A213064 -id:A213064 - cp's OEIS frontend

A292272 a(n) = n - A048735(n) = n - (n AND floor(n/2)).

0, 1, 2, 2, 4, 5, 4, 4, 8, 9, 10, 10, 8, 9, 8, 8, 16, 17, 18, 18, 20, 21, 20, 20, 16, 17, 18, 18, 16, 17, 16, 16, 32, 33, 34, 34, 36, 37, 36, 36, 40, 41, 42, 42, 40, 41, 40, 40, 32, 33, 34, 34, 36, 37, 36, 36, 32, 33, 34, 34, 32, 33, 32, 32, 64, 65, 66, 66, 68, 69, 68, 68, 72, 73, 74, 74, 72, 73, 72, 72, 80, 81, 82, 82, 84, 85, 84, 84, 80, 81, 82, 82, 80, 81

Offset: 0

Antti Karttunen, Sep 16 2017

In binary expansion of n, change those 1's to 0's that have an 1-bit next to them at their left (more significant) side. Only fibbinary numbers (A003714) occur as terms.

			From _Kevin Ryde_, Jun 02 2020: (Start)
     n = 1831 = binary 11100100111
  a(n) = 1060 = binary 10000100100   high 1 of each run
(End)

Antti Karttunen, Table of n, a(n) for n = 0..16383
Index entries for sequences related to binary expansion of n

Cf. A003188, A003714, A003987, A004198, A005940, A048735, A085357, A292371, A292382.

Table[n - BitAnd[n, Floor[n/2]], {n, 0, 93}] (* Michael De Vlieger, Sep 17 2017 *)

a(n) = bitnegimply(n,n>>1); \\ Kevin Ryde, Jun 02 2020

def A292272(n): return n&~(n>>1) # Chai Wah Wu, Jun 29 2022

a(n) = n - A048735(n) = n - (n AND floor(n/2)) = n XOR (n AND floor(n/2)), where AND is bitwise-AND (A004198) and XOR is bitwise-XOR (A003987).
a(n) = n AND A003188(n).
a(n) = A292382(A005940(1+n)).
A059905(a(n)) = A292371(n).
For all n >= 0, A085357(a(n)) = 1.
a(n) = A213064(n) / 2. - Kevin Ryde, Jun 02 2020
a(n) = n AND NOT floor(n/2). - Chai Wah Wu, Jun 29 2022

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

Original entry on oeis.org

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

Alex Ratushnyak, Sep 28 2013

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

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

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

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).

import Data.Bits ((.&.), xor, shiftR)
a229762 n = (n `xor` shiftR n 1) .&. shiftR n 1 :: Int
-- Reinhard Zumkeller, Oct 10 2013

a(n) = bitnegimply(n>>1,n); \\ Kevin Ryde, Feb 27 2021

for n in range(333): print (n ^ (n>>1)) & (n>>1),

def A229762(n): return ~n& n>>1 # Chai Wah Wu, Jun 29 2022

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

A229763 a(n) = (2*n) XOR n AND n, where AND and XOR are bitwise logical operators.

Original entry on oeis.org

0, 1, 2, 1, 4, 5, 2, 1, 8, 9, 10, 9, 4, 5, 2, 1, 16, 17, 18, 17, 20, 21, 18, 17, 8, 9, 10, 9, 4, 5, 2, 1, 32, 33, 34, 33, 36, 37, 34, 33, 40, 41, 42, 41, 36, 37, 34, 33, 16, 17, 18, 17, 20, 21, 18, 17, 8, 9, 10, 9, 4, 5, 2, 1, 64, 65, 66, 65, 68, 69, 66, 65, 72, 73, 74

Offset: 0

Alex Ratushnyak, Sep 28 2013

a(n) is the least significant 1-bit of each run of consecutive 1's in n, and everywhere else 0's. Or equivalently, clear to 0 each 1-bit which has another 1 immediately below. - Kevin Ryde, Feb 27 2021

			From _Kevin Ryde_, Feb 27 2021: (Start)
     n = 1831 = binary 11100100111
  a(n) =  289 = binary   100100001   low 1-bit each run
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 41.

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)).
Cf. A229762 (n XOR floor(n/2) AND floor(n/2), 1-bit below each run).
Cf. A292272 (high 1-bit each run).

import Data.Bits ((.&.), xor, shiftL)
a229763 n = (shiftL n 1 `xor` n) .&. n :: Int
-- Reinhard Zumkeller, Oct 10 2013

Array[BitAnd[BitXor[2 #, #], #] &, 75, 0] (* Michael De Vlieger, Nov 03 2022 *)

a(n) = bitnegimply(n,n<<1); \\ Kevin Ryde, Feb 27 2021

for n in range(333): print (2*n ^ n) & n,

def A229763(n): return n&~(n<<1) # Chai Wah Wu, Jun 29 2022

a(n) = ((2*n) XOR n) AND n = ((2*n) AND n) XOR n.
a(2n) = 2a(n), a(2n+1) = A229762(n). - Ralf Stephan, Oct 07 2013
a(n) = n AND NOT 2n. - Chai Wah Wu, Jun 29 2022
G.f.: x/(1 - x^2) + Sum_{k>=1}(2^k*x^(2^k)/((1 - x)*(1 + x^(2^k))*(1 + x^(2^(k - 1))))). - Miles Wilson, Jan 24 2025

cp's OEIS Frontend

A292272 a(n) = n - A048735(n) = n - (n AND floor(n/2)).

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

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Haskell

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Python

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A229763 a(n) = (2*n) XOR n AND n, where AND and XOR are bitwise logical operators.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Python

Python

Formula