Timeline for Kindda-Perfect number: Is there a sequence of numbers which are equal to the sum of its proper divisors excluding itself as well as 1?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 9, 2017 at 20:05 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| May 9, 2017 at 20:04 | vote | accept | Nikhil Bhavar | ||
| May 9, 2017 at 19:59 | comment | added | Nikhil Bhavar | Oh I missed that. | |
| May 9, 2017 at 19:57 | comment | added | Carlo Beenakker | $2n+1$ 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??? | |
| May 9, 2017 at 19:53 | comment | added | Nikhil Bhavar | For a number n, their sum is equal to 2n+1. | |
| May 9, 2017 at 19:49 | comment | added | Carlo Beenakker | that's what it is, right? you want $n$ to be the sum of all of its divisors excluding 1 and excluding itself. | |
| May 9, 2017 at 19:45 | comment | added | Nikhil Bhavar | 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. | |
| May 9, 2017 at 19:35 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
added 87 characters in body
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| May 9, 2017 at 19:25 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |