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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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  • $\begingroup$ "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...) $\endgroup$ Feb 7, 2010 at 16:22
  • $\begingroup$ aah you mean 2|sigma(n) unless... $\endgroup$ Feb 7, 2010 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, 2010 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). $\endgroup$ Feb 7, 2010 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? $\endgroup$ Feb 9, 2010 at 6:28

3 Answers 3

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

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  • $\begingroup$ 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. $\endgroup$ Feb 6, 2012 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.. $\endgroup$ Jan 27, 2014 at 6:01
  • $\begingroup$ @ThomasBloom I think the paper looks at the sum $S(n)$ of the prime factors of $n$. But this thread is about the quantity $\sigma(n)-n$. For the former, see A039752 and A006145. $\endgroup$ Nov 13, 2021 at 19:29
  • $\begingroup$ In other words, the paper of this answer is about Ruth-Aaron pairs with the correct definition, while this thread is about what happens if you misremember the definition using A001065(n)=A001065(n+1) instead of A001414(n)=A001414(n+1). $\endgroup$ Nov 13, 2021 at 19:53
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The question can be rephrased as asking for sigma(n + 1) = sigma(n) + sigma(1), in line with the "Freshman's Dream."

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    $\begingroup$ 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. $\endgroup$ Feb 7, 2010 at 22:42
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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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