A342100 Abundant numbers k at which the ratio (number of abundant numbers in 1..k)/k reaches a new record high.

12, 18, 20, 24, 40, 42, 56, 60, 72, 80, 84, 88, 90, 102, 104, 108, 112, 114, 354, 366, 368, 372, 380, 384, 392, 396, 400, 402, 464, 468, 476, 480, 492, 500, 504, 552, 560, 564, 572, 576, 580, 582, 650, 654, 836, 840, 945, 948, 952, 954, 1002, 2002, 2004, 2024

Offset: 1

Author

Jon E. Schoenfield, Feb 27 2021

Keywords

• nonn
• fini

Comments

Let rho(k) = (number of abundant numbers in 1..k)/k. According to A302991 ("Decimal expansion of the asymptotic density of abundant numbers"), lim_{k->infinity} rho(k) = 0.247619...

a(115) = 7254; rho(7254) = 1810/7254 = 0.2495175075820...

Conjecture: a(115) is the final term of this sequence.

This sequence is finite since rho(2212) > A302991 and therefore there is a number N such that abs(rho(n) - A302991) < eps for all n > N and for an arbitrarily small eps > 0. Therefore, the number of values of n for which rho(n) > rho(2212) is finite. - Amiram Eldar, Dec 06 2024

Examples

			k=12 is the 1st abundant number, so at k=12, rho(k) increases from 0 to 1/12 = 0.08333..., a record high, so a(1)=12.
k=18 is the 2nd abundant number, so at k=18, rho(k) reaches 2/18 = 1/9 = 0.11111..., the next record high, so a(2)=18.
k=20 is the 3rd abundant number, so at k=20, rho(k) reaches 3/20 = 0.15, the next record high, so a(3)=20.
k=24 is the 4th abundant number, so at k=24, rho(k) reaches 4/24 = 1/6 = 0.16666..., the next record high, so a(4)=24.
k=30 is the 5th abundant number, so at k=30, rho(k) again reaches 5/30 = 1/6; this is not a new record high, so 30 is not a term of the sequence.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..115

Crossrefs

Cf. A005101 (abundant numbers), A302991, A330899.

Programs

Mathematica

s = {}; c = 0; rm = 0; Do[If[DivisorSigma[1, n] > 2*n, c++; If[(r = c/n) > rm, rm = r; AppendTo[s, n]]], {n, 1, 10^4}]; s (* Amiram Eldar, Feb 28 2021 *)

Extensions

Keyword "fini" added by Amiram Eldar, Dec 06 2024