Proof on a conjecture involving $d(N)$

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 prove that – For every number $I$ there exists a number $N$ such that $I$ is the index of beauty of $N$.

Solutions Collecting From Web of "Proof on a conjecture involving $d(N)$"

It turns out that this conjecture is false: there is no number $N$ whose “index of beauty” is $18$.

Note that $d(N) \le 2\sqrt N$ for any number $N$, since the divisors come in pairs $m,N/m$ and one member of each pair is at most $\sqrt N$. Thus $N/d(N) \ge \frac12\sqrt N$. One can compute by brute force that no number less than $(2\cdot 18)^2 = 1296$ has “index of beauty” equal to $18$, and numbers larger than $1296$ all have $N/d(N) \ge \frac12\sqrt N > \frac12\sqrt{1296} = 18$.

The omitted integer values of $N/d(N)$ under $1000$ (found by exhaustive computation, in the above manner) are:

$\{18, 27, 30, 45, 63, 64, 72, 99, 105, 112, 117, 144, 153, 160, 162,
165, 171, 195, 207, 225, 243, 252, 255, 261, 279, 285, 288, 294, 320,
333, 336, 345, 352, 360, 369, 387, 396, 405, 416, 423, 435, 441, 465,
468, 477, 490, 504, 531, 544, 549, 555, 567, 576, 603, 608, 612, 615,
616, 625, 639, 645, 657, 684, 705, 711, 726, 728, 735, 736, 747, 792,
795, 801, 810, 828, 840, 873, 880, 885, 891, 909, 915, 927, 928, 936,
952, 960, 963, 981, 992\}$

My gut feeling is that the set of omitted values in fact has density $1$.