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a(n) = denominator of b(n), where b(1) = 1, b(n+1) = Sum_{k=1..n} b(k)^((-1)^(n-k)).
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%I #14 Sep 30 2017 23:55:24

%S 1,1,1,1,2,4,52,559,2023580,639046564,73885083538076135,

%T 13974134129149036419614094980,

%U 9508386737708519692119190558953351866716894940,167312950453078829361896561420857502596441619698513063185995475418519527687170

%N a(n) = denominator of b(n), where b(1) = 1, b(n+1) = Sum_{k=1..n} b(k)^((-1)^(n-k)).

%H Michael De Vlieger, <a href="/A115587/b115587.txt">Table of n, a(n) for n = 1..19</a>

%e {b(n)} begins 1, 1, 2, 4, 13/2, 43/4, ...

%e So b(7) = 1 + 1 + 1/2 + 4 + 2/13 + 43/4 = 905/52 and therefore a(7) = 52.

%p b[1]:=1: for n from 1 to 14 do b[n+1]:=sum(b[k]^((-1)^(n-k)),k=1..n): a[n]:=denom(b[n]) od: seq(a[n],n=1..14); # _Emeric Deutsch_, Mar 30 2006

%t b[n_] := b[n] = If[n == 1, 1, Sum[b[k]^((-1)^(n - k - 1)), {k, n - 1}]]; Array[Denominator@ b@ # &, 14] (* _Michael De Vlieger_, Sep 30 2017 *)

%Y Cf. A115600, A115601, A115602.

%K frac,nonn

%O 1,5

%A _Leroy Quet_, Mar 13 2006

%E More terms from _Emeric Deutsch_, Mar 30 2006