Best Answer Rainman, 04 March 2014 - 02:27 PM

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- False, as proven by phil1882. A smaller example is 40 < 1+2+4+5+8+10+20.
- True for k>1. For each divisor d
_{i}of n, consider the larger number kd_{i}which is a divisor of kn. Sum(kd_{i}) = k*Sum(d_{i}) = 2kn, but 1 is also a divisor of kn which has not yet been accounted for. So the sum of kn's divisors is at least 2kn+1, which makes kn abundant. - Not sure what you mean. Should it be p
_{1}*p_{2}? In that case, 2*3 = 6 is perfect, all others are deficient. If p_{1}= p_{2}, the divisors of p_{1}*p_{2}are 1, p_{1}, and p_{1}^{2}. The sum 1+p_{1}+p_{1}^{2}< p_{1}+p_{1}+p_{1}^{2}= 2*p_{1}+p_{1}^{2}</= p_{1}*p_{1}+p_{1}^{2}= 2*p_{1}^{2}= 2*p_{1}*p_{2}, so p_{1}*p_{2}is deficient. If p_{1}< p_{2}, the divisors of p_{1}*p_{2}are 1, p_{1}, p_{2}, and p_{1}*p_{2}. The sum 1+p_{1}+p_{2}+p_{1}*p_{2}</= p_{2}+p_{2}+p_{1}*p_{2}= 2p_{2}+p_{1}*p_{2}</= p_{1}*p_{2}+p_{1}*p_{2}= 2*p_{1}*p_{2}, with equality iff p_{1}= 2 and p_{2}= 3. - True. The negation is equivalent to "there is a non-deficient positive integer n and a positive integer k such that kn is deficient", which is easily disproven using the method in my proof for conjecture 2.
- False. 40 and 12 are abundant, but 40+12 = 52 > 1+2+4+13+26 is deficient.
- True. If n is abundant, then nm is abundant for any m.
- False. Every n>1, abundant or otherwise, has a prime factor p, and therefore is of the form pm for some natural m.