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a(n) = denominator((1/n!)*Sum_{k=1..n} 1/(2*k)).
1

%I #9 Oct 29 2024 15:41:32

%S 1,2,8,72,576,14400,28800,470400,22579200,1828915200,18289152000,

%T 2212987392000,26555848704000,4487938430976000,62831138033664000,

%U 942467070504960000,30158946256158720000,8715935468029870080000,52295612808179220480000,18878716223752698593280000

%N a(n) = denominator((1/n!)*Sum_{k=1..n} 1/(2*k)).

%F Conjecture: numerator((1/n!)*Sum_{k=1..n} 1/(2*k)) = A060746(n).

%t a[n_]:=Denominator[Sum[1/(2*k),{k,n}]/n!]; Array[a,20,0]

%o (PARI) a(n) = denominator(sum(k=1, n, 1/(2*k))/n!); \\ _Michel Marcus_, Oct 29 2024

%Y Cf. A000142, A005843, A060746, A377400.

%K nonn,frac

%O 0,2

%A _Stefano Spezia_, Oct 27 2024