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 $\sigma(n)-n = \sigma(n+1)-(n+1)$?

  • $\begingroup$ "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...) $\endgroup$ – Kevin Buzzard Feb 7 '10 at 16:22
  • $\begingroup$ aah you mean 2|sigma(n) unless... $\endgroup$ – Kevin Buzzard Feb 7 '10 at 16:24
  • $\begingroup$ Yes, I was getting ahead of myself a bit. Rephrased to emphasize the point I was trying to make. Thanks, Kevin. $\endgroup$ – Alan Frank Feb 7 '10 at 17:01
  • $\begingroup$ 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). $\endgroup$ – Kevin Buzzard Feb 7 '10 at 18:28
  • $\begingroup$ @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? $\endgroup$ – Bjorn Poonen Feb 9 '10 at 6:28

You should look at Carl Pomerance's followup paper: Ruth-Aaron pairs revisited, http://www.math.dartmouth.edu/~carlp/PDF/paper130.pdf . In his first paper with Erdos they proved a result which showed that the number of RA pairs had asymptotic density 0, but just barely. In the followup Pomerance shows that the the sum of the reciprocals converges (which is muchj stronger).

  • $\begingroup$ 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. $\endgroup$ – Thomas Bloom Feb 6 '12 at 16:07
  • $\begingroup$ It seems to me that the condition about squares and double squares is irrelevant (is wrong)--see my earlier remark under the Question.. $\endgroup$ – Włodzimierz Holsztyński Jan 27 '14 at 6:01

The question can be rephrased as asking for sigma(n + 1) = sigma(n) + sigma(1), in line with the "Freshman's Dream."

  • $\begingroup$ I was hoping to see something about this in Guy's Unsolved Problems In Number Theory, but no such luck. In B13, he discusses sigma(n + 2) = sigma(n) + 2 (other than twin primes, there are only three solutions under 200,000,000). Perhaps the reference P Haukkanen in Math Student 62 (1993) 166-168 will say something, although the review 94j:11006 is not encouraging. B15 in Guy is about sigma(q) + sigma(r) = sigma(q + r) but the discussion does not head in the direction of the question at hand. By the way, I'm the same Gerry as above, just using a different computer. $\endgroup$ – Gerry Myerson Feb 7 '10 at 22:42

Apologies that this isn't a complete answer. The condition $\sigma(n+1)=\sigma(n)+1$ means that $\sigma(n),\sigma(n+1)$ are relatively prime. In the question, you've taken care of the divisibility by $2$ part.

Let $p$ be an odd prime and $q>p$ prime. Let $a_{q,p}=a_q=p$ if $q\equiv 1 \bmod p$ and $a_{q,p}=a_q$ equals the order of $q$ modulo $p$ otherwise. If $\nu_q(n)$, the highest power of $q$ dividing $n$ is, $-1$ mod $a_q$, then $p$ divides $\sigma(n)$. So it is necessary that if $q_1,q_2$ are some primes, then one musn't simultaneously have $\nu_{q_1}(n)\equiv -1 \bmod a_{q_1,p}$ and $\nu_{q_2}(n+1)\equiv -1 \bmod a_{q_2,p}$ for any $p,q_1,q_2$.

For example, if $n=q^2$, where $q$ is prime and $1$ mod $3$, then $n+1$ is $2\bmod 4$ and $3$ divides both $\sigma(n), \sigma(n+1)$, so this is an example of $n$ that must be excluded.


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