Consider segregating positive integers into two categories:

*Even-type*: if its prime factorization has an even number of prime(s), or it equals to \(1\).*Odd-type*: if its prime factorization has an odd number of prime(s).

Define

\(E(N) \) to be the number of positive integers of

*even-type*that are less than or equals to \(N\)\(O(N) \) to be the number of positive integers of

*odd-type*that are less than or equals to \(N\)

I claim that \( E(N) \leq O(N) \) for all positive integers \(N\) greater than \(1\) but less than \(10^9\). Am I right?

**Details and assumptions**:

\(18 = 2 \times 3 \times 3 \), which has \(3\) prime factors, so \(18\) is an

*odd-type*\(19 = 19 \), which has \(1\) prime factor, so \(19\) is an

*odd-type*\(33 = 3 \times 11\), which has \(2\) prime factors, so \(33\) is an

*even-type*If \(N=25\), then \(1,4,6,9,10,14,15,16,21,22,24,25\) are

*even-type*, \(2,3,5,7,8,11,12,13,17,18,19,20,23\) are*odd-type*. So \(E(25)= 12, O(25)=13\). Which means \( E(25) \leq O(25) \), thus it's true for \(N=25\).

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