A346695 Numbers with more divisors than digits in their binary representation.

6, 12, 18, 20, 24, 28, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 105, 108, 110, 112, 114, 120, 126, 132, 140, 144, 150, 156, 160, 162, 168, 176, 180, 192, 196, 198, 200, 204, 208, 210, 216, 220, 224, 225, 228, 234, 240, 252, 260

Offset: 1

Author

Alex Meiburg, Jul 29 2021

Keywords

• nonn
• base
• easy

Comments

Not all terms are perfect or abundant, with 105 being the first deficient term.

There are no primes in the sequence, and 6 is the only semiprime.

By the same comments as those at A175495, this sequence is infinite.

This sequence is a subsequence of A175495.

It is natural to conjecture that this sequence has asymptotic density 0. However, after the first three terms where a(n)/n = 6 -- a function which would increase to infinity if the asymptotic density were zero -- it drops, and it seems to take a long time to get that high again. The first time it gets above 5.0 is at a(30243)=151216. Even as high as a(2188516)=10000000, the density is only ~1/4.57.

The number of terms with m binary digits is Sum_{k>m} A346730(m,k). - Jon E. Schoenfield, Jul 31 2021

Examples

			12 has 6 divisors: {1,2,3,4,6,12}. 12 is written in binary as 1100, which has 4 digits. Since 6 > 4, 12 is in the sequence.
		

Crossrefs

Cf. A000005, A070939.

Cf. A135772 (equal number rather than more).

Cf. A175495 (where "binary digits in n" is replaced by "log_2(n)").

Programs

Mathematica

Select[Range[1000], (DivisorSigma[0, #] > Floor[1 + Log2[#]]) &]

PARI

isok(m) = numdiv(m) > #binary(m); \\ Michel Marcus, Jul 29 2021

Python

from sympy import divisor_count
def ok(n): return divisor_count(n) > n.bit_length()
print(list(filter(ok, range(1, 261)))) # Michael S. Branicky, Jul 29 2021