cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A145831 Log_2 ( A145768( A145830(n))).

Original entry on oeis.org

0, 4, 0, 8, 9, 8
Offset: 1

Views

Author

M. F. Hasler, Oct 20 2008

Keywords

Crossrefs

A145830 = A145768 intersect A000079. A145768(A145830(n)) = 2^a(n). See also A145827-A145829.

Programs

  • PARI
    an=0; for( i=1,10^7, an=bitxor(an,i^2); an & an==1<

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

Original entry on oeis.org

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

Views

Author

Vladimir Reshetnikov, Oct 18 2008

Keywords

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

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.
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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