A330270 a(n) is the least nonnegative integer k such that n XOR k is a square (where XOR denotes the bitwise XOR operator).

0, 0, 2, 2, 0, 1, 2, 3, 1, 0, 3, 2, 5, 4, 7, 6, 0, 1, 2, 3, 4, 5, 6, 7, 1, 0, 3, 2, 5, 4, 7, 6, 4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11, 1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1

Offset

0,3

Comments

This sequence has similarities with A329794 as the XOR operator and the "box" operator defined in A329794 both map (n, n) to 0 for any n (however here we accept 0 as a square).

Links

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Rémy Sigrist, Scatterplot of the ordinal transform of the first 2^20 terms

Rémy Sigrist, Scatterplot of (x, y) such that x XOR y is a square, 0 <= x, y <= 1023

Formula

a(n) = 0 iff n is a square.

a(a(n)) <= n.

Example

For n = 7,
- 7 XOR 0 = 7 (not a square),
- 7 XOR 1 = 6 (not a square),
- 7 XOR 2 = 5 (not a square),
- 7 XOR 3 = 4 = 2^2,
- hence a(7) = 3.

Mathematica

A330270[n_] := Module[{k = -1}, While[!IntegerQ[Sqrt[BitXor[n, ++k]]]]; k];
Array[A330270, 100, 0] (* Paolo Xausa, Feb 19 2024 *)

Prog

(PARI) a(n) = for (k=0, oo, if (issquare(bitxor(n, k)), return (k)))
(Python)
from itertools import count
from sympy.ntheory.primetest import is_square
def A330270(n): return next(k for k in count(0) if is_square(n^k)) # Chai Wah Wu, Aug 22 2023

Crossrefs

See A330271 for the cube variant.

Cf. A000290, A003987, A295520, A329794.

Sequence in context: A101565 A292944 A327188 * A029341 A240181 A079070

Adjacent sequences: A330267 A330268 A330269 * A330271 A330272 A330273

Keyword

nonn,base,easy

Author

Rémy Sigrist, Dec 08 2019

Status

approved