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a(n) = n^2*(n+1)!/(n^tau(n)) where tau(n) is the number of divisors of n.
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%I #12 Aug 03 2024 07:09:32

%S 2,6,24,30,720,140,40320,5670,403200,399168,479001600,300300,

%T 87178291200,6671808000,92990177280,86837751000,6402373705728000,

%U 1158789632000,2432902008176640000,319318388573184,2548754484756480000,53413257724968960000,620448401733239439360000

%N a(n) = n^2*(n+1)!/(n^tau(n)) where tau(n) is the number of divisors of n.

%C The expression n^2*(n+1)!/(n^tau(n)) is always an integer. n^tau(n) is also the product of square divisors of n (cf. A062758).

%H Amiram Eldar, <a href="/A069141/b069141.txt">Table of n, a(n) for n = 1..448</a>

%t a[n_] := (n+1)! / n^(DivisorSigma[0,n] - 2); Array[a, 23] (* _Amiram Eldar_, Aug 03 2024 *)

%o (PARI) a(n) = (n+1)! / n^(numdiv(n)-2); \\ _Amiram Eldar_, Aug 03 2024

%Y Cf. A000005, A062758.

%K easy,nonn

%O 1,1

%A _Benoit Cloitre_, Apr 08 2002

%E a(22)-a(23) from _Amiram Eldar_, Aug 03 2024