login
a(n)=2^(2^n)*sum(k=0,n,1/2^(2^k)).
0

%I #7 Mar 30 2012 18:39:18

%S 1,3,13,209,53505,3506503681,15060318633198616577,

%T 277813843495134114797235287762174738433,

%U 94535152227927400227782074307303551040545228366095741656402842333161034088449

%N a(n)=2^(2^n)*sum(k=0,n,1/2^(2^k)).

%F a(n)=A074854(2^n)=; a(n)=floor(c*2^(2^n)) where c=sum(k>=0, 1/2^(2^k))=0.81642150902...

%F a(n + 1) = 1 + a(n)*2^(2^n), a(0) = 1 [From Peter Moxey (pmoxey(AT)live.com), Mar 14 2010]

%Y Cf. A007404.

%K nonn

%O 0,2

%A _Benoit Cloitre_, Jun 19 2003