cp's OEIS Frontend

Numbers n such that A003961(n) > 2*n.

Numbers n such that A048673(n) > n.

The sequence grows as:

a(10) = 18

a(100) = 192

a(1000) = 1830

a(10000) = 18636

a(100000) = 187350

a(1000000) = 1865226

a(10000000) = 18654333

and the powers of 10 occur at:

a(5) = 10

a(53) = 100

a(536) = 1000

a(5423) = 10000

a(53290) = 100000

a(535797) = 1000000

a(5361886) = 10000000

suggesting that the ratio a(n)/n is converging to an constant and an arbitrary natural number is slightly more likely to be in this sequence than in the complement A246281. See also comments at A246351 and compare to quite a different ratio present in the "inverse" case A246362.

From Antti Karttunen, Aug 27 2020: (Start)

Any perfect number, including all odd perfect numbers (if such numbers exist), must occur in this sequence. See A286385 and A326042 for the reason why.

Like abundancy index (ratio A000203(n)/n), also ratio A003961(n)/n is multiplicative and always > 1 for all n > 1. Thus if the number has a proper divisor that is in this sequence, then the number itself also is. See A337372 for terms included here, but with no proper divisor in this sequence. (End)

For k >= 2, if m * A130789(k) is a term then m * A130789(k-1) is a term. - Peter Munn, Sep 01 2025

Could be called "primeshift-abundant numbers", in analogy with A005101. - Antti Karttunen, Sep 01 2025

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.

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.