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 this_sequence A057841 A135238 A133376

Adjacent sequences: A054871 A054872 A054873 this_sequence A054875 A054876 A054877

easy,nonn

Henry Bottomley (se16(AT)btinternet.com), May 26 2000