A394744 - OEIS

%I #10 Apr 18 2026 16:44:06

%S 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,

%T 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,

%U 0,3,1,0,0,0,2,0,0,2,0,1,1,0,0,0,0,1

%N 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.

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

%C 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

%p N:= 100: # for a(0) .. a(N)

%p V:= Array(0..N):

%p F:= [seq(combinat:-fibonacci(k),k=2..ceil(log[(1+sqrt(5))/2](sqrt(5)*N)))]:

%p nF:= nops(F):

%p for x in F do

%p   for i from 2 to nF do

%p     if x = F[i] then next fi;

%p     for j from 1 to i-1 do

%p       if x = F[j] then next fi;

%p       v:= x^2 + F[i]*F[j];

%p       if v > N then break fi;

%p       V[v]:= V[v]+1

%p od od od:

%p convert(V,list); # _Robert Israel_, Apr 13 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 fibonacciQ[n_] := IntegerQ[Sqrt[5 n^2 + 4]] || IntegerQ[Sqrt[5 n^2 - 4]];

%t c = (fibonacciQ[#1] && fibonacciQ[#2] && fibonacciQ[#3] && #2 < #3 && DuplicateFreeQ[{#1, #2, #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. A000045, A393710, A394740, A394743.

%K nonn

%O 0,8

%A _Clark Kimberling_, Apr 12 2026