A145768 a(n) = the bitwise XOR of squares of first n natural numbers.

0, 1, 5, 12, 28, 5, 33, 16, 80, 1, 101, 28, 140, 37, 225, 0, 256, 33, 357, 12, 412, 37, 449, 976, 400, 993, 325, 924, 140, 965, 65, 896, 1920, 961, 1861, 908, 1692, 965, 1633, 912, 1488, 833, 1445, 668, 1292, 741, 2721, 512, 2816, 609, 2981, 396, 2844, 485

Offset: 0

Author

Vladimir Reshetnikov, Oct 18 2008

Keywords

• easy
• nonn
• base
• look

Comments

Up to n=10^8, a(15) is the only zero term and a(1)=a(9) are the only terms for which a(n)=1. Can it be proved that any number can only appear a finite number of times in this sequence? [M. F. Hasler, Oct 20 2008]

Even terms occur at A014601, odd terms at A042963; A010873(a(n))=A021913(n+1). - Reinhard Zumkeller, Jun 05 2012

If squares occur, they must be at indexes != 2 or 5 (mod 8). - Roderick MacPhee, Jul 17 2017

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• StackExchange, Perfect squares in a XOR-Sum of perfect squares

Crossrefs

Cf. A003815, A145827, A145828, A145829, A145830, A145831. [M. F. Hasler, Oct 20 2008]

Cf. A193232.

Cf. A000290.

Programs

Haskell

import Data.Bits (xor)
a145768 n = a145768_list !! n
a145768_list = scanl1 xor a000290_list  -- Reinhard Zumkeller, Jun 05 2012

Maple

A[0]:= 0:
for n from 1 to 100 do A[n]:= Bits:-Xor(A[n-1],n^2) od:
seq(A[i],i=0..100); # Robert Israel, Dec 08 2019

Mathematica

Rest@ FoldList[BitXor, 0, Array[#^2 &, 50]]

PARI

an=0; for( i=1,50, print1(an=bitxor(an,i^2),",")) \\ M. F. Hasler, Oct 20 2008

PARI

al(n)=local(m);vector(n,k,m=bitxor(m,k^2))

Python

from functools import reduce
from operator import xor
def A145768(n):
    return reduce(xor, [x**2 for x in range(n+1)]) # Chai Wah Wu, Aug 08 2014

Formula

a(n)=1^2 xor 2^2 xor ... xor n^2.