A391373 The largest number with an arithmetic mean of divisors equal to n or 0 is there is no such number.

1, 3, 6, 7, 0, 15, 20, 21, 30, 27, 0, 42, 45, 60, 56, 31, 0, 70, 49, 57, 96, 43, 0, 105, 0, 126, 110, 140, 0, 168, 150, 93, 86, 67, 116, 210, 73, 147, 198, 189, 0, 224, 0, 129, 280, 0, 0, 231, 260, 0, 134, 315, 0, 330, 109, 420, 294, 0, 0, 378, 169, 432, 480

Offset

1,2

Comments

a(n) is the largest k such that sigma(k)/tau(k) = A000203(k)/A000005(k) = n.

Zeros occur if n is in A157847 (not in A157846).

a(n) is finite for all n since sigma(k) >= k+1 and tau(k) <= 2*sqrt(k) implies no qualifying k can exceed 4*n^2.

Links

Table of n, a(n) for n=1..63.

Formula

a(n) = max({k : sigma(k) = n*tau(k)} U {0}).

Example

a(2) = 3 because 3 is the largest k that satisfies sigma(k)/tau(k) = 2.
a(3) = 6 because 6 is the largest k that satisfies sigma(k)/tau(k) = 3.
a(5) = 0 because no k satisfies sigma(k)/tau(k) = 5.

Prog

(Python)
from sympy import divisor_count, divisor_sigma
def a(n):
    max_k = 0
    for k in range(1, 4*n*n + 1):
        if divisor_sigma(k) == n * divisor_count(k): max_k = k
    return max_k
(PARI) a(n) = forstep(k=4*n^2+1, 1, -1, my(f=factor(k)); if (sigma(f)/numdiv(f) == n, return(k))); \\ Michel Marcus, Dec 08 2025

Crossrefs

Cf. A000005, A000203, A162538, A003601, A157847.

Sequence in context: A032338 A382550 A081814 * A133340 A133329 A275696

Adjacent sequences: A391370 A391371 A391372 * A391374 A391375 A391376

Keyword

nonn

Author

Joe Anderson, Dec 07 2025

Status

approved