1,1

It seems that for all n >= 1, Sum_{k=1 .. 2^n} a(k) = 2^(n+3) - 1 = A000225(n+3).

Antti Karttunen, Table of n, a(n) for n = 1..16384

a(n) = A257512(n+1) - A257512(n).

(Scheme) (define (A256490 n) (- (A257512 (+ n 1)) (A257512 n)))

Cf. A000225, A257512, A256489.

Sequence in context: A242968 A021536 A199389 * A197688 A094082 A318378

Adjacent sequences: A256487 A256488 A256489 * A256491 A256492 A256493

nonn

Antti Karttunen, May 03 2015

approved