A350389 a(n) is the largest unitary divisor of n that is an exponentially odd number (A268335).

1, 2, 3, 1, 5, 6, 7, 8, 1, 10, 11, 3, 13, 14, 15, 1, 17, 2, 19, 5, 21, 22, 23, 24, 1, 26, 27, 7, 29, 30, 31, 32, 33, 34, 35, 1, 37, 38, 39, 40, 41, 42, 43, 11, 5, 46, 47, 3, 1, 2, 51, 13, 53, 54, 55, 56, 57, 58, 59, 15, 61, 62, 7, 1, 65, 66, 67, 17, 69, 70, 71

Offset: 1

Author

Amiram Eldar, Dec 28 2021

Keywords

• nonn
• easy
• mult

Links

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

Crossrefs

Cf. A000290, A001222, A007913, A268335, A350387, A350388, A350390.

Programs

Mathematica

f[p_, e_] := If[OddQ[e], p^e, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%2, f[i,1]^f[i,2], 1));} \\ Amiram Eldar, Sep 18 2023

Python

from math import prod
from sympy import factorint
def A350389(n): return prod(p**e if e % 2 else 1 for p, e in factorint(n).items()) # Chai Wah Wu, Feb 24 2022

Formula

Multiplicative with a(p^e) = p^e if e is odd and 1 otherwise.

a(n) = n/A350388(n).

A001222(a(n)) = A350387(n).

a(n) = 1 if and only if n is a positive square (A000290 \ {0}).

a(n) = n if and only if n is an exponentially odd number (A268335).

Sum_{k=1..n} a(k) ~ (1/2)*c*n^2, where c = Product_{p prime} (1 - p/(1+p+p^2+p^3)) = 0.7406196365...

Dirichlet g.f.: zeta(2*s-2) * zeta(2*s) * Product_{p prime} (1 + 1/p^(s-1) - 1/p^(2*s-2) - 1/p^(3*s-1)). - Amiram Eldar, Sep 18 2023