A309952 XOR contraction of binary representation of n.

0, 1, 1, 0, 2, 3, 3, 2, 2, 3, 3, 2, 0, 1, 1, 0, 4, 5, 5, 4, 6, 7, 7, 6, 6, 7, 7, 6, 4, 5, 5, 4, 4, 5, 5, 4, 6, 7, 7, 6, 6, 7, 7, 6, 4, 5, 5, 4, 0, 1, 1, 0, 2, 3, 3, 2, 2, 3, 3, 2, 0, 1, 1, 0, 8, 9, 9, 8, 10, 11, 11, 10, 10, 11, 11, 10, 8, 9, 9, 8, 12, 13, 13

Offset

0,5

Comments

To calculate a(n) write down the binary representation of n. Organize the digits in pairs and calculate the xor of these pairs. The result is a(n) in binary.

Conjecture: The index of the first occurrence of k in a is A000695(k). - Ivan N. Ianakiev, Aug 26 2019

Links

Alois P. Heinz, Table of n, a(n) for n = 0..16383

Index entries for sequences related to binary expansion of n

Formula

a(n) = A292371(n) + A292372(n). - Rémy Sigrist, Aug 25 2019

a(0) = 0, a(4n) = 2*a(n), a(4n+1) = 2*a(n)+1, a(4n+2) = 2*a(n)+1, a(4n+3) = 2*a(n). - Florian Lang, Aug 26 2019

Example

For n=19 we have the binary representation 10011 = 01 00 11. Calculating the xor of the pairs gives 1 0 0 which is 4 in binary and therefore a(19) = 4.

Maple

a:= n-> `if`(n=0, 0, (r-> 2*a((n-r)/4) +r*(3-r)/2)(irem(n, 4))):
seq(a(n), n=0..100);  # Alois P. Heinz, Aug 26 2019

Prog

(Python)
def a(n):
    n = [int(k) for k in bin(n)[2:]]
    if len(n) % 2 != 0:
        n = [0] + n
    result = []
    for i in range(0, len(n), 2):
        result.append(n[i] ^ n[i+1]) #xor
    return int("".join([str(k) for k in result]), 2)
(Python)
from itertools import zip_longest
from operator import xor
def A309952(n): return int(''.join(map(lambda x:str(xor(*x)), zip_longest((s:=tuple(int(d) for d in bin(n)[2:]))[::-2], s[-2::-2], fillvalue=0)))[::-1], 2) # Chai Wah Wu, Jun 30 2022
(PARI) a(n) = {my(b = Vecrev(binary(n)), nb = #b\2, val = fromdigits(Vecrev(vector(nb, i, bitxor(b[2*i-1], b[2*i]))), 2)); if (#b % 2, val += 2^nb); val; } \\ Michel Marcus, Aug 26 2019

Crossrefs

Cf. A000695, A292371, A292372, A292373.

Sequence in context: A096838 A096007 A269043 * A059252 A349319 A251619

Adjacent sequences: A309949 A309950 A309951 * A309953 A309954 A309955

Keyword

base,easy,look,nonn

Author

Florian Lang, Aug 24 2019

Status

approved