Do six consecutive numbers always contain an abundant or perfect number?

Question

Let sigma(n) be the sum of the divisors of n. Take six consecutive numbers. It appears that at least one of the six has sigma(n) >= 2n. Has this been proved?

Answer 1

Any multiple of 6 (more generally, of any perfect number) is abundant.

To answer a question asked in the comments, the asymptotic density of the abundant numbers is between 0.2474 and 0.2480; see Marc Deleglise, Bounds for the density of abundant integers, Experiment. Math. Volume 7, Issue 2 (1998), 137-143. That is, if $A(n)$ is the number of abundant integers less than $n$, $\lim_{n \to \infty} A(n)/n$ exists and is in that interval. Perfect numbers have density zero, so deficient numbers have density just over 0.75. Unfortunately this doesn't give a proof of the result asked for in the original question.

Edited to add: Any six consecutive integers contain an abundant number. So it's natural to ask: Is there any $k$ such that $k$ consecutive integers always contain a deficient number? Erdos showed in 1935 that this is not so. More specifically, there are constants $c_1, c_2$ such that for all large enough $n$, there exist $c_1 \log \log \log n$ consecutive abundant integers less than $n$, but not $c_2 \log \log \log n$ consecutive abundant integers less than $n$. This Google preview of a book by Pickover states that the smallest pair of consecutive abundant numbers is 5775, 5776 and the smallest triplet is 171078830 + {0, 1, 2}. See the sequence A094268 in Sloane. The smallest known quadruplet of abundant numbers begins with $N_4 = 141363708067871564084949719820472453374$. These numbers suggest $c_1, c_2$ are somewhere around $0.37$. In particular, if you conjectured that any six consecutive integers contain a deficient number, the first counterexample would be around $\exp \exp \exp (6 \times 0.37)$ which has thousands of digits.

This is somewhat surprising, I think -- that the deficients are more common but the abundants come in longer runs. But in a way it makes sense. Abundant numbers tend to have lots of small factors, which means that they come in "nice" families; deficient numbers tend to not have lots of small factors, and so are essentially the holes left when the abundant numbers are left out. So perhaps it's reasonable that the holes have less structure than the thing that they're holes in.