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

1,2

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).

Links

Table of n, a(n) for n=1..50.

Example

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.

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 *)

Prog

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

Crossrefs

Cf. A000005, A002183, A036484, A346730.

Sequence in context: A036912 A306371 A333255 * A346311 A097921 A185976

Adjacent sequences: A346726 A346727 A346728 * A346730 A346731 A346732

Keyword

nonn

Author

Jon E. Schoenfield, Jul 30 2021

Status

approved