%I #4 Mar 30 2012 18:51:25
%S 0,1,2,8,2048
%N a(n) = 2^(sum of a(i) where i<n).
%C The next term is too large to include.
%F a(n) = 2^A034797(n-1) = A034797(n) - A034797(n-1)
%e a(4) = 2^(0+1+2+8) = 2^11 = 2048; a(5) = 2^2059>10^619
%Y Cf. A014221, A034797 for partial sum, so a(n) is number of impartial games with value n-1, using natural enumeration of impartial games.
%K easy,nonn
%O 0,3
%A _Henry Bottomley_, May 26 2000