The density of abundant numbers

Let σ₁(n) be the sum of the divisors of n. So, for example, the divisors of 18 are 1, 2, 3, 6, 9, and 18; so σ₁(18) = 1 + 2 + 3 + 6 + 9 + 18 = 39.

If σ₁(n) > 2n, we call n abundant, so 18 is abundant. If σ₁(n) = 2n, we call n perfect; if σ₁(n) < 2n, we call it "deficient".

(If you're wondering about the subscript 1: σ_k(n), more generally, is the sum of kth powers of divisors of n. It turns out that σ₁(n)/n = σ_-1(n), so everything I'm about to say could be rephrased in that terminology.)

A basic result of analytic number theory (for example, see Apostol, Introduction to Analytic Number Theory, Theorem 3.4) is

. If we write

then the previous result says that Σ₁(n) is about n²(π²/12). But Σ₁ is in a sense an "integral" of σ₁; the "average value" of σ₁, for numbers near n, should just be Σ₁'(n), or about n(π²/6). That is, the sum of the divisors of a number near n is, on average, about 1.644n.

So, the logical question to ask is: what can we say about the distribution of σ(n)/n? It seems possible that almost all numbers could have σ(n)/n near π²/6; this would seem to mesh with the fact that large perfect numbers are very rare. But in the 1930's, Davenport and Erdos independently showed that this is in fact not the case, and σ(n)/n does seem to have a well-defined distribution; in particular, that

exists and is continuous. A(x) is just the proportion of numbers for which σ(n)/n is at least x, so A(2) is the proportion of numbers which are abundant. The best result I know of (although I just started thinking about this) is that of Deleglise, which is that A(2) is between 0.2474 and 0.2480. From numerical work one can get approximate A(x) for other x; for example, A(1.5) seems to be about 57%, and A(2.5) seems to be about 8.8%. I don't know the full distribution (although I haven't thoroughly read the last two papers I cite below, and I haven't read the first one at all) but I suspect that it is not one of the familiar ones. The Deleglise paper goes through a lot of work just to compute A(2); it seems like it would require equally much work to compute A(x) for each individual x, when what one would like is of course the entire distribution in one fell swoop.

Here are references for the papers I mentioned:
Davenport, H. "Über numeri abundantes." Sitzungsber. Preuss. Akad. Wiss., Phys.-Math. Kl., No. 6, 830-837, 1933.
Deléglise, M. "Bounds for the Density of Abundant Integers." Exp. Math. 7, 137-143, 1998.
Erdos, P. "On the Density of the Abundant Numbers." J. London Math. Soc. 9, 278-282, 1934.