Is the sum $\sum_{d\mid n}\frac1{d+1}$ never integral?

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!

• oeis.org/A057643 tabulates "Least common multiple of all $(k+1)$'s, where the $k$'s are the positive divisors of $n$." Nov 20, 2018 at 11:04
• Wouldn't this follow from the coprimality of $d+1$ and $n+1$? Nov 20, 2018 at 12:45
• @SylvainJULIEN It will, but these numbers are not necessarily coprime: take $n=9$, $d=3$. Nov 20, 2018 at 12:49
• 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. Nov 20, 2018 at 14:10
• @SylvainJULIEN This is still false! Take $d=3, n=15$ Nov 21, 2018 at 5:19

1 Answer

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 $$\sum_{d|n}\frac{1}{d+m}\not\in\mathbb{Z}.$$

• 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.
– Seva
Dec 3, 2018 at 13:13