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October 19th, 2015, 12:04 AM   #1
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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.
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October 19th, 2015, 04:35 AM   #2
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Having all divisors in arithmetic progression is a really strict condition. I think the only numbers which have it are 0 and the primes.
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October 19th, 2015, 08:10 AM   #3
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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.
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October 19th, 2015, 09:19 AM   #4
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6 is perfect and 12 has 3 as a divisor as well.
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October 19th, 2015, 09:22 AM   #5
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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.
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October 19th, 2015, 10:51 AM   #6
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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.
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October 19th, 2015, 11:38 AM   #7
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Sorry, forgot to mention; we assume k>1, otherwise the problem would be trivial.
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October 19th, 2015, 12:13 PM   #8
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It seems that the answer is "multiples of 6".
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October 19th, 2015, 12:26 PM   #9
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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.
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