A390862 Number of positive integers k such that at least 1/n of the integers from 1 to k are divisors of k.

2, 7, 12, 19, 23, 29, 35, 41, 51, 58, 63, 70, 77, 85, 90, 95, 111, 116, 122, 129, 136, 147, 153, 163, 171, 176, 183, 192, 204, 211, 219, 228, 236, 244, 252, 262, 269, 272, 279, 288, 302, 309, 314, 325, 334, 343, 356, 363, 375, 380, 389, 398, 406, 419, 431, 440, 451, 459, 468, 478, 484, 497

Offset

1,1

Comments

a(n) is finite for all n since d(k) <= 2*sqrt(k) and d(k) >= k/n implies no qualifying k can exceed 4*n^2. Here d(k) = A000005(k) is the number of divisors of k.

Links

Table of n, a(n) for n=1..62.

Formula

a(n) = |{k in Z+ : d(k) >= k/n}|, where d(k) = A000005(k).

Example

For n=2, we need d(k) >= k/2. The qualifying integers are 1, 2, 3, 4, 6, 8, 12 so a(2) = 7.
For n=3, we need d(k) >= k/3. The additional qualifying integers are 5, 9, 10, 18, 24 so a(3) = 12.
For n=4, we need d(k) >= k/4. The additional qualifying integers are 7, 14, 15, 16, 20, 30, 36 so a(4) = 19.

Mathematica

Table[Count[Range[4*n^2], _?(DivisorSigma[0, #]*n >= # &)], {n, 62}] (* Michael De Vlieger, Nov 23 2025 *)

Prog

(Python)
from sympy import divisor_count
def a(n):
    count = 0
    for k in range(1, 4*n*n + 1):
        if divisor_count(k) * n >= k: count += 1
    return count

Crossrefs

Cf. A000005, A051521.

Sequence in context: A099353 A297432 A299401 * A188039 A133459 A023669

Adjacent sequences: A390859 A390860 A390861 * A390863 A390864 A390865

Keyword

nonn,easy

Author

Joe Anderson, Nov 22 2025

Status

approved