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Define sequence x(n) by x(1)=1, thereafter x(n) = (x(n-1)+n)/(1-n*x(n-1)); sequence gives denominator(x(n)).
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%I #15 Aug 09 2023 07:04:58

%S 1,1,1,1,19,73,331,43,281,4511,10873,322921,12179,720817,538759,

%T 87995911,1185403,37171235,46336951,6986985769,2602576465,

%U 243540693677,181777598557,13097400661955,135996437150855,8249498995171439,56213506181241631,601615828819880125,10365435567354511181

%N Define sequence x(n) by x(1)=1, thereafter x(n) = (x(n-1)+n)/(1-n*x(n-1)); sequence gives denominator(x(n)).

%C x(n) = tan( sum_{k=1..n} arctan(k)): see A180657.

%H V. H. Moll, <a href="http://www.tulane.edu/~vhm/papers_html/xn-final.pdf">An arithmetic conjecture on a sequence of arctangent sums</a>, 2012.

%e The x(n) sequence begins 1, -3, 0, 4, -9/19, 105/73, -308/331, 36/43, -423/281, 2387/4511, -26004/10873, ...

%p x:=proc(n) option remember;

%p if n=1 then 1 else (x(n-1)+n)/(1-n*x(n-1)); fi; end;

%p s1:=[seq(x(n),n=1..30)]; # x(n)

%p s2:=map(numer,s1); # A180657

%p s3:=map(denom,s1); # A220447

%t x[n_] := x[n] = If[n == 1, 1, (x[n-1] + n)/(1 - n*x[n-1])];

%t a[n_] := Denominator[x[n]];

%t Table[a[n], {n, 1, 29}] (* _Jean-François Alcover_, Aug 09 2023 *)

%Y For numerators see A180657.

%K nonn,frac

%O 1,5

%A _N. J. A. Sloane_, Dec 22 2012