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a(n) = n!*(n!+1)/2.
2

%I #30 Sep 08 2022 08:45:01

%S 1,1,3,21,300,7260,259560,12703320,812871360,65841128640,

%T 6584096534400,796675481078400,114721266640780800,

%U 19387894024929830400,3800027228319587865600,855006126362753549184000,218881568348707987666944000,63256773252773762936322048000

%N a(n) = n!*(n!+1)/2.

%C a(n) is the number of unordered pairs (not necessarily distinct) of elements in S_n (the symmetric group on n letters). That is, a(n) = binomial(n!,2) + n!. - _Geoffrey Critzer_, Jan 09 2016

%F a(n) + (-n^2-n-3)*a(n-1) + (n-1)*(n^2+2*n-1)*a(n-2) - 2*(n-1)*(n-2)^2*a(n-3) = 0. - _R. J. Mathar_, Mar 21 2013

%F a(n) = Sum_{k=1..n!} k. - _Pedro Caceres_, Mar 10 2018

%F a(n) = A000217(A000142(n)). - _Michel Marcus_, Mar 11 2018

%t Table[n!*(n! + 1)/2, {n, 0, 20}] (* _Vladimir Joseph Stephan Orlovsky_, Jul 07 2011 *)

%o (Magma) [Factorial(n)*(Factorial(n)+1)/2: n in [0..20]]; // _Vincenzo Librandi_, Jan 10 2016

%o (PARI) a(n) = n!*(n!+1)/2; \\ _Altug Alkan_, Jan 10 2015

%Y Cf. A000142, A000217.

%K nonn

%O 0,3

%A _N. J. A. Sloane_, Jul 19 2000