6
$\begingroup$

Perfect number is a positive integer that is equal to the sum of its proper divisors. The smallest perfect number is 6, which is the sum of 1, 2, and 3.

Is there a sequence of numbers which are equal to the sum of its proper divisors excluding itself as well as 1?

If yes, what are they called? Name first few numbers in the sequence.

Is there a general formula for them?

Please provide with some link to read more.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Checked up to 1500, there are none. Several length two cycles, though, like $48\mapsto75\mapsto48$, $140\mapsto195\mapsto140$, $1050\mapsto1925\mapsto1050$, ... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 9, 2017 at 19:18
  • 2
    $\begingroup$ Checked up to $10^6$ in a few seconds with this Sage one-liner and there are none: [n for n in srange(1, 10^6+1) if sigma(n) == 2*n+1]. :) $\endgroup$
    – Federico Poloni
    Commented May 10, 2017 at 8:32

1 Answer 1

Reset to default
20
$\begingroup$

Such an abundant number with abundance 1 is called a quasiperfect number (which is a more professional way to say "kindda-perfect"). None have been found, according to Wikipedia. This 1982 article says that if a quasiperfect number exists, it must be an odd square number greater than $10^{35}$ and have at least seven distinct prime factors. A recent article on this topic is here.

Improve this answer
$\endgroup$
5
  • $\begingroup$ It states, "A quasiperfect number is a natural number n for which the sum of all its divisors is equal to 2n + 1. Equivalently, n is the sum of its non-trivial divisors (that is, its divisors excluding 1 and 'n')." But I am looking for a number for which the sum of most of its divisors is itself. $\endgroup$
    – Nikhil Bhavar
    Commented May 9, 2017 at 19:45
  • 3
    $\begingroup$ that's what it is, right? you want $n$ to be the sum of all of its divisors excluding 1 and excluding itself. $\endgroup$
    – Carlo Beenakker
    Commented May 9, 2017 at 19:49
  • $\begingroup$ For a number n, their sum is equal to 2n+1. $\endgroup$
    – Nikhil Bhavar
    Commented May 9, 2017 at 19:53
  • 5
    $\begingroup$ $2n+1$ is if you do not exclude $n$ as divisor; if you do, the sum is $n+1$, and if you also exclude $1$ (which is what you want), the sum is $n$ --- or am I missing something obvious here??? $\endgroup$
    – Carlo Beenakker
    Commented May 9, 2017 at 19:57
  • $\begingroup$ Oh I missed that. $\endgroup$
    – Nikhil Bhavar
    Commented May 9, 2017 at 19:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .