# Is any odd natural number less than the sum of its factors?

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 $\sigma(n) > 2n$?

• Shouldn't your title say "less than or equal to"? May 9, 2012 at 22:54
• No. As I said, it is well-known that it is unknown whether there are any odd perfect numbers. May 9, 2012 at 23:00
• @Jim: I think he means proper divisors May 9, 2012 at 23:01
• @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. May 9, 2012 at 23:02
• No, 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 $945$ May 9, 2012 at 23:05

Indeed. The smallest one is $945$.
• Yes, 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. Jan 26, 2013 at 19:44