cp's OEIS Frontend

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.

Number of terms of A341614 that divide n.

Claim: a(n) > 0 if and only if A003961(n) > 2*n [i.e., n is in A246282]. That a(n) must be zero when n is in A246281 is obvious, as is also that a(n) > 0 when n is a term of A341614 [as then A378664(n) = n], but that a(n) > 0 for all abundant numbers (A005101) is slightly less clear. So the claim boils down to this: All abundant numbers have at least one (by necessity a proper) divisor d|n such that it is in A341614, i.e., sigma(d) <= 2*d < A003961(d), i.e., that for abundant numbers n, A337345(n) is always strictly greater than A080224(n). Equivalently, of the all nonabundant divisors d of an abundant number, at least one is primeshift-abundant, i.e., A003961(d) > 2*d. This has been proved Dec 11 2024 by Jianing Song in A337372. The claim given in A378658 also follows from that proof.

Term 1 does not occur in this sequence, thus all terms are in A341614, and also in A246282. See A378658, A378662, A378664, A378736 and A337372 for a proof.