A103528 a(n) = Sum_{k = 1..n-1 such that n == k (mod 2^k)} 2^(k-1).

0, 0, 1, 0, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 17, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 17, 34, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8

Offset: 1

Author

N. J. A. Sloane, Mar 22 2005

Keywords

• nonn
• look

Comments

Is there a simpler closed form?

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
• David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.

Crossrefs

Cf. A102371.

Programs

Maple

f:=proc(n) local t1,k; t1:=0; for k from 1 to n-1 do if n mod 2^k = k then t1:=t1+2^(k-1); fi; od: t1; end;

Mathematica

(* b = A102371 (using Alex Ratushnyak's code) *)
b[n_] := b[n] = If[n == 1, 1, BitXor[b[n-1], b[n-1] + n]];
a[n_] := (b[n] + n)/2 - 2^(n-1);
Array[a, 100] (* Jean-François Alcover, Apr 11 2019, after Philippe Deléham *)

PARI

a(n) = sum(k = 1, n-1, if ((n % 2^k) == k, 2^(k-1))); \\ Michel Marcus, May 06 2020

Formula

a(n) = (A102371(n) + n)/2 - 2^(n-1). - Philippe Deléham, Mar 27 2005

G.f.: Sum_{k>=1} 2^(k-1) x^(k+2^k)/(1 - x^(2^k)). - Robert Israel, Jan 21 2017

Conjecture: a(n) = (b(n) - b(n-1) - 1)/2 for n > 1 where b(n) = Sum_{k=0..A000523(n)} c(n-k, k) and c(n, m) = n - (n mod 2^m) with a(1) = 0. - Mikhail Kurkov, Jun 01 2022