A093505 a(n) = floor(A001969(n)/2 + 1/2).

0, 2, 3, 3, 5, 5, 6, 8, 9, 9, 10, 12, 12, 14, 15, 15, 17, 17, 18, 20, 20, 22, 23, 23, 24, 26, 27, 27, 29, 29, 30, 32, 33, 33, 34, 36, 36, 38, 39, 39, 40, 42, 43, 43, 45, 45, 46, 48, 48, 50, 51, 51, 53, 53, 54, 56, 57, 57, 58, 60, 60, 62, 63, 63, 65

Offset

1,2

Comments

The sequence is a quasi-inverse of A161560: For n >= 2, a(A161560(n+1)+1) = n or n-1. - Jaroslav Krizek, Jun 13 2009, corrected Jul 13 2012

Conjecture: Essentially partial sums of A036577 (true for at least 10000 terms). - Sean A. Irvine, Jul 17 2022

From Bartlomiej Pawlik, Apr 19 2026: (Start)

Proof of the conjecture:

Let z(n) be the position of the n-th 0 in the binary Thue-Morse sequence A010060, with z(0) = 0. Since A001969 lists the indices of 0's in A010060, we have z(n) = A001969(n+1). Also, A036577 gives the number of 1's between successive 0's in A010060, so A036577(n+1) = z(n+1) - z(n) - 1.

Therefore, for n >= 1, Sum_{k=1..n-1} A036577(k) = Sum_{k=0..n-2} (z(k+1) - z(k) - 1) = z(n-1) - (n-1) = A001969(n) - n + 1 = a(n).

Hence a(n) is exactly the sequence of partial sums of A036577, with initial term a(1) = 0.

a(k) is the number of 1's before the (k+1)-th 0 in binary Thue-Morse sequence A010060. (End)

Links

Table of n, a(n) for n=1..65.

Formula

a(n) = A001969(n) - n + 1 for n >= 2. - Jaroslav Krizek, Jun 13 2009

Example

a(5) = 5 because the fifth evil number = 9 and floor(9/2 + 1/2) = 5.

Prog

(PARI) a(n)=(2*n+hammingweight(n-1)%2-1)\2 \\ Charles R Greathouse IV, Mar 22 2013

Crossrefs

Cf. A001969, A000069, A036577.

Sequence in context: A010761 A320840 A161172 * A238527 A146071 A167411

Adjacent sequences: A093502 A093503 A093504 * A093506 A093507 A093508

Keyword

easy,nonn

Author

Michael Joseph Halm, May 13 2004

Extensions

Corrected and edited by Jaroslav Krizek, Jun 13 2009

Status

approved