smallest ODD number n whose sum of divisors (including n) is at least 3n As usual, for a positive integer $n$, let $\sigma(n)$ denote the sum of all positive divisors of $n$ (including 1 and $n$).
What is the smallest ODD number such that $\sigma(n) \ge 3n$?
For comparison, the answers to some related questions are:
Smallest $n$ of any parity satisifying $\sigma(n) \ge 3n$ is $n=120=2^3\cdot3\cdot5$. We have $\sigma(120)=360$.
Smallest $n$ of any parity satisifying $\sigma(n) > 3n$ is $n=180=2^2\cdot3^2\cdot5$. We have $\sigma(180)=546$.
Smallest ODD $n$ satisifying $\sigma(n) \ge 2n$ is $n=945=3^3\cdot5\cdot7$. We have $\sigma(945)=1920$.
I suspect that the answer to my above question is
$$
1310112879075 = 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29
$$
but I can't quite prove it.
I can prove that the answer must be divisible by each of the primes from 3 to 23, and then I can use trial and error to compare various candidates (e.g. compare multiplication by 29 with multiplication by $3^3$, and such "playing around").

Nope, the smallest is 
$$
n=1018976683725=3^3 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29,
$$
which has $\frac{\sigma(n)}{n}=\frac{34283520}{11350339}\approx 3.02048423399513$.
Just for the record, I found this by using your observation that any candidate must be a multiple of $m:=3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23$.  Your answer is $9765m$, and by running a for loop over all smaller (odd) multiples, I came upon the (unique) smaller answer of $n=9135m$.
