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%I #14 Oct 22 2022 16:19:14
%S 1,605,51005,5837732,599407380,60462121402
%N (10^n)-th primitive antiharmonic number.
%C We conjecture that lim_{n->oo} a(n)/100^n = lim_{n->oo} A228023(n)/n^2 = 6. This is supported by the values a(n)/(10^n)^2 = 6.05, 5.10, 5.84, 5.99, 6.05 for n = 1..5, as well as by the values of A228023(n)/n^2.
%F a(n) = A228023(10^n).
%Y Cf. A020487, A227771, A228023, A228024.
%K nonn
%O 0,2
%A _Jonathan Sondow_ and _Charles R Greathouse IV_, Aug 04 2013
%E a(5) from _Charles R Greathouse IV_, Sep 03 2013
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