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

Author

Florian Lang, Aug 24 2019

Keywords

• base
• easy
• look
• nonn

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

Examples

			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.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..16383
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000695, A292371, A292372, A292373.

Programs

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

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

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

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