FUN WITH THE σ(n) FUNCTION

Andrew Feist

1. Notation (Standard and Otherwise). The function σ(n) is one of the basic number-theoretic arithmetic functions. It is defined as:

σ(n) = ∑_{d|n} d.

Some values of σ(n) for small n can be found in [2, sequence A000203]. (Note: It is known that σ(n) is also multiplicative, i.e., if j and k have no factors in common other than 1, σ(jk) = σ(j)σ(k).) The Dirichlet convolution of two arithmetic functions f(n) and g(n), itself a function of n, is defined as follows.

f ∗ g = ∑_{d|n} f(d)g(\frac{n}{d}).

We will use ⊥ to denote relative primality.

2. Sigma–Primes. A number n is called sigma–prime if and only if n ⊥ σ(n). The sigma–prime numbers below 100 can be found in [2, sequence A014567]. Two rather straightforward theorems are the following.

\underline{Theorem 1}. All powers of primes are sigma–prime.

\underline{Theorem 2}. No perfect numbers are sigma–prime.

To build this theory, we shall, in the time–honored tradition of mathematics, start with the simple examples and move up. If a number n is the product of two primes, say p and q, then σ(n) = 1 + p + q + pq = σ(p)σ(q). Now, the only divisors of pq are p and q. Clearly, p ⊥ σ(p). Thus, p ⊥ σ(pq) if and only if p ⊥ σ(q). Similarly, q ⊥ σ(pq) if and only if q ⊥ σ(p). Assuming p < q, we can see that (unless p = 2 and q = 3), p + 1 < q and from that p + 1 ⊥ q. Note also that in the case of the above exception, q + 1 ⊥̸ p. Thus, we can generalize and say that n = pq is sigma–prime if and only if q + 1 ⊥ p. This easily extends to the following theorem.

\underline{Theorem 3}. If n = p_{1}p_{2} ⋯ p_{k}, where p_{1} < p_{2} < ⋯ < p_{k}, and each of p_{1}, …, p_{k} is a prime, then n is sigma–prime if and only if p_{i} ⊥ 1 + p_{j} whenever i < j.

This also leads to the following corollary.

\underline{Corollary 1}. If k is odd and greater than 1, then 2k is not sigma–prime.

\underline{Proof}. Since k is odd, it has at least one odd prime factor p; thus, 2k has 2 and p for prime factors. But, 2 ⊥̸ 1 + p.

Now we come to the cases where n has prime powers as factors. Thus, if n = ∏_{i} p_{i}^{e_{i}}, then σ(n) = ∏_{i} σ(p_{i}^{e_{i}}). Then if n is to be sigma–prime, each of the factors of the first product must be relatively prime to each of the factors of the second product. Unfortunately, the trick we used in the single-power case will not work here; p_{i} < p_{j} will not imply that p_{i}^{e_{i}} < p_{j}^{e_{j}}.

\underline{Theorem 4}. A number n = ∏_{i} p_{i}^{e_{i}} is sigma–prime if and only if for every i, j,

p_{i} ⊥ ∑_{k=0}^{e_{j}} p_{j}^{k}.

We close this section with a surprising theorem and a conjecture.

\underline{Theorem 5}. The square of an even perfect number is sigma–prime.

\underline{Proof}. According to a famous result of Euler, every even perfect number is of the form (2^{p−1}(2^{p} − 1)), where 2^{p} − 1 is prime. Thus, the square of an even perfect number is of the form (2^{2p−2}(2^{p}−1)^{2}), where 2 and 2^{p}−1 are its prime factors. Thus, we have two things to check: (1) 2 ⊥ 1+(2^{p}−1)+(2^{2p}−2^{p+1}+1) = 2^{2p}−2^{p+1}+2^{p}+1. This is obvious, as the left-hand side is 2 and the right-hand side is odd; and (2) 2^{p} − 1 ⊥ 1 + 2 + 2^{2} + ⋯ + 2^{2p−2} = 2^{2p−1} − 1. We will prove this by contradiction. Assume that, in fact, 2^{p}−1|2^{2p−1}−1. We know that 2^{p}−1|(2^{p}−1)^{2} = 2^{2p}−2^{p+1}+1. By our assumption, we also know that 2^{p} −1|2^{2p} −2 (by multiplying the right-hand side by 2). Then 2^{p} − 1 must divide the (absolute) difference of these two numbers, which is 2^{p+1} − 3. But this is 2(2^{p} − 1) − 1, and this implies that 2^{p} − 1|1. This is a contradiction (as 2^{p} − 1 ≥ 3), and the theorem is proved.

\underline{Conjecture}. The natural density of the set of sigma–prime numbers is zero.

This seems a reasonable conjecture to make; the set of prime powers has density zero and the set of sigma–prime numbers is not much larger. However, no proof has been forthcoming.

3. Dirichlet Inverse of σ(n). To discuss an inverse, we must first have an identity. Looking at the definition of Dirichlet convolution, after a little thought we see that the value of the identity function must be one at n = 1 and zero elsewhere.

This tells us immediately that σ^{−1}(1) = 1/σ(1) = 1. It has been shown that the inverse of a multiplicative function is itself multiplicative (for a proof, see [1]), so we need only concern ourselves with the prime powers.

\underline{Theorem 6}. σ^{−1}(p) = −p − 1, where p is a prime.

\underline{Proof}. Since p > 1, the identity value under Dirichlet convolution has the value 0 at p. Then, 0 = σ(p)σ^{−1}(1) + σ(1)σ^{−1}(p) = (p + 1) + σ^{−1}(p), and therefore, σ^{−1}(p) = −p − 1.

\underline{Theorem 7}. σ^{−1}(p^{2}) = p, where p is a prime. \underline{Proof}. Again, the identity value is 0. Then,

0 = σ(p^{2})σ^{−1}(1) + σ(p)σ^{−1}(p) + σ(1)σ^{−1}(p^{2})
= (1 + p + p^{2}) + (p + 1)(−p − 1) + σ^{−1}(p^{2})
= p^{2} + p + 1 − p^{2} − 2p − 1 + σ^{−1}(p^{2})
= −p + σ^{−1}(p^{2}).

Thus,

p = σ^{−1}(p^{2}).

\underline{Theorem 8}. σ^{−1}(p^{k}) = 0, where p is a prime, for all integer k ≥ 3. \underline{Proof}. We will prove this using induction, so let’s start with k = 3.

0 = σ(p^{3})σ^{−1}(1) + σ(p^{2})σ^{−1}(p) + σ(p)σ^{−1}(p^{2}) + σ(1)σ^{−1}(p^{3})

= (1 + p + p^{2} + p^{3}) + (1 + p + p^{2})(−p − 1) + (1 + p)(p) + σ^{−1}(p^{3})

= (1 + p + p^{2} + p^{3}) + (−1 − p − p^{2} − p^{3} − p − p^{2}) + (p + p^{2}) + σ^{−1}(p^{3})

= σ^{−1}(p^{3}).

That takes care of the base case; let’s assume that σ^{−1}(p^{n}) = 0 for all n, 3 ≤ n < k and see what happens.

0 = σ(p^{k})σ^{−1}(1) + σ(p^{k−1})σ^{−1}(p) + σ(p^{k−2})σ^{−1}(p^{2}) + ⋯ + σ(1)σ^{−1}(p^{k})

= (1 + p + p^{2} + ⋯ + p^{k}) + (1 + p + p^{2} + ⋯ + p^{k−1})(−p − 1)

+ (1 + p + p^{2} + ⋯ + p^{k−2})(p) + σ^{−1}(p^{k})

= (1 + p + p^{2} + ⋯ + p^{k}) − (1 + 2p + 2p^{2} + ⋯ + 2p^{k−1} + p^{k})

+ (p + p^{2} + ⋯ + p^{k−1}) + σ^{−1}(p^{k})

= σ^{−1}(p^{k}).

These are all the cases; thus, we can generate the sigma inverse function for all n. If we write n in canonical prime factorization form, n = ∏_{i} p_{i}^{e_{i}}, and define the sequence {a_{i}} as

a_{i} = \{ \begin{matrix}−p − 1, & if e_{1} = 1 \\ p, & if e_{1} = 2 \\ 0, & if e_{i} > 2 \end{matrix} .

Then, σ^{−1}(n) = ∏_{i} a_{i}.

\underline{Note}. The values σ^{−1}(n), starting 1, −3, −4, 2, −6, 12, …, have now been added to Sloane’s Encyclopedia [2, sequence A046692].

Acknowledgement. The author was partially supported by NFS grant DMS-9983320.

\underline{References}

1. T. M. Apostol, Introduction to Analytic Number Theory, Springer–Verlag, New York, 1976.

2. N. J. A. Sloane, The On–Line Version of the Encyclopedia of Integer Sequences, http://www.research.att.com/˜ njas/sequences/eisonline.html

Andrew Feist Department of Mathematics Duke University Durham, NC 27708–0320 email: andrewf@math.duke.edu