A372381 The number of divisors of the largest divisor of n whose number of divisors is a power of 2.

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 8, 4, 4, 2, 8, 4, 4, 4

Offset: 1

Author

Amiram Eldar, Apr 29 2024

Keywords

• nonn
• easy
• mult

Comments

First differs from A286324 at n = 32, and from A331109 at n = 64.

Also, the number of infinitary divisors of the largest divisor of n whose number of divisors is a power of 2.

Links

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

Crossrefs

Cf. A000005, A004709, A036537, A037445, A372379, A372380.

Cf. A286324, A331109.

Programs

Mathematica

f[p_, e_] := 2^Floor[Log2[e + 1]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = vecprod(apply(x -> 2^exponent(x+1), factor(n)[, 2]));

Python

from math import prod
from sympy import factorint
def A372381(n): return prod(1<<(e+1).bit_length()-1 for e in factorint(n).values()) # Chai Wah Wu, Apr 30 2024

Formula

Multiplicative with a(p^e) = 2^floor(log_2(e+1)).

a(n) = A000005(A372379(n)).

a(n) = A037445(A372379(n)).

a(n) = A000005(n) if and only if n is in A036537.

a(n) <= A372380(n), with equality if and only if n is cubefree (A004709).