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 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
CROSSREFS
KEYWORD
AUTHOR
Vladimir Reshetnikov, Oct 18 2008
STATUS
approved