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%I #12 Aug 01 2020 15:45:18
%S 60,140,210,280,315,360,462,504,616,630,693,728,770,792,819,910,924,
%T 936,990,1001,1092,1144,1170,1287,1320,1386,1430,1530,1560,1584,1638,
%U 1683,1716,1870,1872,1989,2002,2090,2142,2145,2210,2244,2288,2310,2431,2448,2470,2508
%N Numbers k that have 3 divisors d1, d2, d3 such that d1 < d2 < d3 < 2*d1 and are pairwise coprime and d1*d2*d3 = k.
%C (k/4)^(1/3) < d1 < k^(1/3). Proof: as k = d1 * d2 * d3 < d1 * (2*d1) * (2*d1) = 4*d1^3 we have (k/4)^(1/3) < d1 and as k = d1 * d2 * d3 > d1 * d1 * d1 = d1^3 we have k^(1/3) > d1. Q.e.d.
%e 210 is in the sequence because 5*6*7 = 210 and each of these factors are pairwise coprime and 5 < 6 < 7 < 2*5 = 10.
%Y Cf. A333966, A336629.
%K nonn
%O 1,1
%A _David A. Corneth_ and _Amiram Eldar_, Jul 28 2020
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