11
$\begingroup$

Recall that a positive integer $n$ is a perfect number if and only if $$\frac{\sigma(n)}n=\sum_{d\mid n}\frac1d=2.$$

QUESTION: Is my following conjecture true?

Conjecture. (i) We have $\sum_{d\mid n}\frac1{d+1}\not\in\mathbb Z$ for all $n=1,2,3,\ldots$. Moreover, for any positive integers $k$ and $m$, all the numbers $$\sum_{d\mid n}\frac1{(d+m)^k}\ \ (n=1,2,3,\ldots)$$ have pairwise distinct fractional parts, and none of them is an integer.

(ii) For any integer $k>1$, all the numbers $$\sum_{d\mid n}\frac1{d^k}\ \ (n=1,2,3,\ldots)$$ have pairwise distinct fractional parts.

I formulated this conjecture in October 2015 on the basis of my computation. Your comments are welcome!

$\endgroup$
7
  • 1
    $\begingroup$ oeis.org/A057643 tabulates "Least common multiple of all $(k+1)$'s, where the $k$'s are the positive divisors of $n$." $\endgroup$ Nov 20, 2018 at 11:04
  • $\begingroup$ Wouldn't this follow from the coprimality of $ d+1 $ and $ n+1 $? $\endgroup$ Nov 20, 2018 at 12:45
  • 3
    $\begingroup$ @SylvainJULIEN It will, but these numbers are not necessarily coprime: take $n=9$, $d=3$. $\endgroup$ Nov 20, 2018 at 12:49
  • 1
    $\begingroup$ In 2015 I checked $\sum_{d\mid n}\frac1{d+1}\not\in\mathbb Z$ for all $n\le2\times10^5$ and found no counterexample. $\endgroup$ Nov 20, 2018 at 14:10
  • 3
    $\begingroup$ @SylvainJULIEN This is still false! Take $d=3, n=15$ $\endgroup$ Nov 21, 2018 at 5:19

1 Answer 1

-1
$\begingroup$

For a given set of primes $Q=\{q_1,\dots,q_k\}$, to each prime $p\not\in Q$ we may associate the lattice $$ L=L_{q_1,\dots,q_k,p}=\{(a_1,\dots,a_k)\in\mathbb{Z}^k: \prod_{i=1}^kq_i^{a_i}\equiv 1 \bmod p\}. $$ and the coset $$ H_m=H_{m;q_1,\dots,q_k,p}=\{(a_1,\dots,a_k)\in\mathbb{Z}^k: \prod_{i=1}^kq_i^{a_i}\equiv -m \bmod p\}. $$ Then for all $n$ of the form $$ n=\prod_{i=1}^kq_i^{e_i}, $$ $e_i>0$ for all $1\leq i\leq k$, if there exists a prime $p$ such that the box $$ E_n=\{(a_1,\dots,a_k)\in\mathbb{Z}^k: 0\leq a_i\leq e_i, \mbox{for all }1\leq i\leq k\}, $$ intersects $H_m$ at exactly one element, (thus exactly one divisor $d$ of $n$ satisfies $p|(d+m)$), such $n$ satisfies $$ \sum_{d|n}\frac{1}{d+m}\not\in\mathbb{Z}. $$

$\endgroup$
1
  • 5
    $\begingroup$ It is not a problem to find $n$ such that the sum in question is not an integer. The problem is to show that it is never an integer. $\endgroup$
    – Seva
    Dec 3, 2018 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.