Talk:Proper divisors
Dupe?
The explanation on the J implementation here makes a good point: is this task just a trivial change of Factors of an integer? --Mwn3d (talk) 17:09, 16 December 2014 (UTC)
- Yep, it is allied, but deficient, perfect, abundant number classifications as well as Amicable pairs are based on them. --Paddy3118 (talk) 18:34, 16 December 2014 (UTC)
- It seems like it'd be easier to just use the "proper divisors" definition in other tasks where appropriate rather than having a separate task with essentially the same code (except for maybe esoteric languages). Especially since any definition with "factors of an integer" could be made using "proper divisors" with one or two extra additions or subtractions. It just doesn't seem worth it. It looks to me like having a task to "print the number 3" and another task to "print the number before 4". --Mwn3d (talk) 19:11, 16 December 2014 (UTC)
- OK. I understand your point. I have just added the third of the triplet of tasks Abundant, deficient and perfect number classifications, they do form a tight triplet, but I am probably pushing it cos I started it. Could it stay guys? (I am definitely not impartial - would be nice to have one champion though in what I hope will be a debate) --Paddy3118 (talk) 19:26, 16 December 2014 (UTC)
- Yes I see those, but like I said, those could easily be made using the sum of the old factors task definition and "2n" on the right side of the formulas. Seems like "2 pi" and "tau" to me, but let's see if anyone else has an opinion on whether this task should stay. --Mwn3d (talk) 19:37, 16 December 2014 (UTC)
- OK. I understand your point. I have just added the third of the triplet of tasks Abundant, deficient and perfect number classifications, they do form a tight triplet, but I am probably pushing it cos I started it. Could it stay guys? (I am definitely not impartial - would be nice to have one champion though in what I hope will be a debate) --Paddy3118 (talk) 19:26, 16 December 2014 (UTC)
Definition
I don't believe the definition used is correct. In particular "always includes 1" doesn't follow from the definition given at the linked site, or MathWorld or OEIS.
A simple definition from Mathworld: "A positive proper divisor is a positive divisor of a number n, excluding n itself." Danaj (talk) 21:09, 16 December 2014 (UTC)
- From which we see that 1 will always divide an integer without remainder. I make a point about mentioning 1 as a reference stated that sometimes the same term "proper divisors" is used when 1 is excluded. --Paddy3118 (talk) 18:25, 16 December 2014 (UTC)
- I guess my point was about the input 1. The proper divisors of 1 should be the empty set, while all the current implementations are using the "always includes 1" to mean it should be {1}. Another interesting case is the input 0, but I think we can ignore that. Danaj (talk) 21:09, 16 December 2014 (UTC)