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October 19th, 2015, 12:04 AM  #1 
Newbie Joined: Oct 2015 From: US Posts: 5 Thanks: 0  Semiperfect numbers with divisors in arithmetic progression
I have been tasked with finding all semiperfect numbers with divisors in arithmetic progression. I think all numbers are in form 6n, but I have no proof. Help? Thank you. 
October 19th, 2015, 04:35 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Having all divisors in arithmetic progression is a really strict condition. I think the only numbers which have it are 0 and the primes.

October 19th, 2015, 08:10 AM  #3 
Newbie Joined: Oct 2015 From: US Posts: 5 Thanks: 0 
Numbers such as 6 (1+2+3) and 12 (2+4+6) work. as a note: semiperfect numbers are numbers that have some, but not all, of its divisors add up to itself Last edited by Dabes10101; October 19th, 2015 at 08:21 AM. 
October 19th, 2015, 09:19 AM  #4 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
6 is perfect and 12 has 3 as a divisor as well.

October 19th, 2015, 09:22 AM  #5 
Newbie Joined: Oct 2015 From: US Posts: 5 Thanks: 0 
I worded the problem poorly, I guess. The problem that I am solving is that I have a positive integer n that has divisors d1, d2, ..., dk. I am looking for all positive integers n such that n=d1 + d2 + ... + dk, d1, d2, ..., dk are in arithmetic progression, and d1, d2, ..., dk do not have to be all of the divisors of n. 
October 19th, 2015, 10:51 AM  #6 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
So you're looking for all positive integers n such that there exists a subset of size k of divisors of n that are in arithmetic progress and sum to n? If so, all positive integers have this property for k = 1; choose n itself. 
October 19th, 2015, 11:38 AM  #7 
Newbie Joined: Oct 2015 From: US Posts: 5 Thanks: 0 
Sorry, forgot to mention; we assume k>1, otherwise the problem would be trivial.

October 19th, 2015, 12:13 PM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
It seems that the answer is "multiples of 6". 
October 19th, 2015, 12:26 PM  #9 
Newbie Joined: Oct 2015 From: US Posts: 5 Thanks: 0 
I think so as well; however, I cannot think of a proof for this. All I know is that such a number cannot only have two of its divisors add up to it and be in arithmetic progression.


Tags 
arithmetic, divisors, numbers, progression, semiperfect 
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