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a(n) = A056622(n!).
3

%I #25 Jul 27 2024 15:03:27

%S 1,1,1,1,1,12,12,12,36,720,720,480,480,1680,3024,12096,12096,145152,

%T 145152,7257600,345600,1900800,1900800,136857600,684288000,4447872000,

%U 4447872000,435891456000,435891456000,3138418483200,3138418483200,6276836966400,190207180800

%N a(n) = A056622(n!).

%C Previous name "Square root of largest unitary square divisor of n!" was incorrect. See A374989 for the correct sequence with this name. - _Amiram Eldar_, Jul 26 2024

%F a(n) = A055772(n)/A055230(n) = A000188(n!)/A055229(n!).

%F a(n) = A056622(n!). - _Michel Marcus_, Aug 16 2020

%F a(n) = sqrt(A056628(n)). - _Amiram Eldar_, Jul 08 2024

%e a(12) = A056622(12!) = A000188(12!)/A055229(12!) = 1440/3 = 480.

%t f[p_, 1] := 1; f[p_, e_] := If[EvenQ[e], p^(e/2), p^((e-3)/2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 33] (* _Amiram Eldar_, Jul 08 2024 *)

%o (PARI)

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

%o A008833(n) = n/core(n) \\ _Michael B. Porter_, Oct 17 2009

%o A056623(n) = (A008833(n)/(A055229(n)^2)); \\ _Antti Karttunen_, Nov 19 2017

%o a(n) = sqrtint(A056623(n!)); \\ _Michel Marcus_, Aug 16 2020

%Y Cf. A055772, A055230, A000188, A055229, A056622, A056628, A374989.

%K nonn

%O 1,6

%A _Labos Elemer_, Aug 08 2000

%E More terms from _Michel Marcus_, Aug 16 2020

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