A370574 a(n) is the least k > 0 such that n^2 XOR k^2 is a positive square number (where XOR denotes the bitwise XOR operator), or -1 if no such k exists.
-1, -1, 4, 3, 3, 8, 24, 6, 23, 6, 44, 16, 43, 48, 36, 12, 111, 46, 180, 12, 72, 88, 9, 7, 7, 86, 20, 59, 20, 72, 56, 24, 479, 222, 20, 15, 20, 360, 15, 24, 183, 144, 13, 11, 11, 18, 943, 14, 1103, 14, 747, 172, 405, 40, 159, 31, 492, 40, 28, 144, 28, 112, 31, 48, 660
Offset: 1
Examples
For n = 6: we have: k 6^2 XOR k^2 Positive square? -- ----------- ---------------- 1 37 No 2 32 No 3 45 No 4 52 No 5 61 No 6 0 No 7 21 No 8 100 Yes So a(6) = 8.
Links
- Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000
- Interactive scatterplot of (x, y, z) such that x^2 XOR y^2 = z^2 and x, y, z < 2^13 [provided your web browser supports the Plotly library, you should see icons on the top right corner of the page: if you choose "Orbital rotation", then you will be able to rotate the plot alongside three axes]
- Rémy Sigrist, Logarithmic scatterplot of the first 140000 terms
Crossrefs
Cf. A055527.
Programs
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Maple
f:= proc(n) local k; for k from 1 by 1 do if k <> n and issqr(MmaTranslator[Mma]:-BitXor(n^2, k^2)) then return k fi od end proc: f(1):= -1: f(2):= -1: map(f, [$1..100]); # Robert Israel, Mar 01 2024
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Mathematica
A370574[n_] := If[n <= 2, -1, Block[{k = 0}, While[++k == n || !IntegerQ[ Sqrt[BitXor[n^2, k^2]]]]; k]]; Array[A370574, 100] (* Paolo Xausa, Mar 01 2024 *)
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PARI
a(n) = { if (n <= 2, return (-1), for (k = 1, oo, if (k!=n && issquare( bitxor(n^2, k^2)), return (k)););); } -
Python
from sympy import integer_nthroot def A370574(n): if n <= 2: return -1 k, n_2 = 1, n**2 while True: if (integer_nthroot(n_2 ^ k**2, 2))[1] and k != n: return k k += 1 # Karl-Heinz Hofmann, Mar 03 2024
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