A368523 -id:A368523 - cp's OEIS frontend

A354768 Numbers k such that d(k)/k >= d(m)/m for all m > k, where d(k) is the number-of-divisors function A000005(k).

1, 2, 4, 6, 8, 12, 18, 24, 30, 36, 48, 60, 72, 84, 90, 120, 144, 180, 240, 252, 360, 420, 480, 504, 540, 720, 840, 900, 1008, 1080, 1260, 1440, 1680, 1800, 2520, 2640, 2880, 3360, 3780, 3960, 5040, 5280, 5400, 5460, 5544, 6300, 7560, 7920, 8400, 10080, 10920, 12600, 15120, 15840, 16380, 18480

Offset: 1

N. J. A. Sloane, Jun 21 2022

nonn

Because of the bound d(m) <= 2*sqrt(m), in order for k to be in the sequence it suffices that d(k)/k >= d(m)/m for k < m < (2*k/d(k))^2. - Robert Israel, Jan 23 2023

David desJardins, Posting to Math Fun Mailing List, Jun 21 2022.

Cf. A000005, A066523, A354769, A368523.

N:= 10^6:
Q:= [seq(numtheory:-tau(k)/k, k=1..N)]:
V:= Vector(10^6):
r:= 2/10^3:
for n from 10^6 to 1 by -1 do
r:= max(Q[n],r);
V[n]:= r;
od:
select(i -> Q[i] >= V[i+1], [$1..10^6-1]); # Robert Israel, Jan 23 2023

More terms from Robert Israel, Jan 23 2023

A323383 Proper divisors of 24.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 12

Offset: 1

Keith J. Bauer, Jan 12 2019

These are the only 7 positive integers k which meet the constraint tau(k) >= k/2, where tau(k) = A000005(k). Inductively, there can only be a finite number of integers which meet this constraint. 1 has a perfect tau(k) / k ratio at 1. Every time a j-th power of a prime is multiplied by it, its ratio is multiplied by (j + 1)/p^j. Although 2 also achieves a perfect score, the scores must degrade after 2 because the above ratio is less than 1 otherwise.

			tau(1) = 1 >= 0.5
tau(2) = 2 >= 1
tau(3) = 2 >= 1.5
tau(4) = 3 >= 2
so 1, 2, 3, 4 are in the sequence.
tau(5) = 2 < 2.5
so 5 is not in the sequence.

Cf. A000005, A368523.

Equals A018253 without 24.

Select[Range[10^3], 2 DivisorSigma[0, #] >= # &] (* Michael De Vlieger, Jan 20 2019 *)

for (n = 1, 100, if (sigma(n, 0) >= n / 2, print1(n, ", ")));

Renamed by Andrey Zabolotskiy, Apr 15 2025

cp's OEIS Frontend

A354768 Numbers k such that d(k)/k >= d(m)/m for all m > k, where d(k) is the number-of-divisors function A000005(k).

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