%I #12 Oct 22 2019 21:19:45
%S 1,6,35,190,1015,5304,27417,142142,736782,3816852,19774690,102446730,
%T 530743749,2749606626,14244797910
%N a(n) is the least k such that the average number of unitary divisors of {1..k} is >= n.
%C The unitary version of A085829.
%F Lim_{n->oo} a(n+1)/a(n) = exp(zeta(2)) = exp(Pi^2/6) = 5.180668... (since A064608(n) ~ n*log(n)/zeta(2)).
%e a(2) = 6 because the average number of unitary divisors of {1..6} is A064608(6)/6 = 13/6 > 2.
%t seq={}; s = 0; k = 1; Do[While[s += 2^PrimeNu[k]; s < k*n, k++]; AppendTo[seq, k]; k++, {n, 1, 10}]; seq
%Y Cf. A013661, A034444, A064608, A085829.
%K nonn,more
%O 1,2
%A _Amiram Eldar_, Oct 22 2019