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%I #13 Jun 20 2021 08:30:41
%S 4,6,12,60,5040,293318625600
%N Chain of 6 highly composite numbers generated when subject to the recurrence relation tau(a(n+1)) = a(n), with a(0)=3, where tau(n) is the number-of-divisors function A000005(n).
%C We omit the seed a(0) from the sequence and keep the offset at 1, because 3 is not highly composite. - _R. J. Mathar_, Jun 20 2021
%e Since 4 is the HCN with 3 divisors, we have tau(4) = 3 and therefore a(1)=4; the HCN with 4 divisors is 6, so that tau(6) = 4 and hence a(2)=6; the HCN with 6 divisors is 12 so that tau(12) = 6, implying a(3)=12, ...
%Y Cf. A002182.
%Y A finite subsequence of A009287.
%K fini,full,nonn
%O 1,1
%A _Lekraj Beedassy_, Dec 22 2007
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