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a(n) satisfies the equation n/(n-1) + a(n)/n! = H(n), where H(n) is the n-th harmonic number.
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%I #30 May 02 2020 02:42:08

%S -1,2,18,124,900,7188,63504,618336,6596640,76635360,963895680,

%T 13056819840,189581333760,2938083321600,48416639846400,

%U 845487698227200,15598004134809600,303161985274982400,6191998554470400000,132599321499875328000,2970952207377960960000

%N a(n) satisfies the equation n/(n-1) + a(n)/n! = H(n), where H(n) is the n-th harmonic number.

%F a(n) = n! * (H(n) - n/(n-1)). - _Alois P. Heinz_, Jul 13 2017

%t Table[n!*(HarmonicNumber[n] - n/(n - 1)), {n, 2, 22}] (* _Michael De Vlieger_, Jul 13 2017 *)

%o (Python)

%o from sympy import factorial, harmonic

%o def a(n): return factorial(n-2)*(harmonic(n)*(n-1) - n)*n

%o print([a(n) for n in range(2, 26)]) # _Indranil Ghosh_, Jul 14 2017

%Y Cf. A001008, A002805.

%K sign

%O 2,2

%A _Joseph Wheat_, Jul 12 2017

%E More terms from _Alois P. Heinz_, Jul 13 2017