%I #6 May 14 2021 02:54:00
%S 1,6,24,105,385,1554,6063,23688,92610,362112,1416360,5539296,21663378,
%T 84725487,331362185,1295952084,5068450464,19822658688
%N a(n) is the least k such that the average number of infinitary divisors of {1..k} is >= n.
%F Lim_{n->oo} a(n+1)/a(n) = exp(1/(2*c)) = 3.9109891037..., where c is A327576.
%e a(2) = 6 since the average number of infinitary divisors of {1..6} is A327573(6)/6 = 13/6 > 2.
%t f[p_, e_] := 2^DigitCount[e, 2, 1]; idivnum[1] = 1; idivnum[n_] := Times @@ (f @@@ FactorInteger[n]); seq={}; s = 0; k = 1; Do[While[s = s + idivnum[k]; s < k*n, k++]; AppendTo[seq, k]; k++, {n, 1, 10}]; seq
%Y Cf. A037445, A327573, A327576.
%Y The infinitary version of A085829.
%Y Similar sequences: A328331, A336304, A338891, A338943, A344272, A344273.
%K nonn,more
%O 1,2
%A _Amiram Eldar_, May 13 2021