TITLE: Is the set of all numbers which divide a specific function of their prime factors, infinite?
QUESTION [10 upvotes]: 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!

REPLY [5 votes]: As a complement to Will Jagy's fine analysis I'll just add the result of an exhaustive search up to $4\cdot 10^9$ followed by a search restricted to odd numbers $n$ whose prime factors $p^k$ verify $p\le 73,\ p^k\le 11^3$ using Will Jagy's method :
$\begin{array} {rl|l}
 n & \text{factor(n)} & \text{factor(f(n))} \\
\hline \\
 1 & 1 & 1 \\
 819 & 3^2\cdot 7\cdot 13 & 2^8\cdot 3^2\cdot 7\cdot 13 \\
 2941029 & 3^5\cdot 7^2\cdot 13\cdot 19 & 2^{10}\cdot 3^5\cdot 5\cdot 7^2\cdot 13\cdot 19 \\
 6517665 & 3^4\cdot 5\cdot 7\cdot 11^2\cdot 19 & 2^{12}\cdot 3^4\cdot 5^2\cdot 7\cdot 11^2\cdot 19 \\
 14705145 & 3^5\cdot 5\cdot 7^2\cdot 13\cdot 19 & 2^{13}\cdot 3^6\cdot 5\cdot 7^2\cdot 13\cdot 19 \\
 1010238075 & 3^4\cdot 5^2\cdot 7\cdot 11^2\cdot 19\cdot 31 & 2^{17}\cdot 3^4\cdot 5^3\cdot 7\cdot 11^2\cdot 19\cdot 31 \\
 2279297475 & 3^5\cdot 5^2\cdot 7^2\cdot 13\cdot 19\cdot 31 & 2^{18}\cdot 3^6\cdot 5^2\cdot 7^2\cdot 13\cdot 19\cdot 31 \\
\hline \\
528901498125 & 3^5\cdot 5^4\cdot 7^3\cdot 11\cdot 13\cdot 71 & 2^{20}\cdot 3^5\cdot 5^4\cdot 7^3\cdot 11\cdot 13\cdot 71 \\ 
9046757919375 & 3^4\cdot 5^4\cdot 11^3\cdot 31\cdot 61\cdot 71 & 2^{20}\cdot 
3^5\cdot 5^4\cdot 7\cdot 11^3\cdot 31\cdot 61\cdot 71 \\
63327305435625 & 3^4\cdot 5^4\cdot 7\cdot 11^3\cdot 31\cdot 61\cdot 71 & 2^{24}\cdot 3^6\cdot 5^4\cdot 7\cdot 11^3\cdot 31\cdot 61\cdot 71 \\
\end{array}$
(people finding other solutions should feel free to edit this table!).
I must add Robert Israel's 'Far Away' solution (without $3$ factor) : $5^{21}\cdot 7^7\cdot 11^4\cdot 23^2\cdot 29\cdot 41\cdot 71\cdot 79\cdot 83\cdot 89\cdot 139\cdot 167\cdot 179\cdot 521\cdot 571\cdot 601\cdot 761\cdot 1201 $ $\cdot 1523\cdot 12207031\cdot 9137\cdot 3221\cdot 5281$
Your sequence 819,2941029,... appears unknown to The On-Line Encyclopedia of Integer Sequences™. You could propose to add your sequence there (perhaps with some motivation that we too could appreciate :-)).
EDIT : I see no clear regularity in this sequence (except that only relatively small primes seem involved in $n$ and only very small ones in $\frac{f(n)}n$) and would conjecture that the sequence is infinite...

UPDATE: Since this problem was given at a calculator-free contest and to answer Sarah's questions let's try to find solutions without computer (using more or less Robert Israel's method). In fact finding solutions this way will be much faster and easier than using brute force.
The first thing to observe is that $f$ defines what is called a 'multiplicative function' in number theory : $$f(m\cdot n)=f(m)\cdot f(n)\ \ \text{for}\ \ (m,n)=1$$
This implies that we will only need to examine $f(p^k)$ for $p$ prime and $k\in \mathbb{N}$. To save some place I'll replace the $f$ function by the function $F$ obtained by removing the powers of $2$ and make a little table of $F(p^k)$ for small values of $p$ and $k$ (the first thing to do when you don't know what to do!) :
$
\begin{array} {r|ccccc}
 p & k=1 & 2 & 3 & 4 & 5\\
\hline \\
 3 & 1 & 13 & 5 & 11^2 & 7\cdot 13 \\
 5 & 3 & 31 & 3\cdot 13 & 11\cdot 71 & 3^2\cdot 7\cdot 31 \\
 7 & 3 & 3^2\cdot 19 & 3\cdot 5^2\\
 11 & 3\cdot 5 \\
 13 & 3\cdot 7 \\
 17 & 3^2 \\
 19 & 3^2\cdot 5 \\
\end{array}
$
$F(3^1)=\frac{3^2-1}8=1$ doesn't look interesting (we lost the initial $3$ and got nothing)
$F(3^2)=\frac{3^3-1}2=13$ seems more interesting so let's search (not at all inspired by the neatly ordered table above...)  

$F(13)=\frac{168}8=3\cdot 7$ (an additional $3$ and a new prime $7$) so let's search :  
$F(7)=3$ and we got our first solution :
$F(3^2\cdot 7\cdot 13)= 3^2\cdot 7 \cdot 13$

Let's try this again with the second interesting value of the table :
$F(3^3)=5$ so let's search $F(5)$ :

$F(5)= 3 \cdots$ not enough $3$ to continue... Next try :

$F(3^4)=11^2$

$F(11^2)=5\cdot 7\cdot 19\\  $  we got $F(5)=F(7)=3$  so let's search
$F(19)= 3^2\cdot 5$
and we found our second solution :
$F(3^4\cdot 5\cdot 7\cdot 11^2\cdot 19)=3^4\cdot 5\cdot 7\cdot 11^2\cdot 19$

I'll let you find other solutions too!