My Math Forum Semiperfect numbers with divisors in arithmetic progression

 Number Theory Number Theory Math Forum

 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.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post raymondbello New Users 1 August 4th, 2014 12:48 PM jiasyuen Algebra 1 May 1st, 2014 03:40 AM luke97 Algebra 2 April 23rd, 2014 05:04 PM jareck Algebra 3 July 6th, 2012 06:38 AM Algebra 2 April 5th, 2012 09:46 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top