Let us denote as $d(n)$ some proper divisor of $n$ such that $n$ is odd. I am trying to prove that $$\sum_{d\leq\sqrt{n}}d\left(n\right)\nmid n$$
That is, I conjecture that the sum of the proper divisors of some odd number $n$ which are less than the square root of $n$ is not a proper divisor of $n$.
That is not true for even numbers; for example, $$\sum_{d\leq\sqrt{6}}d\left(6\right)=1+2$$ $$1+2\mid 6$$
¿Some idea?
Thanks in advance!
Let $$ f(n) = \sum_{\substack{d \le \sqrt n \\ d\, \mid\, n}} d. $$
By a computer search, the odd numbers $n < 1\;000\;000$ for which $f(n) \ne 1$ (meaning $n$ isn't prime) and $f(n) \mid n$ are $$45, 117, 475, 775, 2009, 3321, 7605, 10575, 13189, 20025, 43681, 78319, 88723, \\ 123201, 488961, 682051, 796797, 836675, 886075, 892769.$$