36 is abundant as is any power of a perfect or abundant number. I suspect that any multiple of a perfect or abundant, but I have proved it only for a multiple which has no divisor in common with the given number. I haven’t tried to prove the general case, only what I could do without pencil and paper.
In investigating this, let s(n) be the sum of all divisors of n and note that n is perfect iff s(n) = 2n and is abundant iff s(n) > 2n. Then know that if m and n have no divisors in common, then s(nm) = s(n)s(m).
Yes, any proper multiple of a perfect or abundant number is abundant: taking F(n) = s(n)/n, we have that F(n) is the sum of the reciprocals of the divisors of n. Of course, if m is a proper multiple of n, then the divisors of m are a proper superset of the divisors of n, establishing that F(m) > F(n), with the desired result as a corollary.
Which seems to suggest that deficient numbers should get pretty scarce, as you increase in size. Oh, they won’t go away entirely, of course, since all primes are deficient, but is there any theorem analogous to the Prime Number Theorem about the distribution of deficient numbers?
This goes hand-in-hand with the fact that the asymptotic average value of F(n) is less than 2. Specifically, it is the sum of the reciprocal squares (thus, it equals π[sup]2[/sup]/6 ~= 1.6). For F(n) is the sum over all i of [1 if i divides n and 0 otherwise]/i, and the bracketed component has average value 1/i.
[Those who worry about my commuting of limits and infinite sums in the last post [as I did for a bit] can re-assure themselves by observing that the average of [1 if i divides n and 0 otherwise] over a range differs from 1/i by at most 1/(length of range); thus, the difference between the average of the first n values of F and the sum of the first n reciprocal squares is at most 1/n * the sum of the first n reciprocals, which is of order log(n)/n and thus vanishes in the limit, assuring us that the asymptotic average value of F is indeed the sum of the reciprocal squares.]