Is the set of all numbers which divide a specific function of their prime factors, infinite? Consider a number $n$ with prime factorization $n=p_1^{k_1} \cdot p_2^{k_2} \dots \cdot p_z^{k_z}$. We define a function $f(n)$ to be $f(n)=(p_1^{k_1+1}-1) \cdot (p_2 ^{k_2 +1}-1) \dots \cdot (p_z^{k_z+1}-1)$. Let $S$ be the set that contains all odd integers $n > 2$ such that $n|f(n)$. What properties of $S$ can be deduced? More specifically:
Is $S$ finite or infinite? What are some numbers that can be part of $S$? Is there a general form to these numbers?
I really dont know where to start, at all.  Thanks for the help!
 A: The main thing is that the number $n$ cannot be squarefree, as it has a largest prime factor $r$ which is at least 2 larger than any other prime factor $p,$ so $r$ does not divide any $p_j^2 - 1 = (p_j + 1)(p_j -1).$ 
Anyway, I wrote out a little grid by hand, and I get another example with
$$ n = 3^4 \cdot 5 \cdot 7 \cdot 11^2 \cdot 19  = 6,517,665.  $$  with
$$f(n) = 2^{12} \cdot 3^4 \cdot 5^2 \cdot 7 \cdot 11^2 \cdot  19.  $$ 
Note that your function is a little bit peculiar. Your $f(n) = \sigma(n) \cdot (p_1 - 1)(p_2 - 1)\cdots (p_z - 1),$ where $\sigma(n)$ is the sum of the divisors of $n.$ Where did you get this problem?
Another example with
$$ n = 3^4 \cdot 5^4 \cdot 11^3 \cdot 31 \cdot 61 \cdot 71  = 9,046,757,919,375.  $$  with
$$f(n) = 2^{20} \cdot 3^5 \cdot 5^4 \cdot 7 \cdot 11^3 \cdot 31 \cdot 61 \cdot 71.$$ 
EDIT, Saturday morning: as an "operations research" problem, this has a fairly clear approach in terms of a computer program. Not the same ordering Raymond used in his exhaustive search, yet exhaustive nevertheless. Fix one bound on primes used, call it $B,$ so all primes $p_j < B.$ Fix a second bound $M \geq B^2$ which bounds exponents as in $p^{k+1} < M.$ So, for each $p < B,$ the exponent bound $k_p$ will be
$$ k \; \leq \; \; k_p = -1 + \left\lfloor \frac{\log M}{\log p}   \right\rfloor  $$
Next, for each pair $(p,k)$ with $p < B$ and $p^{k+1} < M,$ perform a strictly limited small factoring of $p^{k+1} - 1,$ do trial division by only those primes $q$ that are also smaller than $B.$ The point here is that we do not care what prime factors there might be that are larger than $B,$ we would ignore them anyway. So the factoring step is lightning fast. 
Suppose there are $b$ odd primes up to $B.$ So, for each pair $(p,k),$ save a list/vector, for each odd prime $q < B$ the entry is the exponent $v_q (p^{k+1} - 1).$  
The search is now, for each odd $p < B,$ choose an exponent $0 \leq k \leq  k_p,$ retrieve the list of all the $v_q,$ add on, resulting in a complete factorization of a candidate number $n$ and a factorization, of $f(n)$ using only primes $q < B.$ This is enough information to decide whether $n | f(n).$ 
I did something like this by hand, first with $B=20,$ later with $B=75.$ However, I was making selections by eye. There may be, as there usually are, ways to build in selection criteria to avoid checking all possible $b$-tuples. Looking at Raymond's list again, I would also suggest varying the bounds per prime, we might as well allow $3^k$ for pretty large $k,$ but beginning with $5$ we can be more strict with the bounds, so as to cut down on the final search space.
In case it is possible to have two solutions that are relatively prime, the product of two such is a new solution. So far, though, it appears that 3 is a factor of all solutions, which makes this comment likely superfluous.
