A346729 Maximum number of divisors among n-bit numbers.

1, 2, 4, 6, 8, 12, 16, 20, 24, 32, 40, 48, 64, 80, 96, 120, 144, 168, 200, 240, 288, 360, 432, 504, 600, 720, 864, 1008, 1152, 1344, 1600, 1920, 2304, 2688, 3072, 3584, 4096, 4800, 5760, 6720, 7680, 8640, 10080, 11520, 13824, 16128, 18432, 20736, 23040, 27648

Offset: 1

Author

Jon E. Schoenfield, Jul 30 2021

Keywords

• nonn

Comments

a(n) is the maximum value of tau(k)=A000005(k) for k in the interval [2^(n-1), 2^n - 1]. For n >= 3, that smallest k at which tau(k) is maximized in that interval is A036484(n).

No term is repeated: for n >= 1, if k is the number in [2^(n-1), 2^n - 1] at which tau(k) is maximized (i.e., tau(k) = a(n)), then 2k, which will be a number in [2^n, 2^(n+1) - 1], will have more divisors than k has, so a(n+1) >= tau(2k) > tau(k) = a(n).

Examples

			There are four 3-bit numbers: 4 = 100_2, 5 = 101_2 = 5, 6 = 110_2, 7 = 111_2. 5 and 7 are both prime, so each has 2 divisors; 4 = 2^2 has 3 divisors (1, 2, and 4), and 6 = 2*3 has 4 divisors (1, 2, 3, and 6). Thus, the maximum number of divisors among 3-bit numbers is A000005(6) = 4, so a(3)=4.
		

Crossrefs

Cf. A000005, A002183, A036484, A346730.

Programs

Mathematica

a[n_]:=Max[Table[DivisorSigma[0,k],{k,2^(n-1),2^n-1}]]; Table[a[n],{n,23}] (* Stefano Spezia, Aug 02 2021 *)

PARI

a(n) = vecmax(apply(numdiv, [2^(n-1)..2^n-1])); \\ Michel Marcus, Aug 03 2021

Python

from sympy import divisors
def a(n): return max(len(divisors(n)) for n in range(2**(n-1), 2**n))
print([a(n) for n in range(1, 18)]) # Michael S. Branicky, Aug 02 2021