A054874 a(n) = 2^(Sum_{i<n} a(i)).

0, 1, 2, 8, 2048

Offset

0,3

Comments

The next term is too large to include.

Links

Table of n, a(n) for n=0..4.

Formula

a(n) = 2^A034797(n-1) = A034797(n) - A034797(n-1).

From Jianing Song, Feb 10 2026: (Start)

a(n) = a(n-1) * 2^a(n-1) for n >= 2, a(1) = 1.

The sequence {b(n)=2^a(n)} satisfies b(n) = b(n-1)^b(n-1) for n >= 2, b(1) = 2. (End)

Example

a(4) = 2^(0+1+2+8) = 2^11 = 2048; a(5) = 2^2059>10^619.

Crossrefs

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: A306241 A073630 A027733 * A174736 A324567 A135238

Adjacent sequences: A054871 A054872 A054873 * A054875 A054876 A054877

Keyword

easy,nonn

Author

Henry Bottomley, May 26 2000

Status

approved