0,3
The next term is too large to include.
a(n) = 2^A034797(n-1) = A034797(n) - A034797(n-1)
a(4) = 2^(0+1+2+8) = 2^11 = 2048; a(5) = 2^2059>10^619
Cf. A014221, A034797 for partial sum, so a(n) is number of impartial games with value n-1, using natural enumeration of impartial games.
Sequence in context: A121015 A073630 A027733 * A174736 A135238 A133376
Adjacent sequences: A054871 A054872 A054873 * A054875 A054876 A054877
easy,nonn
Henry Bottomley (se16(AT)btinternet.com), May 26 2000
