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Number of numbers <= n with same number of distinct prime factors as n.
10

%I #11 Jul 13 2019 17:52:58

%S 1,1,2,3,4,1,5,6,7,2,8,3,9,4,5,10,11,6,12,7,8,9,13,10,14,11,15,12,16,

%T 1,17,18,13,14,15,16,19,17,18,19,20,2,21,20,21,22,22,23,23,24,25,26,

%U 24,27,28,29,30,31,25,3,26,32,33,27,34,4,28,35,36,5,29,37,30,38,39,40,41

%N Number of numbers <= n with same number of distinct prime factors as n.

%H Rémy Sigrist, <a href="/A067003/b067003.txt">Table of n, a(n) for n = 1..10000</a>

%F a(A002110(n)) = 1.

%e a(11)=8 since 2,3,4,5,7,8,9,11 each have one distinct prime factor. a(12)=3 since 6,10,12 each have two distinct prime factors.

%e From _Gus Wiseman_, Dec 28 2018: (Start)

%e Column n lists the a(n) positive integers less than or equal to n with the same number of distinct prime factors as n:

%e 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

%e ---------------------------------------------------------------------

%e 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

%e 2 3 4 5 7 8 6 9 10 11 12 14 13 16 15 17 18

%e 2 3 4 5 7 8 6 9 10 12 11 13 14 16 15

%e 2 3 4 5 7 8 6 10 9 11 12 13 14

%e 2 3 4 5 7 6 8 9 10 11 12

%e 2 3 4 5 7 8 6 9 10

%e 2 3 4 5 7 8 6

%e 2 3 4 5 7

%e 2 3 4 5

%e 2 3 4

%e 2 3

%e 2

%e (End)

%t Table[Length[Select[Range[n],PrimeNu[#]==PrimeNu[n]&]],{n,100}] (* _Gus Wiseman_, Dec 28 2018 *)

%o (PARI) a(n) = my(nb = #factor(n)~); sum(k=1, n, #factor(k)~ == nb); \\ _Michel Marcus_, Jul 13 2019

%Y Positions of 1's are A002110.

%Y Cf. A001221, A008479, A058933, A067004. Inverse of A000961, A007774, A033992, A033993, A051270 etc.

%Y Cf. A000010, A006049, A061142, A294277, A294278, A302242, A322837, A322841.

%K nonn,look

%O 1,3

%A _Henry Bottomley_, Dec 21 2001