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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).
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%I #10 Apr 29 2026 18:16:30

%S 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,

%T 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,

%U 3,4,4,2,2,3,3,0,4,0,2,4,1,0,3,2,4,6

%N 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).

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

%H Robert Israel, <a href="/A394745/b394745.txt">Table of n, a(n) for n = 0..10000</a>

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

%p f:= proc(n) local x,y,z;

%p add(nops(select(t -> max(t,(n-x^2)/t) < x, NumberTheory:-Divisors(n-x^2))),x=1..ceil(sqrt(n))-1)

%p end proc:

%p map(f, [$0..100]); # _Robert Israel_, Apr 23 2026

%t t[n_, c_] := Module[{r}, r = Flatten[Table[If[n - x^2 <= 0, {},

%t Map[({x, #, Quotient[n - x^2, #]} &),

%t Select[Divisors[n - x^2], Divisible[n - x^2, #] &]]], {x, 1,

%t Floor[Sqrt[n - 1]]}], 1]; Select[r, Apply[c, #] &]];

%t c = (#1 > Max[#2, #3] &);Table[{n, t[n, c]}, {n, 1, 30}]

%t Join[{0}, Table[Length[t[n, c]], {n, 1, 130}]]

%t (* _Peter J. C. Moses_, Mar 29 2026 *)

%Y Cf. A393710.

%K nonn

%O 0,12

%A _Clark Kimberling_, Apr 13 2026