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%I #21 Sep 27 2014 16:08:52
%S 35,22446139,4481106818619089
%N The smallest positive k such that ( sopfr(k)*tau(k) )^n = sigma(k) where sopfr is the sum of prime factors with multiplicity (A001414).
%C 35 is the only solution for n=1.
%C Incorrect, there are three solutions < 10^10 for n = 1: 35, 42 and 68. - _Donovan Johnson_, Jun 11 2013
%C a(3) = 14844221560107739 (conjectured) is most likely minimal but it hasn't been proved. No solutions have been found (minimal or otherwise) where the number was not squarefree.
%C a(3) <= 4481106818619089. - _Donovan Johnson_, Jun 10 2013
%H Mersenne Forum, <a href="http://www.mersenneforum.org/showthread.php?p=87743#post87743">Mersenne forum thread</a>
%F min {k : (A001414(k)*A000005(k))^n = A000203(k)}. - _R. J. Mathar_, Jun 04 2013
%e 22446139 factors as: 31*67*101*107=k, sopfr(k) = sum of prime factors of k = 31+67+101+107 = 306. tau(k) = num of divisors of k = 2^4 = 16. sigma(k) = sum of divisors of k = (31+1)*(67+1)*(101+1)*(107+1) = 23970816. (306*16)^2 = 23970816. As this k turns out to be minimal, a(2)=22446139.
%Y Cf. A126028, A000005, A000203, A001414, A226479, A226480.
%K hard,nonn,bref
%O 1,1
%A _Fred Schneider_, Dec 14 2006
%E New name from _R. J. Mathar_, Jun 04 2013
%E a(3) from _Hiroaki Yamanouchi_, Sep 26 2014