%I

%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