A229762 a(n) = (n XOR floor(n/2)) AND floor(n/2), where AND and XOR are bitwise logical operators.

0, 0, 1, 0, 2, 2, 1, 0, 4, 4, 5, 4, 2, 2, 1, 0, 8, 8, 9, 8, 10, 10, 9, 8, 4, 4, 5, 4, 2, 2, 1, 0, 16, 16, 17, 16, 18, 18, 17, 16, 20, 20, 21, 20, 18, 18, 17, 16, 8, 8, 9, 8, 10, 10, 9, 8, 4, 4, 5, 4, 2, 2, 1, 0, 32, 32, 33, 32, 34, 34, 33, 32, 36, 36, 37, 36, 34, 34, 33

Offset

0,5

Comments

a(n) has a 01 bit pair in place of each 10 bit pair in n, and everywhere else 0 bits. Or equivalently a(n) has a 1-bit immediately below each run of 1's in n, but excluding a run ending at the least significant bit since below that is below the radix point. - Kevin Ryde, Feb 27 2021

Links

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191

Formula

a(n) = (n XOR floor(n/2)) AND floor(n/2) = (n AND floor(n/2)) XOR floor(n/2).

a(n) = floor(n/2) AND NOT n. - Chai Wah Wu, Jun 29 2022

Example

From Kevin Ryde, Feb 27 2021: (Start)
     n = 7267 = binary 1110001100011
  a(n) =  528 = binary   01000010000   1-bit below each run
(End)

Prog

(Python)
for n in range(333): print (n ^ (n>>1)) & (n>>1),
(Python)
def A229762(n): return ~n& n>>1 # Chai Wah Wu, Jun 29 2022
(Haskell)
import Data.Bits ((.&.), xor, shiftR)
a229762 n = (n `xor` shiftR n 1) .&. shiftR n 1 :: Int
-- Reinhard Zumkeller, Oct 10 2013
(PARI) a(n) = bitnegimply(n>>1, n); \\ Kevin Ryde, Feb 27 2021

Crossrefs

Cf. A003188 (n XOR floor(n/2)).

Cf. A048724 (n XOR (n*2)).

Cf. A048735 (n AND floor(n/2)).

Cf. A213370 (n AND (n*2)).

Cf. A213064 (n XOR (n*2) AND (n*2), 1-bit above each run).

Cf. A229763 ((2*n) XOR n AND n, low 1-bit each run).

Sequence in context: A116389 A216344 A332011 * A062110 A122896 A191348

Adjacent sequences: A229759 A229760 A229761 * A229763 A229764 A229765

Keyword

nonn,base,easy

Author

Alex Ratushnyak, Sep 28 2013

Status

approved