A perfect number is an integer $n$ greater than $1$ that equals the sum of its factors, excluding $n$ itself. For example, $6 = 1 + 2 +3 $ so $6$ is perfect. It is unknown whether there are any odd perfect numbers. My question is, are there any odd integers $n$ greater than $1$ such that the sum of all of $n$'s factors, excluding $n$, is greater than $n$? In number-theoretic language, does there exist odd $n$ with $\sigma(n) > 2n$?
Is any odd natural number less than the sum of its factors?
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elementary-number-theory
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0Shouldn't your title say "less than or equal to"? – 2012-05-09
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1No. As I said, it is well-known that it is unknown whether there are any odd perfect numbers. – 2012-05-09
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0@Jim: I think he means *proper* divisors – 2012-05-09
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1@user20520: my point is that if the answer to your question *as stated in the title* was "yes", then you would have solved a famous unsolved problem, so you probably want to include the possibility of equality. – 2012-05-09
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0No, I have not solved that famous problem! It just seemed unlikely that any odd number could be less than the sum of its proper divisors, if you think about small numbers like $15$, $35$, etc. See the answer below - the smallest odd number with the desired property is $945$ – 2012-05-09
2 Answers
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Look up "abundant numbers" in Wikipedia and at https://oeis.org/A005101
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3The [odd abundant numbers](https://oeis.org/A006038) have an entry of their own. – 2012-05-09
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1Actually A006038 are the odd *primitive* abundant numbers. The odd abundant numbers are https://oeis.org/A005231 – 2012-05-09
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Indeed. The smallest one is $945$.
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2Did you know that offhand? – 2012-05-09
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5@AlexBecker: Yes. Unfortunately, given boundedness of my memory, it just means that there are a lot more important things I do not have space for. – 2012-05-09
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3@André Nicolas: Just because your memory is bounded does not mean you have reached its limit. I'll wager that your memory is at most 1-1/e full. – 2012-05-10
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0Can you come up with solution by using number theory ? – 2013-01-26
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0Yes, and getting that $945$ is the first is easy, but identifying say the $10$-th involves some fiddling. Start from the standard expression for $\frac{\sigma(n)}{n}$ in terms of the prime factorization. We want to get past $2$. Of course we will use the *first* few odd primes, and the exponents must be non-increasing. We have to weigh increasing the exponent of a small prime versus adding a new prime. Sometimes it can be close. – 2013-01-26
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0Oh I couldn't follow that.Would you please clarify more and add it to the answer ? – 2013-01-28