TITLE: How many integers are of the form $n/d(n)$, where $d(n)$ is the number of divisors of $n$?
QUESTION [5 upvotes]: This was post by me on   Maths SE: but it did not get any solution.
Some months ago I made the following conjecture -
Let $d(n)$ denote the number of divisors of $n $.
Then let $N$ be a number such that $d(N)$ divides $N$ . Also let $I= \frac{N}{d(N)}$ which is defined as the "Index of Beauty of $N$ ".
Then, For every number $I$ there exists a number $N$ such that $I$ is the index of beauty of $N$.
This conjecture was proved false by Greg Martin here.
 He said that it can be showed by exaustive computation that the following $I$ fail the conjecture under $1000$ are  $\{18, 27, 30, 45, 63, 64, 72, 99, 105, 112, 117, 144, 153, 160, 162,
165, 171, 195, 207, 225, 243, 252, 255, 261, 279, 285, 288, 294, 320,
333, 336, 345, 352, 360, 369, 387, 396, 405, 416, 423, 435, 441, 465,
468, 477, 490, 504, 531, 544, 549, 555, 567, 576, 603, 608, 612, 615,
616, 625, 639, 645, 657, 684, 705, 711, 726, 728, 735, 736, 747, 792,
795, 801, 810, 828, 840, 873, 880, 885, 891, 909, 915, 927, 928, 936,
952, 960, 963, 981, 992\}$
Now what I am interested is that sequence of $I$ that fails.

(i) Is this sequence infinite?How? 
(ii)Is there any approximation which can tell the number of such failed $I$ less than a fixed $x$ 
(iii)If the sequence is infinite then are there  canonical forms in which all of the values are in our list.

REPLY [16 votes]: Here's an elementary argument proving that the set of numbers $I$ that fail the conjecture is infinite. 
Claim. $p^{17} \in I$ for all primes $p > 19$.  
Proof. Suppose that $p^{17}=\frac{N}{d(N)}$ for some $N$.  Write $N=p^{17+k}n$ where $p$ does not divide $n$.  Then, $d(N)=(18+k)d(n)$, and so $p^kn=(18+k)d(n)$. Since $p>19$ and $n \geq d(n)$ this can only hold if $k=0$.  But now $\frac{n}{d(n)}=18$, which Greg Martin has shown is impossible.