A394745 a(n) = number of triples (x, y, z) such that x^2 + y*z = n, where x,y,z are positive integers satisfying x > max(y,z).

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 1, 2, 2, 1, 0, 2, 0, 0, 1, 1, 2, 2, 3, 0, 2, 0, 2, 1, 0, 0, 3, 2, 2, 3, 3, 2, 0, 2, 1, 2, 0, 2, 0, 1, 4, 3, 3, 2, 4, 2, 2, 1, 2, 0, 5, 0, 0, 2, 2, 2, 4, 3, 4, 4, 2, 2, 3, 3, 0, 4, 0, 2, 4, 1, 0, 3, 2, 4, 6

Offset

0,12

Comments

Is 2262 the greatest n for which a(n) = 0? There are no others <= 10^5. - Robert Israel, Apr 24 2026

Links

Robert Israel, Table of n, a(n) for n = 0..10000

Example

a(51) = 4 counts these triples : (6, 3, 5), (6, 5, 3), (7, 1, 2), (7, 2, 1).

Maple

f:= proc(n) local x, y, z;
   add(nops(select(t -> max(t, (n-x^2)/t) < x, NumberTheory:-Divisors(n-x^2))), x=1..ceil(sqrt(n))-1)
end proc:
map(f, [$0..100]); # Robert Israel, Apr 23 2026

Mathematica

t[n_, c_] := Module[{r}, r = Flatten[Table[If[n - x^2 <= 0, {},
       Map[({x, #, Quotient[n - x^2, #]} &),
       Select[Divisors[n - x^2], Divisible[n - x^2, #] &]]], {x, 1,
       Floor[Sqrt[n - 1]]}], 1]; Select[r, Apply[c, #] &]];
c = (#1 > Max[#2, #3] &); Table[{n, t[n, c]}, {n, 1, 30}]
Join[{0}, Table[Length[t[n, c]], {n, 1, 130}]]
(* Peter J. C. Moses, Mar 29 2026 *)

Crossrefs

Cf. A393710.

Sequence in context: A219558 A279210 A232243 * A034876 A091393 A359735

Adjacent sequences: A394742 A394743 A394744 * A394746 A394747 A394748

Keyword

nonn

Author

Clark Kimberling, Apr 13 2026

Status

approved