · 3 · 5^{α} ∈ S∗_{\frac{3}{2}} for 1 ≤ α < k,

\begin{aligned}s_{\frac{3}{2}} (2^{3} · 3 · 5^{k}) & = σ(2^{3} · 3 · 5^{k}) − 2^{3} · 3 · 5^{k} − 2^{2} · 3(5 + 5^{2} + … + 5^{k}) \\ & = 36 · 5^{k} = \frac{3}{2} · 2^{3} · 3 · 5^{k}, \end{aligned}

implying that 2^{3} · 3 · 5^{k} ∈ S∗_{\frac{3}{2}} . This completes the proof of Theorem 5.

Because of the recursive nature of the set S, it is generally difficult to establish if a given integer belongs to S or not. Nevertheless, as we mentioned in our opening remarks, we have found the first three consecutive integers which are not in S. This is the object of our last theorem and it serves as an introduction to our third problem.

Theorem 6. The smallest three consecutive integers not in S are 171078830, 171078831 and 171078832.

Proof . Clearly any three consecutive integers not in S are necessarily abundant. But using a computer one can establish that the smallest abundant integer n such that both n + 1 and n + 2 are also abundant is 171078830. For these three numbers we have

n = 171078830 = 2 · 5 · 13 · 23 · 29 · 1973; σ(n) − 2n = 16004900;

n + 1 = 171078831 = 3^{3} · 7 · 11 · 19 · 61 · 71; σ(n + 1) − 2(n + 1) = 677538;

n + 2 = 171078832 = 2^{4} · 31 · 344917; σ(n + 2) − 2(n + 2) = 992.

Among the 63 proper divisors of n, only 5 are abundant, namely 2990, 3770, 86710, 5899270 and 7438210. The other 58 are therefore all in S. Since the sum of these 5 numbers is 13430950, which is smaller that 16004900, this proves that n ∉ S. All of the 127 proper divisors of n + 1 being deficient, they are all in S, and their sum exceeds n + 1, which proves that n + 1 ∉ S. Finally, among the 19 proper divisors of n + 2, only one, namely 496, is abundant. The other 18 are therefore all in S and their sum exceeds n + 2 by 992 − 496 = 496, which proves that n + 2 ∉ S. This ends the proof of Theorem 6.

Using the Chinese Remainder Theorem and the fact that ∏_{p}(1 + \frac{1}{p}) diverges, one can easily show that for each integer k ≥ 2, there exist infinitely many k-tuples (n, n + 1,…, n + k − 1) made of non-deficient numbers. In particular, if a_{1} < a_{2} < … stands for the sequence of deficient numbers, then lim sup_{r→∞}(a_{r+1}−a_{r}) = +∞. Note that if follows trivially from the fact that the density of deficient numbers exceeds 1/2 that lim inf_{r→∞}(a_{r+1} − a_{r}) = 1. Similarly if s_{1} < s_{2} < … stand for the elements of S, then since the lower density of S is > 1/2, then lim inf_{r→∞}(s_{r+1}− s_{r}) = 1. The following problem is about the value of lim sup_{r→∞} (s_{r+1} − s_{r}).

Problem 3. Let s_{1} < s_{2} < … stand for the elements of S. Prove that

lim sup_{r→∞} (s_{r+1} − s_{r}) = +∞.

Acknowledgement. We wish to thank A. Granville and C. Pomerance for valuable remarks.

References

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Jean-Marie De Koninck (Received 30 03 1998) Département de Mathématiques (Revised 12 06 1998) Université Laval G1K 7P4, Québec Canada

jmdk@mat.ulaval.ca

Aleksandar Ivić Katedra Matematike RGF-a Universiteta u Beogradu -Dušina 7, 11000 Beograd Serbia (Yugoslavia)

aleks@ivic.matf.bg.ac.yu
eivica@ubbg.etf.bg.ac.yu