Quote:
Originally Posted by mart_r
For example, for large enough odd sigma(n)n, it's quite easy to find a possible n. It doesn't take long to find a prime p such that q=(sigma(n)n1p) is also prime and n=p*q is found.
The problem now is determining whether this is the smallest solution (mostly it isn't), and how efficiently can it be found?

Clearly I'm no expert, but I have made a script to look backwards before, that is:
y=sigma(n)n knowing y solve for possible n so I agree solving for any n can be done simple enough ( though pari can only go so far it seems) really solving for smallest n should be possible as soon as we can limit down the amount of partitions of y that are usable ( as I have tried once they have to have properties of a list of proper divisors so primes need to be in the list or it doesn't work, 1 is a divisor of every natural so you can take it to partitions of y1 that don't include 1,that include naturals only divisible by the primes in that partition, but don't include powers of those primes that will bring the top number to less than n ( and clearly the top prime is less than y to fit into a partition of y).