A377562 Numbers that have twice as many infinitary divisors as noninfinitary divisors.

4, 9, 12, 18, 20, 25, 28, 32, 44, 45, 49, 50, 52, 60, 63, 68, 72, 75, 76, 84, 90, 92, 96, 98, 99, 108, 116, 117, 121, 124, 126, 132, 140, 147, 148, 150, 153, 156, 160, 164, 169, 171, 172, 175, 188, 198, 200, 204, 207, 212, 220, 224, 228, 234, 236, 242, 243, 244, 245

Offset: 1

Author

Amiram Eldar, Nov 01 2024

Keywords

• nonn
• easy

Comments

Numbers k such that A037445(k) = 2 * A348341(k).

Numbers k such that A037445(k)/A000005(k) = 2/3. For numbers k in A036537, all the divisors are infinitary divisors, so A037445(k)/A000005(k) = 1. For numbers k that are not in A036537, the largest possible ratio A037445(k)/A000005(k) is 2/3.

Numbers whose prime factorization has exactly one exponent of the form 3*2^k-1, with k >= 0, and the rest of the exponents, if there are any, are of the form 2^k-1, with k >= 1.

The asymptotic density of this sequence is d * Sum_{p prime} (Sum_{k>=0} 1/p^(3*2^k-1))/(1 + Sum_{k>=1} 1/p^(2^k-1)) = 0.23171917985739378623..., where d = A327839.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000225, A036537, A037445, A060687 (unitary analog), A077609, A083329, A327839, A335832, A348341, A377563.

Programs

Mathematica

f[p_, e_] := 2^DigitCount[e, 2, 1]/(e + 1); q[1] = False; q[n_] := Times @@ f @@@ FactorInteger[n] == 2/3; Select[Range[250], q]

PARI

is(n) = {my(f = factor(n)); vecprod(apply(x -> (1 << hammingweight(x)) / (x+1), f[, 2])) == 2/3;}