A255070 (1/2)*(n minus number of runs in the binary expansion of n): a(n) = (n - A005811(n)) / 2 = A236840(n)/2.

0, 0, 0, 1, 1, 1, 2, 3, 3, 3, 3, 4, 5, 5, 6, 7, 7, 7, 7, 8, 8, 8, 9, 10, 11, 11, 11, 12, 13, 13, 14, 15, 15, 15, 15, 16, 16, 16, 17, 18, 18, 18, 18, 19, 20, 20, 21, 22, 23, 23, 23, 24, 24, 24, 25, 26, 27, 27, 27, 28, 29, 29, 30, 31, 31, 31, 31, 32, 32, 32, 33, 34, 34, 34, 34, 35

Offset: 0

Author

Antti Karttunen, Feb 14 2015

Keywords

• nonn
• base

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8192
• Helmut Prodinger and Friedrich J. Urbanek, Infinite 0-1-Sequences Without Long Adjacent Identical Blocks, Discrete Mathematics, Volume 28, Number 3, 1979, pages 277-289. Also first author's copy. See theorem 3.6, n_1(k) = a(k).

Crossrefs

Least inverse: A091067 (also the positions of records).

Greatest inverse: A255068.

Run lengths: A106836.

Cf. A005811, A060833, A236840, A255054, A255056, A255072, A269363, A269367.

Programs

Mathematica

a[n_] := (n - Length@ Split[IntegerDigits[n, 2]])/2; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Jul 16 2023 *)

Scheme

(define (A255070 n) (/ (A236840 n) 2))

Formula

a(n) = A236840(n) / 2 = (n - A005811(n)) / 2.

Other identities:

a(A091067(n)) = n for all n >= 1.

a(A255068(n)) = n for all n >= 0.

a(A269363(n)) = A269367(n). - Antti Karttunen, Aug 12 2019