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A377505 a(n) is the number of positive integers that have Omega(n) prime factors and these are all <= n.

%I A377505 #21 Dec 24 2024 13:49:10
%S A377505 1,1,2,3,3,6,4,20,10,10,5,35,6,21,21,126,7,84,8,120,36,36,9,495,45,45,
%T A377505 165,165,10,220,11,3003,66,66,66,1001,12,78,78,1365,13,455,14,560,560,
%U A377505 105,15,11628,120,680,120,680,16,3876,136,3876,136,136,17,4845,18
%N A377505 a(n) is the number of positive integers that have Omega(n) prime factors and these are all <= n.
%C A377505 If drawing at random with replacement from the primes <= n as many as n has prime factors, 1/a(n) is the probability that the product of the prime numbers drawn is equal to n.
%H A377505 Felix Huber, <a href="/A377505/b377505.txt">Table of n, a(n) for n = 1..10000</a>
%H A377505 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/FundamentalTheoremofArithmetic.html">Fundamental theorem of arithmetic</a>
%F A377505 a(n) = binomial(pi(n) + Omega(n) - 1, Omega(n)) where pi = A000720 and Omega = A001222.
%F A377505 a(p) = pi(p) for prime p.
%e A377505 a(4) = 3 because 3 positive integers have Omega(4) = 2 prime factors <= 4: 4 = 2*2, 6 = 2*3, 9 = 3*3.
%e A377505 a(6) = 6 because 6 positive integers have Omega(6) = 2 prime factors <= 6: 4 = 2*2, 6 = 2*3, 9 = 3*3, 10 = 2*5, 15 = 3*5, 25 = 5*5.
%e A377505 a(7) = 4 because 4 positive integers have Omega(7) = 1 prime factor <= 7: 2, 3, 5, 7.
%p A377505 with(NumberTheory):
%p A377505 A377505:=n->binomial(pi(n)+Omega(n)-1,Omega(n));
%p A377505 seq(A377505(n),n=1..61);
%t A377505 Table[Binomial[PrimePi[n]+PrimeOmega[n]-1,PrimeOmega[n]],{n,61}] (* _James C. McMahon_, Dec 24 2024 *)
%Y A377505 Cf. A000040, A000720, A001222, A007318, A037031, A377537.
%K A377505 nonn
%O A377505 1,3
%A A377505 _Felix Huber_, Dec 20 2024