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

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?

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12n <= 1 + 6n + 2 + 3n + 3 + 2n + 6 + n , so this should help answer the question. Is it a homework question? Gerhard "Ask Me About System Design" Paseman, 2010.03.23 – Gerhard Paseman Mar 23 '10 at 21:24
Not homework. Just an observation. – tdnoe Mar 23 '10 at 21:26
Similar questions might be better for this forum, e.g. what is the maximum number possible of abundant integers in any closed interval of length 5? Gerhard "Ask Me About System Design" Paseman, 2010.03.23 – Gerhard Paseman Mar 23 '10 at 21:29
I think you have enough now to make your own proof. Good conjecture on your part, now see what you can do to improve it. Gerhard "Ask Me About System Design" Paseman, 2010.03.23 – Gerhard Paseman Mar 23 '10 at 21:30
Thank you all for answering this obvious question. Next time I will have a cup of coffee first! – tdnoe Mar 24 '10 at 4:23

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.

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This doesn't add anything to Gerhard Paseman's comments, does it? – Gerry Myerson Mar 23 '10 at 22:01
Since you answered (and since matrices aren't entertaining me enough right now), can you comment on the density of deficient numbers? Cf. my train of thought in the comments to the question. Gerhard "Ask Me About System Design" Paseman, 2010.03.23 – Gerhard Paseman Mar 23 '10 at 22:01
Gerry, I think it does add to the comments. Gerhard Paseman gave a hint that seemed a little bit opaque to me, probably because he believed this was a homework question; I was attempting a bit more transparency. – Michael Lugo Mar 23 '10 at 22:09
Thank you for answering. I would have been happy for a comment. I'll give you my first up vote. – Gerhard Paseman Mar 23 '10 at 22:10
Thanks again for your extension. I will look up Erdos' proof and see if it resembles the sketch I gave above that there exists arbitrarily long sequences which do not contain a deficient number. – Gerhard Paseman Mar 24 '10 at 0:37