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Numbers k whose only divisor in A246282 is k itself, i.e., A003961(k) > 2k, but for none of the proper divisors d|k, dA003961(d) > 2d.

Question: Do the odd terms in A326134 all occur here? Answer is yes, if the following conjecture holds: This is a subsequence of A263837, nonabundant numbers. In other words, we claim that any abundant number k (A005101) has A337345(k) > 1 and thus is a term of A341610. (The conjecture indeed holds. See the proof below).

From Antti Karttunen, Dec 06 2024: (Start)

Observation 1: The thirteen initial terms (4, 6, 9, ..., 69, 91) are only semiprimes in A246282, all other semiprimes being in A246281 (but none in A341610), and there seems to be only 678 terms m with A001222(m) = 3, from a(14) = 125 to the last one of them, a(2691) = 519963. There are more than 150000 terms m with A001222(m) = 4. In general, there should be only a finite number of terms m for any given k = A001222(m). Compare for example with A287728.

Observation 2: The intersection with A005101 (and thus also with A091191) is empty, which then implies the claims made in the sequences A378662, A378664, from which further follows that there are no 1's present in any of these sequences: A378658, A378736, A378740.

(End)

Proof of the latter observation by Jianing Song, Dec 11 2024: (Start)

Let's write p' for the next prime after the prime p. Also, write Q(n) = A003961(n)/sigma(n) which is multiplicative.

Proposition: For n > 1 not being a prime nor twice a prime, n has a factor p such that Q(n) > p'/p.

This implies that if n is abundant [including any primitively abundant n in A091191], then n has a factor p such that A003961(n/p)/(n/p) = (A003961(n)/n)/(p'/p) > sigma(n)/n [which is > 2 because n is abundant], so n/p is in A246282, meaning that n cannot be in this sequence.

Proof. We see that 1 <= Q(p) <= Q(p^2) <= ..., which implies that if n verifies the proposition, then every multiple of n also verifies it. Since n = p^2 > 4 and n = 8 verify the proposition, it suffices to consider the case where n = pq is the product of two distinct odd primes. Suppose WLOG that p < q, so q >= p', then using q/(q+1) >= p'/(p'+1) we have

Q(n) = p'q'/((p+1)(q+1)) >= p'^2*q'/(q(p+1)(p'+1)) > (p'^2-1)*q'/(q(p+1)(p'+1)) = (p'-1)/(p+1) * q'/q >= q'/q.

(End)

Number of terms of A246282 that divide n.

Number of divisors d of n for which A048673(d) > d.

Numbers k for which A337345(k) > 1, or equally, for which A337346(k) > 0.

Obviously A337346(n) = 0 for any noncomposite and for any semiprime, thus this is a subsequence of A033942. The first term of A033942 not present here is 125, as A337345(125) = 1.

Empirically checked: in range 1 .. 2^31, all abundant numbers are found in this sequence. For proving this, we should concentrate only on checking A091191, as the set A005101 \ A091191 (non-primitive abundant numbers) is certainly included, as for any divisor d for which sigma(d) > 2*d (or even sigma(d) >= 2*d), we also have A003961(d) > 2*d.

Number of terms of A337381 less than n that divide n.

Terms k of A263837 for which A337345(k) > 1 (or equally, for which A337346(k) > 0).

Numbers k such that sigma(k) <= 2k and k has at least two distinct divisors d such that 2d < A003961(d). By necessity, one of these divisors is then k itself.