A330271 a(n) is the least nonnegative integer k such that n XOR k is a cube (where XOR denotes the bitwise XOR operator).
0, 0, 2, 2, 4, 4, 6, 6, 0, 1, 2, 3, 4, 5, 6, 7, 11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4, 32, 32, 34, 34, 36, 36, 38, 38, 32, 33, 34, 35, 36, 37, 38, 39, 43, 42, 41, 40, 47, 46, 45, 44, 35, 34, 33, 32, 39, 38, 37, 36, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Offset: 0
Examples
For n = 4: - 4 XOR 0 = 4 (not a cube), - 4 XOR 1 = 5 (not a cube), - 4 XOR 2 = 6 (not a cube), - 4 XOR 3 = 7 (not a cube), - 4 XOR 4 = 0 = 0^3, - hence a(4) = 4.
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 cube, 0 <= x, y <= 1023
Crossrefs
Programs
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Mathematica
A330271[n_] := Module[{k = -1}, While[!IntegerQ[CubeRoot[BitXor[n, ++k]]]]; k]; Array[A330271, 100, 0] (* Paolo Xausa, Feb 20 2024 *)
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PARI
a(n) = for (k=0, oo, if (ispower(bitxor(n,k),3), return (k)))
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Python
from itertools import count from sympy import integer_nthroot def A330271(n): return next(k for k in count(0) if integer_nthroot(n^k,3)[1]) # Chai Wah Wu, Aug 23 2023
Formula
a(n) = 0 iff n is a cube.
Comments