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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 Fibonacci numbers.
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%I #13 May 16 2026 22:38:50

%S 0,0,1,2,2,2,4,4,1,4,4,4,4,2,6,2,2,6,1,4,2,0,4,0,2,6,2,4,4,2,6,2,0,4,

%T 2,6,0,0,4,0,4,4,0,6,2,0,4,0,2,4,1,4,0,0,0,0,2,0,0,4,0,0,0,0,6,4,4,6,

%U 2,6,2,0,6,2,4,0,0,4,0,2,4,0,0,0,0,2

%N a(n) = number of triples (x, y, z) such that x^2 + y*z = n, where x,y,z are positive Fibonacci numbers.

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

%e a(17) = 6 counts these triples: (1, 2, 8), (1, 8, 2), (2, 1, 13), (2, 13, 1), (3, 1, 8), (3, 8, 1).

%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 fibonacciQ[n_] := IntegerQ[Sqrt[5 n^2 + 4]] || IntegerQ[Sqrt[5 n^2 - 4]];

%t c = (fibonacciQ[#1] && fibonacciQ[#2] && fibonacciQ[#3] &);

%t 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, A394740, A394744.

%K nonn

%O 0,4

%A _Clark Kimberling_, Apr 09 2026