A379953 Largest k >= 0 such that (n*k)^3/(n^3+k^3) is an integer.

0, 2, 6, 4, 0, 12, 0, 8, 18, 10, 0, 24, 0, 42, 30, 16, 0, 36, 0, 20, 42, 22, 0, 48, 0, 26, 54, 84, 0, 60, 0, 32, 66, 34, 0, 72, 0, 38, 78, 40, 0, 210, 0, 44, 90, 46, 0, 96, 0, 50, 102, 52, 0, 108, 0, 168, 456, 58, 0, 120, 0, 62, 126, 64, 260, 132, 0, 68, 138, 1330, 0, 144, 0, 74, 150, 76, 0, 1794, 0, 80, 162, 82, 0, 420

Offset: 1

Author

Antti Karttunen, Jan 16 2025

Keywords

• nonn

Comments

For all n, a(n) < n^2, as for k^3 + n^3 to divide k^3 * n^3, it must also divide n^3 * (k^3 + n^3) - k^3*n^3 = n^6, so k^3 <= n^6 - n^3, and in particular k < n^2. - Robert Israel, Jan 16 2025

Not every a(n) is a multiple of n: a(231) = 616 is the first case where n does not divide a(n).

First odd terms are a(1474) = 6633, a(1628) = 2849 and a(1860) = 5115.

For all odd n, a(n) is even.

If n = p^j where p is a prime >= 5, then a(n) = 0 (see link for proof). - Robert Israel, Jan 16 2025

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Robert Israel, Proof of comment (which applies to A119612, A379953 and A379954)

Crossrefs

Cf. A119612, A379954.

Programs

Maple

f:= proc(n) local k;
  for k from n^2 by -1 do
    if (n*k)^3 mod (n^3 + k^3) = 0 then return k fi
  od
end proc:
map(f, [$1..100]); # Robert Israel, Jan 16 2025

Mathematica

A379953[n_] := Module[{k = n^2}, While[PowerMod[--k*n, 3, # + k^3] > 0] & [n^3]; k];
Array[A379953, 100] (* Paolo Xausa, Jan 19 2025 *)

PARI

A379953(n) = forstep(k=n^2, 0, -1, if(!(((n*k)^3)%(k^3+n^3)), return(k)));