# Are there pairs of consecutive integers with the same sum of factors?

## Background/Motivation

I was planning to explain Ruth-Aaron pairs to my son, but it took me a few moments to remember the definition. Along the way, I thought of the mis-definition, a pair of consecutive numbers with the same sum of divisors. Well, that's actually two definitions, depending on whether you are looking only at proper divisors. Suppose all divisors. I quickly found (14,15) which both have a divisor sum (sigma function) of 24. Some more work provided (206,207) and then a search on OEIS gave sequence A002961.

What about proper divisors only? (2,3) comes quickly, but then nothing for a while. Noting that the parity of this value ($\sigma(n) -n$) is the same as that of $n$ unless $n$ is a square or twice a square, any solution pair must include one number of that form. With that much information in hand, I posted this problem at the reference desk on Wikipedia. User PrimeHunter determined that there were no solutions up to $10^{12}$, but there were no general responses.

Aside from the parity issue, I haven't found other individual constraints that would filter the candidates--the number of adjacent values identical modulo $p$ for other small primes is at least as great as would be expected by chance, and there are a fair number of pairs that are arithmetically close.

Other than (2,3), are there pairs of consecutive integers such that $\sigma(n)-n = \sigma(n+1)-(n+1)$?

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"Noting that 2|(sigma(n)-n) unless n is a square or twice a square". This is nonsense, right? (assuming sigma(n) is the sum of the divisors of n...) –  Kevin Buzzard Feb 7 '10 at 16:22
aah you mean 2|sigma(n) unless... –  Kevin Buzzard Feb 7 '10 at 16:24
Yes, I was getting ahead of myself a bit. Rephrased to emphasize the point I was trying to make. Thanks, Kevin. –  Alan Frank Feb 7 '10 at 17:01
You've now edited the post to say " the parity of sigma (n) is the same as that of n unless n is a square or twice a square", and this is still wrong (try n=3). Either that or I've misunderstood what you mean by sigma(n) (which you also didn't define, but which usually means the sum of (all) divisors of n). –  Kevin Buzzard Feb 7 '10 at 18:28
@Alan: Instead of saying "the parity of this value (sigma(n)-n) is the same as that of n", why not simply say "2|sigma(n)" as Kevin suggested? –  Bjorn Poonen Feb 9 '10 at 6:28

Don't both of these follow easily for the problem at hand, given the observation that at least one of the pairs must be a square or twice a square? Then the density of such integers up to $N$ must be $\ll \sqrt{N}$, so the sum of reciprocals converges by partial summation. –  Thomas Bloom Feb 6 '12 at 16:07