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%I #19 Nov 18 2020 10:27:41
%S 1,21,165,1274,9435,69720,515230,3807265,28132035,207869515,
%T 1535959665,11349295155
%N a(n) is the least number k such that the average number of odd divisors of {1..k} is >= n.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/OddDivisorFunction.html">Odd Divisor Function</a>.
%F a(n+1)/a(n) approaches e^2.
%e a(5) = 9435 because the average number of odd divisors of {1..9435} is >= 5.
%t m = 1; sum = 0; s = {}; Do[sum += DivisorSigma[0, k/2^IntegerExponent[k, 2]]; If[sum >= m*k, AppendTo[s, k]; m++], {k, 1, 10^6}]; s (* _Amiram Eldar_, Nov 15 2020 *)
%o (PARI) a(n) = my(s=1, k=1); while(s<k*n, k++; s=s+numdiv(k>>valuation(k, 2))); k; \\ _Michel Marcus_, Nov 14 2020
%Y Cf. A001227, A060831, A072334, A085829, A328331, A338943.
%K nonn,more
%O 1,2
%A _Ilya Gutkovskiy_, Nov 14 2020
%E a(11)-a(12) from _Amiram Eldar_, Nov 16 2020