A372379 The largest divisor of n whose number of divisors is a power of 2.

1, 2, 3, 2, 5, 6, 7, 8, 3, 10, 11, 6, 13, 14, 15, 8, 17, 6, 19, 10, 21, 22, 23, 24, 5, 26, 27, 14, 29, 30, 31, 8, 33, 34, 35, 6, 37, 38, 39, 40, 41, 42, 43, 22, 15, 46, 47, 24, 7, 10, 51, 26, 53, 54, 55, 56, 57, 58, 59, 30, 61, 62, 21, 8, 65, 66, 67, 34, 69, 70

Offset: 1

Author

Amiram Eldar, Apr 29 2024

Keywords

• nonn
• easy
• mult

Comments

First differs from A350390 at n = 32.

The largest term of A036537 dividing n.

The largest divisor of n whose exponents in its prime factorization are all of the form 2^k-1 (A000225).

Links

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

Crossrefs

Cf. A000225, A000523, A036537, A138302, A162643, A282940, A350390, A353897, A372380, A372381.

Programs

Mathematica

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

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i, 1]^(2^exponent(f[i, 2]+1)-1));}

Python

from math import prod
from sympy import factorint
def A372379(n): return prod(p**((1<<(e+1).bit_length()-1)-1) for p, e in factorint(n).items()) # Chai Wah Wu, Apr 30 2024

Formula

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

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

a(A162643(n)) = A282940(n).

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 0.7907361848... = Product_{p prime} (1 + Sum_{k>=1} (p^f(k) - p^(f(k-1)+1))/p^(2*k)), f(k) = 2^floor(log_2(k))-1 for k >= 1, and f(0) = 0.