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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 * ... .
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%I #19 Oct 15 2022 08:09:49

%S 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,

%T 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,

%U 2,1,1,1,12,1,1,1,2,1,1,1,8,3,1,1,2,1,1,1,4,1

%N 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 * ... .

%H Seiichi Manyama, <a href="/A305762/b305762.txt">Table of n, a(n) for n = 0..10000</a>

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

%F a(n) = gcd(24, n/gcd(6,n)). - _Andrew Howroyd_, Jul 24 2018

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

%t a[n_] := GCD[24, n/GCD[6, n]]; Array[a, 100, 0] (* _Amiram Eldar_, Oct 15 2022 *)

%o (PARI) a(n)=gcd(24, n/gcd(6,n)) \\ _Andrew Howroyd_, Jul 24 2018

%o (Ruby)

%o require 'prime'

%o def A305762(n)

%o return 24 if n == 0

%o s = 1

%o s *= 3 if n % 9 == 0

%o n.prime_division.each{|i|

%o s *= 2 ** [3, (i[1] - 1)].min if i[0] == 2

%o }

%o s

%o end

%o p (0..144).map{|i| A305762(i)}

%Y Cf. A305756.

%K nonn,mult

%O 0,1

%A _Seiichi Manyama_, Jun 10 2018

%E Keyword:mult added by _Andrew Howroyd_, Jul 24 2018