A394744 a(n) = number of triples (x, y, z) such that x^2 + y*z = n, where x,y,z are distinct nonzero Fibonacci numbers and y < z.

0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 2, 1, 0, 1, 0, 1, 3, 0, 2, 0, 0, 1, 0, 0, 3, 0, 2, 2, 0, 1, 1, 0, 1, 0, 1, 0, 0, 2, 0, 1, 2, 0, 3, 1, 0, 1, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 3, 0, 2, 3, 0, 3, 1, 0, 0, 0, 2, 0, 0, 2, 0, 1, 1, 0, 0, 0, 0, 1

Offset

0,8

Comments

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

The sequence is unbounded, because for any m, F(2*m-1-2*i)^2 + F(2*i)*F(4*m-2-2*i) = F(2*m-1)^2 is the same for 1 <= i < m, where F = A000045. - Robert Israel, Apr 13 2026

Links

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

Maple

N:= 100: # for a(0) .. a(N)
V:= Array(0..N):
F:= [seq(combinat:-fibonacci(k), k=2..ceil(log[(1+sqrt(5))/2](sqrt(5)*N)))]:
nF:= nops(F):
for x in F do
  for i from 2 to nF do
    if x = F[i] then next fi;
    for j from 1 to i-1 do
      if x = F[j] then next fi;
      v:= x^2 + F[i]*F[j];
      if v > N then break fi;
      V[v]:= V[v]+1
od od od:
convert(V, list); # Robert Israel, Apr 13 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, #] &]];
fibonacciQ[n_] := IntegerQ[Sqrt[5 n^2 + 4]] || IntegerQ[Sqrt[5 n^2 - 4]];
c = (fibonacciQ[#1] && fibonacciQ[#2] && fibonacciQ[#3] && #2 < #3 && DuplicateFreeQ[{#1, #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. A000045, A393710, A394740, A394743.

Sequence in context: A341978 A193527 A127505 * A194923 A321103 A358133

Adjacent sequences: A394741 A394742 A394743 * A394745 A394746 A394747

Keyword

nonn

Author

Clark Kimberling, Apr 12 2026

Status

approved