A090079 In binary expansion of n: reduce contiguous blocks of 0's to 0 and contiguous blocks of 1's to 1.

0, 1, 2, 1, 2, 5, 2, 1, 2, 5, 10, 5, 2, 5, 2, 1, 2, 5, 10, 5, 10, 21, 10, 5, 2, 5, 10, 5, 2, 5, 2, 1, 2, 5, 10, 5, 10, 21, 10, 5, 10, 21, 42, 21, 10, 21, 10, 5, 2, 5, 10, 5, 10, 21, 10, 5, 2, 5, 10, 5, 2, 5, 2, 1, 2, 5, 10, 5, 10, 21, 10, 5, 10, 21, 42, 21, 10, 21, 10, 5, 10, 21, 42, 21

Offset: 0

Author

Reinhard Zumkeller, Nov 20 2003

Keywords

• nonn
• base

Comments

a(a(n))=a(n); a(n)=A090078(A090077(n))=A090077(A090078(n)).

All terms are without consecutive equal binary digits: a(A000975(n)) = A000975(n) and a(m) <> A000975(n) for m < A000975(n). - Reinhard Zumkeller, Feb 16 2013

Examples

			100 -> '1100100' -> [11][00][1][00] -> [1][0][1][0] -> '1010' ->
10=a(100).
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A007088, A090077, A090078, A090080.

Cf. A030308, A005811.

Programs

Haskell

a090079 = foldr (\b v -> 2 * v + b) 0 . map head . group . a030308_row
-- Reinhard Zumkeller, Feb 16 2013

Mathematica

Table[FromDigits[#, 2] &@ Map[First, Split@ IntegerDigits[n, 2]], {n, 0, 83}] (* Michael De Vlieger, Dec 12 2016 *)
FromDigits[Split[IntegerDigits[#,2]][[All,1]],2]&/@Range[0,90] (* Harvey P. Dale, Oct 10 2017 *)

Python

from itertools import groupby
def a(n): return int("".join(k for k, g in groupby(bin(n)[2:])), 2)
print([a(n) for n in range(84)]) # Michael S. Branicky, Jul 23 2022

Formula

Conjecture: a(n) = (2^(A005811(n)+1) + (1-(-1)^n)/2 - 2)/3. - Velin Yanev, Dec 12 2016