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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.

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    $\begingroup$ This question is too basic for mathoverflow. You may find oeis.org/A036763 useful, though. $\endgroup$
    – Charles
    Jun 2, 2014 at 18:31
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    $\begingroup$ I disagree - I think this is unlikely to be resolved except through research-level methods, yet it is quite possible that it can be resolved, at least partially. So I think the question fits the parameters for mathoverflow. $\endgroup$
    – Greg Martin
    Jun 2, 2014 at 18:37
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    $\begingroup$ This questions is not too basic for mathoverflow. How can it be so when I can barely even understand the question? $\endgroup$
    – user51538
    Jun 2, 2014 at 18:57
  • $\begingroup$ @GregMartin: In that case I'll answer it. $\endgroup$
    – Charles
    Jun 2, 2014 at 19:15
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    $\begingroup$ Since this question now has 4 close votes, I wish to say that I agree with Greg Martin that the question seems perfectly fine for MO. I don't know how many exceptional $I$ there are up to $x$, and it seems not easy. Naturally there may be many opinions on the interest of the question, but perhaps that applies to many problems on MO. $\endgroup$
    – Lucia
    Jun 2, 2014 at 23:50

2 Answers 2

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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.

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    $\begingroup$ Here is a variant of your construction which produces many excluded numbers. I claim that if $p$ (a prime) is large then $15p$ fails the conjecture. For if $n=15pd(n)$ and $p$ is large then $p$ can divide $n$ only to the power $1$ (use $d(n)\le n^{1/3}$ for large $n$). If now $n=pm$ with $(p,m)=1$ then we must have $m=30d(m)$, contradicting Greg Martin's result that $30$ is excluded. In fact, inspecting his table it looks like $15p$ is excluded for all primes $p>5$. $\endgroup$
    – Lucia
    Jun 3, 2014 at 2:16
  • $\begingroup$ @Lucia extremely good proof $\endgroup$
    – Shivam Patel
    Jun 3, 2014 at 2:25
  • $\begingroup$ @Lucia can you prove that if the number in the list is in the form $15n$ then $n$ is a prime. $\endgroup$
    – Shivam Patel
    Jun 3, 2014 at 2:26
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    $\begingroup$ @ShivamPatel: Not true -- $810$, $840$, $960$ etc. I'm afraid I don't know any more about this sequence! $\endgroup$
    – Lucia
    Jun 3, 2014 at 2:28
  • $\begingroup$ @Lucia sorry ..I did not see the list ...just as a curiosity so can you tell me the distribution of numbers in form $10n$ in the sequence $\endgroup$
    – Shivam Patel
    Jun 3, 2014 at 2:30
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This is a simple consequence of the density of refactorable numbers, http://oeis.org/A033950, and the growth of the divisor function. Basically, $n/d(n)>x$ if $x>n^{1+\frac{0.7}{\log\log n}}$ (where the constant could be anything greater than $\log 2$) with only finitely many exceptions. But of these numbers only

$$ O\left(\frac{(\log\log x)^{k^3-1}}{\log x}\right)=\\ O\left(\frac{(\log(\log n+\frac{0.7\log n}{\log\log n}))^{k^3-1}}{\log n+\frac{0.7\log n}{\log\log n}}\right)=\\ O\left(\frac{(\log\log n)^{k^3-1}}{\log n}\right) $$

are refactorable (for every $k>1$), so there must be infinitely many exceptions. Further, these 'exceptions' have density 1 (though the constants are nasty).

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  • $\begingroup$ Hmm. n/ something bigger than 1 is bigger than x if x is bigger than n times something else bigger than 1. One of us is doing something wrong. $\endgroup$
    – The Masked Avenger
    Jun 2, 2014 at 19:38
  • $\begingroup$ I'm not sure I understand your argument. What exactly are you claiming about the set of numbers of the form $n/d(n)$? $\endgroup$
    – Lucia
    Jun 2, 2014 at 19:44
  • $\begingroup$ All the $+$ should be $-$. $\endgroup$
    – Emil Jeřábek
    Jun 2, 2014 at 22:55
  • $\begingroup$ And the second inequality needs to read the other way round, that is, $x<n^{1-0.7/\log\log n}$. $\endgroup$
    – Emil Jeřábek
    Jun 2, 2014 at 23:01
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    $\begingroup$ @Charles: I think your argument is unclear and possibly flawed -- at least I haven't understood it. Could you write it carefully -- it has several typos, and the second part of it seems unclear to me. $\endgroup$
    – Lucia
    Jun 2, 2014 at 23:28

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