A391368 - OEIS

%I #9 Dec 14 2025 13:55:51

%S 1,1,2,1,0,3,2,1,3,2,0,4,1,3,4,1,0,5,2,1,5,1,0,5,0,2,4,4,0,6,2,1,1,1,

%T 1,8,1,2,3,3,0,8,0,1,9,0,0,5,3,0,2,3,0,8,1,5,5,0,0,9,1,5,10,2,3,3,0,1,

%U 1,7,0,10,0,1,2,2,0,5,1,2,5,1,0,12,0,0

%N Number of positive integers k with an arithmetic mean of divisors equal to n.

%C a(n) is the number of integers k such that sigma(k)/tau(k) = A000203(k)/A000005(k) = n.

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

%C 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.

%F a(n) = |{k in Z+ : sigma(k) = n*tau(k)}|.

%e For n=2, we need sigma(k)/tau(k) = 2. The only qualifying integer is 3 so a(2) = 1.

%e For n=6, we need sigma(k)/tau(k) = 6. The qualifying integers are 11, 14, 15 so a(6) = 3.

%o (Python)

%o from sympy import divisor_count, divisor_sigma

%o def a(n):

%o     count = 0

%o     for k in range(1, 4*n*n + 1):

%o         if divisor_sigma(k) == n * divisor_count(k): count += 1

%o     return count

%Y Cf. A000005, A000203, A162538, A157847, A157846.

%K nonn

%O 1,3

%A _Joe Anderson_, Dec 07 2025

%E More terms from _Sean A. Irvine_, Dec 14 2025