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a(1)=1; for n >= 2, a(n) = floor(n!/(Sum_{k=1..n-1} a(k))).
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%I #14 Apr 26 2020 06:31:31

%S 1,2,2,4,13,32,93,274,861,2830,9707,34662,128442,492747,1952714,

%T 7978537,33552502,145002884,643093018,2923285048,13604173759,

%U 64747674282,314856179539,1562985778041,7914087230121,40843626440195,214695804264578,1148729745111018,6252380887804219

%N a(1)=1; for n >= 2, a(n) = floor(n!/(Sum_{k=1..n-1} a(k))).

%H Vaclav Kotesovec, <a href="/A130797/b130797.txt">Table of n, a(n) for n = 1..800</a>

%p a[1]:=1: for n from 2 to 27 do a[n]:=floor(factorial(n)/(sum(a[k],k=1..n-1))) end do: seq(a[n], n = 1 .. 27); # _Emeric Deutsch_, Jul 22 2007

%t a={1};For[n=2,n<50,n++, AppendTo[a, Floor[n!/Sum[a[[i]], {i, 1, n - 1}]]]]; a (* _Stefan Steinerberger_, Jul 16 2007 *)

%t f[l_] := Append[l, Floor[(Length[l] + 1)!/(Plus @@ l)]], Nest[f, {1}, 26] (* _Robert G. Wilson v_ *)

%K nonn

%O 1,2

%A _Leroy Quet_, Jul 15 2007

%E More terms from _Stefan Steinerberger_, _Emeric Deutsch_ and _Joshua Zucker_, Jul 16 2007