A394743 a(n) = number of triples (x, y, z) such that x^2 + y*z = n, where x,y,z are positive Fibonacci numbers.

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, 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, 2, 6, 2, 0, 6, 2, 4, 0, 0, 4, 0, 2, 4, 0, 0, 0, 0, 2

Offset

0,4

Links

Table of n, a(n) for n=0..85.

Example

a(6) = 4 counts these triples: (1,1,5), (1,5,1), (2,1,2), (2,2,1).
a(17) = 6 counts these triples: (1, 2, 8), (1, 8, 2), (2, 1, 13), (2, 13, 1), (3, 1, 8), (3, 8, 1).

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, #] &]];
fibonacciQ[n_] := IntegerQ[Sqrt[5 n^2 + 4]] || IntegerQ[Sqrt[5 n^2 - 4]];
c = (fibonacciQ[#1] && fibonacciQ[#2] && fibonacciQ[#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, A394740, A394744.

Sequence in context: A125915 A071472 A367120 * A064135 A332476 A179651

Adjacent sequences: A394740 A394741 A394742 * A394744 A394745 A394746

Keyword

nonn

Author

Clark Kimberling, Apr 09 2026

Status

approved