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a(n) = A000188(n)/A055229(n).
6

%I #41 Jul 27 2024 09:42:30

%S 1,1,1,2,1,1,1,1,3,1,1,2,1,1,1,4,1,3,1,2,1,1,1,1,5,1,1,2,1,1,1,2,1,1,

%T 1,6,1,1,1,1,1,1,1,2,3,1,1,4,7,5,1,2,1,1,1,1,1,1,1,2,1,1,3,8,1,1,1,2,

%U 1,1,1,3,1,1,5,2,1,1,1,4,9,1,1,2,1,1,1,1,1,3,1,2,1,1,1,2,1,7,3,10,1,1,1,1

%N a(n) = A000188(n)/A055229(n).

%C Previous name: "Square root of largest unitary square divisor of n." The previous name was incorrect for numbers that have an odd exponent in their prime factorization that is larger than 3. For the correct square root of largest unitary square divisor of n see A071974. - _Amiram Eldar_, Jul 26 2024

%C Multiplicative because quotient of two multiplicative sequences. - _Christian G. Bower_, May 16 2005

%H Antti Karttunen, <a href="/A056622/b056622.txt">Table of n, a(n) for n = 1..16384</a>

%F Multiplicative with a(p^e) = p^(e/2) if e even, a(p) = 1, and a(p^e) = p^((e-3)/2) for odd e > 1. - _Amiram Eldar_, Sep 14 2020

%F Dirichlet g.f.: zeta(2*s-1) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s-1) + 1/p^(3*s)). - _Amiram Eldar_, Dec 18 2023

%F a(n) = sqrt(A056623(n)). - _Amiram Eldar_, Jul 26 2024

%e For n = 125: A000188(125) = 5, A055229(125) = 5, so a(125) = 1.

%e For n = 360: A000188(360) = 6, A055229(360) = 2, so a(360) = 3.

%t f[p_, e_] := If[EvenQ[e], p^(e/2), If[e == 1, 1, p^((e - 3)/2)]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Sep 14 2020 *)

%o (PARI)

%o A000188(n) = core(n, 1)[2]; \\ _Michel Marcus_, Feb 27 2013

%o A055229(n) = { my(c=core(n)); gcd(c, n/c); }; \\ _Charles R Greathouse IV_, Nov 20 2012

%o A056622(n) = (A000188(n)/A055229(n)); \\ _Antti Karttunen_, Nov 19 2017

%Y Cf. A000188, A055229, A034444, A056623, A071974.

%K nonn,easy,mult

%O 1,4

%A _Labos Elemer_, Aug 08 2000

%E Name replaced with a formula by _Amiram Eldar_, Jul 26 2024