A088435 1/2 + half of the (n+1)-st component of the continued fraction expansion of sum(k>=1,1/3^(2^k)).

3, 2, 2, 1, 2, 3, 2, 1, 3, 2, 1, 2, 2, 3, 2, 1, 3, 2, 2, 1, 2, 3, 1, 2, 3, 2, 1, 2, 2, 3, 2, 1, 3, 2, 2, 1, 2, 3, 2, 1, 3, 2, 1, 2, 2, 3, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 3, 2, 1, 2, 2, 3, 2, 1, 3, 2, 2, 1, 2, 3, 2, 1, 3, 2, 1, 2, 2, 3, 2, 1, 3, 2, 2, 1, 2, 3, 1, 2, 3, 2, 1, 2, 2, 3, 1, 2, 3, 2, 2, 1, 2, 3, 2, 1, 3

Offset: 1

Author

Benoit Cloitre, Nov 08 2003

Keywords

• nonn

Comments

To construct the sequence use the rule : a(1)=3, then a(a(1)+a(2)+...+a(n)+1)=2+(-1)^n and fill in any undefined place with 2.

Examples

			Example to illustrate the comment : a(a(1)+1) = a(4) = 2+(-1)^1 = 1 and a(2), a(3) are undefined. The rule forces a(2) = a(3) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..8192 (computed from the b-file of A004200 provided by _Harry J. Smith_)

Crossrefs

Cf. A088431.

Formula

a(n) = (1/2) * (1+A004200(n+1)).

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