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Number of subsets of {1..n} whose elements have the same number of prime factors, counted with multiplicity.
3

%I #16 Mar 20 2021 17:51:16

%S 1,2,3,5,6,10,12,20,21,25,33,49,51,83,99,131,132,196,200,328,336,400,

%T 528,784,786,1042,1554,1570,1602,2114,2178,3202,3203,4227,6275,10371,

%U 10375,12423,20615,36999,37007,41103,41231,49423,49679,50191,82959,99343,99345,164881,165905,296977,299025,331793,331809,593953,593985,1118273,2166849,2232385

%N Number of subsets of {1..n} whose elements have the same number of prime factors, counted with multiplicity.

%H Sebastian Karlsson, <a href="/A339511/b339511.txt">Table of n, a(n) for n = 0..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeFactor.html">Prime Factor</a>

%F a(n) = 1 + Sum_{k=1..n} 2^A335097(k). - _Sebastian Karlsson_, Feb 23 2021

%e a(6) = 12 subsets: {}, {1}, {2}, {3}, {4}, {5}, {6}, {2, 3}, {2, 5}, {3, 5}, {4, 6} and {2, 3, 5}.

%t Array[Count[Subsets@ Range[#], _?(SameQ @@ PrimeOmega[#] &)] &, 16, 0] (* _Michael De Vlieger_, Feb 24 2021 *)

%o (Python)

%o from sympy import primeomega

%o def test(n):

%o if n<2: return n-1

%o return primeomega(n)

%o def a(n):

%o tests = [test(i) for i in range(n+1)]

%o return sum(2**tests.count(v)-1 for v in set(tests))

%o print([a(n) for n in range(60)]) # _Michael S. Branicky_, Dec 07 2020

%Y Cf. A001222, A319169, A339512, A339514.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Dec 07 2020

%E a(25)-a(48) from _Michael S. Branicky_, Dec 07 2020