A003815 -id:A003815 - cp's OEIS frontend

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

Vladimir Reshetnikov, Oct 18 2008

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

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

Cf. A003815, A145827, A145828, A145829, A145830, A145831. [M. F. Hasler, Oct 20 2008]
Cf. A193232.
Cf. A000290.

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

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

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

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

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

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

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

A003816 a(0) = 0, a(n) = a(n-1) XOR -n.

Original entry on oeis.org

0, 1, -1, 4, 0, 5, -1, 8, 0, 9, -1, 12, 0, 13, -1, 16, 0, 17, -1, 20, 0, 21, -1, 24, 0, 25, -1, 28, 0, 29, -1, 32, 0, 33, -1, 36, 0, 37, -1, 40, 0, 41, -1, 44, 0, 45, -1, 48, 0, 49, -1, 52, 0, 53, -1, 56, 0, 57, -1, 60, 0, 61, -1, 64, 0, 65, -1, 68

Offset: 0

Marc LeBrun

Index entries for linear recurrences with constant coefficients, signature (0,1,0,1,0,-1).

Cf. A003815.

[(n-(n+1)*(-1)^n+(-1)^((2*n+1-(-1)^n) div 4))/2 : n in [0..100]]; // Wesley Ivan Hurt, May 03 2016

A003816:=n->(n-(n+1)*(-1)^n+(-1)^((2*n+1-(-1)^n)/4))/2: seq(A003816(n), n=0..150); # Wesley Ivan Hurt, May 03 2016

CoefficientList[Series[x*(1 - x + 3 x^2 + x^3)/((x^2 - 1)^2*(x^2 + 1)), {x, 0, 100}], x] (* Wesley Ivan Hurt, May 01 2016 *)
nxt[{n_,a_}]:={n+1,BitXor[a,-n-1]}; -#&/@(NestList[nxt,{0,0},70][[All,2]]) (* or *) LinearRecurrence[{0,1,0,1,0,-1},{0,1,-1,4,0,5},70] (* Harvey P. Dale, Oct 16 2019 *)

a(n)=([0,1,0,0,0,0; 0,0,1,0,0,0; 0,0,0,1,0,0; 0,0,0,0,1,0; 0,0,0,0,0,1; -1,0,1,0,1,0]^n*[0;1;-1;4;0;5])[1,1] \\ Charles R Greathouse IV, May 02 2016

G.f.: x*(1-x+3x^2+x^3)/((x^2-1)^2*(x^2+1)).
|a(n)| = n-(-1)^n*|a(n-1)|. - Vladeta Jovovic, Mar 13 2003
a(4n)=0, a(4n+1)=4n+1, a(4n+2)=-1, a(4n+3)=4n+4, n>=0.
From Wesley Ivan Hurt, May 01 2016, May 03 2016: (Start)
a(n) = a(n-2)+a(n-4)-a(n-6) for n>5.
a(n) = (-1)^floor((n-1)/2) * Sum_{i=0..n} i*(-1)^floor(i/2).
a(n) = (n-(n+1)*(-1)^n+(-1)^((2*n+1-(-1)^n)/4))/2. (End)
E.g.f.: (cos(x) + (-1 + 2*x)*cosh(x) - sin(x) + sinh(x))/2. - Ilya Gutkovskiy, May 03 2016

A126084 a(n) = XOR of first n primes.

Original entry on oeis.org

0, 2, 1, 4, 3, 8, 5, 20, 7, 16, 13, 18, 55, 30, 53, 26, 47, 20, 41, 106, 45, 100, 43, 120, 33, 64, 37, 66, 41, 68, 53, 74, 201, 64, 203, 94, 201, 84, 247, 80, 253, 78, 251, 68, 133, 64, 135, 84, 139, 104, 141, 100, 139, 122, 129, 384, 135, 394, 133, 400, 137, 402, 183, 388, 179

Offset: 0

Esko Ranta, Mar 02 2007

The values at odd positive indices are even and the values at even positive indices are odd.
Does this sequence contain any zeros for n > 0? Probabilistically, one would expect so; but none in first 10000 terms. - Franklin T. Adams-Watters, Jul 17 2011
None below 1.5 * 10^11: any prime p such that a(pi(p)) = 0 is 43 bits or longer. Heuristic chances that a prime below 2^100 yields 0 are about 45%. Note that an n-bit prime can yield 0 only if a(pi(p)) is odd, where p is the smallest n-bit prime. That is, for n > 1, there are no zeros from pi(2^n) to pi(2^(n+1)) if A007053(n) is even. - Charles R Greathouse IV, Jul 17 2011

			a(4) = 3 because ((2 XOR 3) XOR 5) XOR 7 = (1 XOR 5) XOR 7 = 4 XOR 7 = 3
[Or, in base 2]
((10 XOR 11) XOR 101) XOR 111 = (1 XOR 101) XOR 111 = 100 XOR 111 = 11

Franklin T. Adams-Watters, Table of n, a(n) for n = 0..10000

Cf. A003815, A004767, A112591.

Module[{nn=70,prs},prs=Prime[Range[nn]];Table[BitXor@@Take[prs,n],{n,0,nn}]] (* Harvey P. Dale, Jun 23 2016 *)

al(n)=local(m);vector(n,k,m=bitxor(m,prime(k))) /* Produces a vector without a(0) = 0; Franklin T. Adams-Watters, Jul 17 2011 */

v=primes(300); for(i=2,#v,v[i]=bitxor(v[i],v[i-1])); concat(0, v) \\ Charles R Greathouse IV, Aug 26 2014

q=0; forprime(p=2, 313, print1(q, ","); q=bitxor(q, p)) /* Klaus Brockhaus, Mar 06 2007; adapted by Rémy Sigrist, Oct 23 2017 */

from operator import xor
from functools import reduce
from sympy import primerange, prime
def A126084(n): return reduce(xor,primerange(2,prime(n)+1)) if n else 0 # Chai Wah Wu, Jul 09 2022

a(0) = 0; a(n) = a(n-1) XOR prime(n).

More terms from Klaus Brockhaus, Mar 06 2007
Edited by N. J. A. Sloane, Oct 22 2017 (merging old entry A193174 with this)
Edited by Rémy Sigrist, Oct 23 2017

A077140 a(1) = 1 and then add n to the previous term if n is coprime to the previous term, otherwise subtract n from the previous term. a(1) = 1 and a(n) = a(n-1) + n if gcd(n, a(n-1)) = 1, otherwise a(n) = a(n-1) - n.

Original entry on oeis.org

1, 3, 0, -4, 1, 7, 0, -8, 1, 11, 0, -12, 1, 15, 0, -16, 1, 19, 0, -20, 1, 23, 0, -24, 1, 27, 0, -28, 1, 31, 0, -32, 1, 35, 0, -36, 1, 39, 0, -40, 1, 43, 0, -44, 1, 47, 0, -48, 1, 51, 0, -52, 1, 55, 0

Offset: 1

Amarnath Murthy, Oct 30 2002

a(2k+1) = (k+1) (mod 2), a(4k) = -4k, a(4k+2) = 4k+3. Proof: If a(4k+3)=0 then a(4k+4) = -4k-4, a(4k+5)=1, a(4k+6) = 1+4k+6 and again, a(4k+7)=0. - Ralf Stephan, Mar 18 2003
Abs(a(n)) = A003815(n). - Reinhard Zumkeller, Oct 09 2007
With different signs, it can be obtained as abs(a(n)) = abs(Sum_{i=0..n} (-1)^h(i)*i)  where h(i) is the Hamming weight of i, A000120, the number of 1s in base 2. - Olivier Gérard, Jul 30 2012

			From _Ilya Gutkovskiy_, Dec 21 2015: (Start)
a(1) = 1;
a(2) = 1 + 2 = 3;
a(3) = 1 + 2 - 3 = 0;
a(4) = 1 + 2 - 3 - 4 = -4;
a(5) = 1 + 2 - 3 - 4 + 5 = 1;
a(6) = 1 + 2 - 3 - 4 + 5 + 6 = 7;
a(7) = 1 + 2 - 3 - 4 + 5 + 6 - 7 = 0;
a(8) = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 = -8; etc. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A077141.

CoefficientList[Series[(x^2-2x-1)/((x^2+1)^2*(x-1)),{x,0,100}],x] (* Vincenzo Librandi, Jul 30 2012 *)
nxt[{n_,a_}]:={n+1,If[CoprimeQ[a,n+1],a+n+1,a-n-1]}; NestList[nxt,{1,1},60][[All,2]] (* Harvey P. Dale, Jul 26 2020 *)

v=vector(100);v[1]=1;for(k=2,100,if(gcd(v[k-1],k)>1,v[k]=v[k-1]-k,v[k]=v[k-1]+k));print(v)

a(1) = 1 and a(n) = a(n-1) + n if gcd(n, a(n-1)) = 1, otherwise a(n) = a(n-1) - n.
G.f.: x(x^2-2x-1)/((x^2+1)^2*(x-1)). - Ralf Stephan, Mar 18 2003
Abs(a(n)) = ((n+1) mod 2)*n + (floor((n+(n mod 2))/2) mod 2). - Tj Wrenn (tjwrenn(AT)cs.utexas.edu), Apr 07 2005
a(n) = Sum_{k = 1..n} (-(-1)^((2*k - (-1)^k + 1)/4)*k). - Ilya Gutkovskiy, Dec 21 2015

More terms from Ralf Stephan, Mar 18 2003

A193232 Bitwise XOR of first n triangular numbers.

Original entry on oeis.org

0, 1, 2, 4, 14, 1, 20, 8, 44, 1, 54, 116, 58, 97, 8, 112, 248, 97, 202, 116, 166, 65, 188, 424, 132, 449, 158, 484, 114, 449, 16, 480, 1008, 449, 914, 484, 894, 449, 804, 40, 796, 65, 966, 116, 938, 1953, 920, 2032, 872, 1953, 858, 1652, 790, 1665, 844, 1352

Offset: 0

Franklin T. Adams-Watters, Jul 18 2011

nonn

John Tyler Rascoe, Table of n, a(n) for n = 0..10000

Cf. A000217, A145768, A003815.

a:= proc(n) option remember; `if`(n=0, 0,
      Bits[Xor](a(n-1), n*(n+1)/2))
    end:
seq(a(n), n=0..55);  # Alois P. Heinz, Feb 19 2023

Module[{nn=60,trs},trs=Accumulate[Range[nn]];Table[BitXor@@Take[trs,n],{n,0,nn}]] (* Harvey P. Dale, Dec 15 2017 *)

al(n) = local(m); vector(n,k,m=bitxor(m,k*(k+1)\2))

from operator import xor
from functools import reduce
def A193232(n): return reduce(xor, (x*(x+1) for x in range(n+1)))//2 # Chai Wah Wu, Dec 16 2021

cp's OEIS Frontend

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

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A003816 a(0) = 0, a(n) = a(n-1) XOR -n.

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A126084 a(n) = XOR of first n primes.

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A077140 a(1) = 1 and then add n to the previous term if n is coprime to the previous term, otherwise subtract n from the previous term. a(1) = 1 and a(n) = a(n-1) + n if gcd(n, a(n-1)) = 1, otherwise a(n) = a(n-1) - n.

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A193232 Bitwise XOR of first n triangular numbers.

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Programs

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Mathematica

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Python