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

0,3

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

Formula

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

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]]

Prog

(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))
(Haskell)
import Data.Bits (xor)
a145768 n = a145768_list !! n
a145768_list = scanl1 xor a000290_list  -- Reinhard Zumkeller, Jun 05 2012
(Python)
from operator import xor
def A145768(n):
....return reduce(xor, [x**2 for x in range(n+1)]) # Chai Wah Wu, Aug 08 2014

Crossrefs

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

Cf. A193232.

Cf. A000290.

Sequence in context: A128439 A240187 A172426 * A162778 A160807 A038376

Adjacent sequences: A145765 A145766 A145767 * A145769 A145770 A145771

Keyword

easy,nonn,base,look

Author

Vladimir Reshetnikov, Oct 18 2008

Status

approved