A377563 -id:A377563 - cp's OEIS frontend

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

Amiram Eldar, Nov 01 2024

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.

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

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

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]

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

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI