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

1,1

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

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]

Prog

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

Crossrefs

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

Sequence in context: A059269 A369209 A081619 * A336594 A304365 A038109

Adjacent sequences: A377559 A377560 A377561 * A377563 A377564 A377565

Keyword

nonn,easy

Author

Amiram Eldar, Nov 01 2024

Status

approved