A390862 - OEIS

%I #18 Nov 28 2025 22:57:48

%S 2,7,12,19,23,29,35,41,51,58,63,70,77,85,90,95,111,116,122,129,136,

%T 147,153,163,171,176,183,192,204,211,219,228,236,244,252,262,269,272,

%U 279,288,302,309,314,325,334,343,356,363,375,380,389,398,406,419,431,440,451,459,468,478,484,497

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

%C 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.

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

%e For n=2, we need d(k) >= k/2. The qualifying integers are 1, 2, 3, 4, 6, 8, 12 so a(2) = 7.

%e For n=3, we need d(k) >= k/3. The additional qualifying integers are 5, 9, 10, 18, 24 so a(3) = 12.

%e 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.

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

%o (Python)

%o from sympy import divisor_count

%o def a(n):

%o     count = 0

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

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

%o     return count

%Y Cf. A000005, A051521.

%K nonn,easy

%O 1,1

%A _Joe Anderson_, Nov 22 2025