A305762 a(0) = 24, a(n) = 2^(max(0, min(3, p - 1))) * 3^(max(0, min(1, q - 1))) where n = 2^p * 3^q * 5^r * ... .

24, 1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 1, 8, 1, 3, 1, 2, 1, 1, 1, 4, 1, 1, 3, 2, 1, 1, 1, 8, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 2, 3, 1, 1, 8, 1, 1, 1, 2, 1, 3, 1, 4, 1, 1, 1, 2, 1, 1, 3, 8, 1, 1, 1, 2, 1, 1, 1, 12, 1, 1, 1, 2, 1, 1, 1, 8, 3, 1, 1, 2, 1, 1, 1, 4, 1

Offset: 0

Author

Seiichi Manyama, Jun 10 2018

Keywords

• nonn
• mult

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A305756.

Programs

Mathematica

a[n_] := GCD[24, n/GCD[6, n]]; Array[a, 100, 0] (* Amiram Eldar, Oct 15 2022 *)

PARI

a(n)=gcd(24, n/gcd(6,n)) \\ Andrew Howroyd, Jul 24 2018

Ruby

require 'prime'
def A305762(n)
  return 24 if n == 0
  s = 1
  s *= 3 if n % 9 == 0
  n.prime_division.each{|i|
    s *= 2 ** [3, (i[1] - 1)].min if i[0] == 2
  }
  s
end
p (0..144).map{|i| A305762(i)}

Formula

a(n+144) = a(n).

a(n) = gcd(24, n/gcd(6,n)). - Andrew Howroyd, Jul 24 2018

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 77/36. - Amiram Eldar, Oct 15 2022

Extensions

Keyword:mult added by Andrew Howroyd, Jul 24 2018