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a(n) = H(2n)*(2n)!/n! where H(n) = sum{k=1 to n} 1/k, the n-th harmonic number.
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%I #7 Apr 09 2014 10:16:40

%S 0,3,25,294,4566,88572,2064504,56243184,1754322480,61664980320,

%T 2412077832000,103928089910400,4890979939104000,249630539076979200,

%U 13734482066534476800,810355993296898406400,51040766703826377676800

%N a(n) = H(2n)*(2n)!/n! where H(n) = sum{k=1 to n} 1/k, the n-th harmonic number.

%t f[n_] := HarmonicNumber[2n](2n)!/n!; Table[f@n, {n, 0, 16}] - _Robert G. Wilson v_, Nov 26 2006

%Y Cf. A124078, A124079.

%K easy,nonn

%O 0,2

%A _Leroy Quet_, Nov 23 2006

%E More terms from _Robert G. Wilson v_, Nov 26 2006