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Not squarefree, not a nontrivial prime power and not in {squarefree} times {nontrivial prime powers}.

Numbers k such that A056170(k) > 1. The asymptotic density of this sequence is 1 - (6/Pi^2) * (1 + A154945) = 0.05668359058... - Amiram Eldar, Nov 01 2020

From Amiram Eldar, Nov 01 2020: (Start)

Also, numbers with more than two non-unitary prime divisors, i.e., numbers k such that A056170(k) > 2, or equivalently, numbers divisible by the squares of three distinct primes.

The complement of the union of A005117, A190641 and A338539.

The asymptotic density of this sequence is 1 - 6/Pi^2 - (6/Pi^2)*A154945 - (3/Pi^2)*(A154945^2 - A324833) = 0.0033907041... (End)

Numbers k such that A348039(k) > 1.

Numbers k such that A348037(k) < A003557(k).

Numbers k such that A327564(k) > A348038(k).

Differs from A036785 and A338539 for the first time at n=20, where a(n) = 450, as A036785(20) = A338539(20) = 441 is not included in this sequence.

Numbers k such that A056170(k) = A001221(A057521(k)) = 3.

Numbers divisible by the squares of exactly three distinct primes.

Subsequence of A318720 and first differs from it at n = 123.

The asymptotic density of this sequence is (eta_1^3 - 3*eta_1*eta_2 + 2*eta_3)/Pi^2 = 0.0032920755..., where eta_j = Sum_{p prime} 1/(p^2-1)^j (Pomerance and Schinzel, 2011).

Numbers k such that A056170(k) = A001221(A057521(k)) = 4.

Numbers divisible by the squares of exactly four distinct primes.

The asymptotic density of this sequence is (eta_1^4 - 6*eta_1^2*eta_2 + 3*eta_2^2 + 8*eta_1*eta_3 - 6*eta_4)/(4*Pi^2) = 0.0000970457..., where eta_j = Sum_{p prime} 1/(p^2-1)^j (Pomerance and Schinzel, 2011).

Numbers k such that A056170(k) = A001221(A057521(k)) = 5.

Numbers divisible by the squares of exactly five distinct primes.

The asymptotic density of this sequence is (eta_1^5 - 10*eta_1^3*eta_2 + 15*eta_1*eta_2^2 + 20*eta_1^2*eta_3 - 20*eta_2*eta_3 - 30*eta_1*eta_4 + 24*eta_5)/(20*Pi^2) = 0.0000015673..., where eta_j = Sum_{p prime} 1/(p^2-1)^j (Pomerance and Schinzel, 2011).

All terms k are multiples of 36 and A056170(k) = 2. Proof: Imagine k had a non-unitary prime divisor p^e, with p > 3 and e > 1. Then p^e divides A003968(k) only if k has also another non-unitary prime divisor q^h (with h > 1), such that p divides (q+1), which implies that q > p. But then q^h divides A003968(k) only if there is yet another non-unitary prime divisor r^i, such that r > q (and i > 1), and so on, which is clearly impossible by reductio ad infinitum. Therefore we should consider only the cases p=2 and p=3, because they are only primes that can occur as non-unitary prime factors in k, and at least either of them must occur with exponent larger than one, because every k is nonsquarefree. Let e = A007814(k) and h = A007949(k), so that 2^e and 3^h are the highest powers of 2 and 3 that divide k. Because A003968 changes "extra" 2's to 3's and extra 3's to 4's, it must follow that e >= h > e/2. Therefore, if e >= 2 (k is a multiple of 4), h must be at least 2. On the other hand, if h >= 2, then e also must be at least 2. In other words, if k is a multiple of 4, it must then also be a multiple of 9, and vice versa, thus k is a multiple of 36 and k has exactly two non-unitary prime divisors (2^e and 3^h, with e, h > 1), therefore this is a subsequence of A338539.