A224915 a(n) = Sum_{k=0..n} n XOR k where XOR is the bitwise logical exclusive-or operator.

0, 1, 5, 6, 22, 23, 27, 28, 92, 93, 97, 98, 114, 115, 119, 120, 376, 377, 381, 382, 398, 399, 403, 404, 468, 469, 473, 474, 490, 491, 495, 496, 1520, 1521, 1525, 1526, 1542, 1543, 1547, 1548, 1612, 1613, 1617, 1618, 1634, 1635, 1639, 1640, 1896, 1897, 1901, 1902, 1918

Offset: 0

Author

Alex Ratushnyak, Apr 19 2013

Keywords

• nonn
• easy

Examples

			a(2) = (0 xor 2) + (1 xor 2) = 2 + 3 = 5.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A001196 (bit doubling).

Row sums of A051933.

Other sums: A222423 (AND), A350093 (OR), A265736 (IMPL), A350094 (CNIMPL), A004125 (mod).

Programs

Maple

read("transforms"):
A051933 := proc(n,k)
    XORnos(n,k) ;
end proc:
A224915 := proc(n)
    add(A051933(n,k),k=0..n) ;
end proc: # R. J. Mathar, Apr 26 2013
# second Maple program:
with(MmaTranslator[Mma]):
seq(add(BitXor(n,i),i=0..n),n=0..60); # Ridouane Oudra, Dec 09 2020

Mathematica

Array[Sum[BitXor[#, k], {k, 0, #}] &, 53, 0] (* Michael De Vlieger, Dec 09 2020 *)

PARI

a(n) = sum(k=0, n, bitxor(n, k)); \\ Michel Marcus, Jun 08 2019

PARI

a(n) = (3*fromdigits(binary(n),4) - n) >>1; \\ Kevin Ryde, Dec 17 2021

Python

for n in range(59):
    s = 0
    for k in range(n):  s += n ^ k
    print(s, end=',')

Python

def A224915(n): return 3*int(bin(n)[2:],4)-n>>1 # Chai Wah Wu, Aug 21 2023

Formula

a(n) = Sum_{j=1..n} 4^(v_2(j)), where v_2(j) is the exponent of highest power of 2 dividing j. - Ridouane Oudra, Jun 08 2019

a(n) = n + 3*Sum_{j=1..floor(log_2(n))} 4^(j-1)*floor(n/2^j), for n>=1. - Ridouane Oudra, Dec 09 2020

From Kevin Ryde, Dec 17 2021: (Start)

a(2*n+b) = 4*a(n) + n + b where b = 0 or 1.

a(n) = (A001196(n) - n)/2.

a(n) = A350093(n) - A222423(n), being XOR = OR - AND.

(End)