|
%I #19 Sep 14 2019 14:50:56
%S 2,3,4,5,3,7,8,9,5,11,6,13,7,7,16,17,9,19,10,10,11,23,12,25,13,27,14,
%T 29,10,31,32,16,17,17,18,37,19,19,20,41,14,43,22,22,23,47,24,49,25,25,
%U 26,53,27,27,28,28,29,59,20,61,31,31,64,32,22,67,34,34,23
%N a(n) = floor(n / omega(n)) where omega = A001221.
%H Hauke Löffler, <a href="/A325943/b325943.txt">Table of n, a(n) for n = 2..10001</a>
%e a(2) = 2; 2 has one distinct prime divisor {2}, so a(2) = floor(2/1) = 2.
%e a(10) = 5; 10 has two distinct prime divisors {2,5}, so a(10) = floor(10/2) = 5.
%e a(15) = 7; 15 has two distinct prime divisors {3,5}, so a(15) = floor(15/2) = 7.
%o (SageMath)
%o [ n // (len(prime_divisors(n))) for n in range(2, 20) ]
%K nonn
%O 2,1
%A _Hauke Löffler_, Sep 09 2019
|
