|
%I #45 Nov 22 2020 01:32:12
%S 1,4,32,2178,416417176
%N a(n) is the least number k such that the average number of prime divisors of {1..k} counted with multiplicity is >= n.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeFactor.html">Prime Factor</a>.
%e a(1) = 4 since the average number of prime divisors of {1..4} counted with multiplicity equals (0 + 1 + 1 + 2)/4 = 1 which is >= 1 and this is the least such number.
%e a(3) = 2178 because the average number of prime divisors of {1..2178} counted with multiplicity is >= 3 and this is the least such number.
%t s[n_] := Module[{m = 0, c = 0, k = 1, sum = 0, seq = {}}, While[c < n, sum += PrimeOmega[k]; If[sum >= m*k, c++; AppendTo[seq, k]; m++]; k++]; seq]; s[4] (* _Amiram Eldar_, Nov 18 2020 *)
%o (PARI) a(n)=my(m=0,k=1);while(k>0, m+=bigomega(k); if(m>=k*n,break);k++);k \\ _Derek Orr_, Nov 18 2020
%Y Cf. A001222, A022559, A085829, A328331, A338891, A338943.
%K nonn,more
%O 0,2
%A _Ilya Gutkovskiy_, Nov 18 2020
|
