Background:

The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.

Challenge

Given a positive integer through any standard input format, output whether it is not perfect.

A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.

Test Cases:

Imperfect:
1,12,13,18,20,1000,33550335

Perfect:
6,28,496,8128,33550336,8589869056

Rules

• Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.


• Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Truthy/Falsey, please make sure to specify in your answer.
  
  
  • This means your values don't have to be literally Truthy/Falsey. Your Truthy output may evaluate to false in your language and vice-versa.
  
  

Japt -!, 4 bytes

¥â¬x

For some reason ¦ doesnt work on tio so I need to use the -! flag and ¥ instead

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R, 33 bytes

!2*(n=scan())-sum(which(!n%%1:n))

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Returns TRUE for perfect numbers ans FALSE for imperfect ones.

CJam, 17 bytes

ri_,(;1$%!,:+=

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Javascript, 62

n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)

Explanation (although it's pretty simple)

n=> //return function that takes n
 n== //and returns if n is equal to
 [...Array(n).keys()] //an array [0..(n-1)]...
 .filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
 .reduce((a,b)=>a+b) //summed up

Thanks to Jo King for the improvement!