A259561 Numbers k such that k^2+2 is the product of a Fibonacci number and a Lucas number.

0, 1, 2, 3, 4, 10, 11, 29, 40, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596

Offset

1,3

Comments

Except for a(6)=10 and a(9)=40, it seems that a(n) is a Lucas number.

Except for a(3)=2, a(6)=10 and a(9)=40, it seems that a(n)^2+2 is a Lucas number.

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (3,-1).

Formula

a(n+6) = A002878(n) for n>=4.

a(n) = 3*a(n-1)-a(n-2) for n>11. - Colin Barker, Jun 30 2015

G.f.: x^2*(11*x^9-15*x^8-36*x^7+6*x^6-15*x^5+x^4-3*x^3-2*x^2-x+1) / (x^2-3*x+1). - Colin Barker, Jun 30 2015

Example

0^2+2 = 2 = 1*2 = F(1)*L(0);
1^2+2 = 3 = 1*3 = F(1)*L(2);
2^2+2 = 6 = 2*3 = F(3)*L(2);
3^2+2 = 11 = 1*11 = F(1)*L(5);
4^2+2 = 18 = 1*18 = F(1)*L(6);
10^2+2 = 102 = 34*3 = F(9)*L(2).

Maple

with(combinat, fibonacci):nn:=100:lst:={}:
a:=n->2*fibonacci(n-1)+fibonacci(n):
for i from 0 to nn do:
for j from 0 to nn do:
  x:=sqrt(a(i)*fibonacci(j)-2):
  if x=floor(x) then lst:=lst union {x}:
  else fi:
od:
od:
print(lst):

Mathematica

LinearRecurrence[{3, -1}, {0, 1, 2, 3, 4, 10, 11, 29, 40, 76, 199}, 50] (* Paolo Xausa, Jul 15 2024 *)

Prog

(PARI) concat(0, Vec(x^2*(11*x^9-15*x^8-36*x^7+6*x^6-15*x^5+x^4-3*x^3-2*x^2-x+1)/(x^2-3*x+1) + O(x^50))) \\ Colin Barker, Jun 30 2015

Crossrefs

Cf. A002878, A000032, A000045.

Sequence in context: A115897 A116019 A352148 * A087460 A082866 A085701

Adjacent sequences: A259558 A259559 A259560 * A259562 A259563 A259564

Keyword

nonn,easy

Author

Michel Lagneau, Jun 30 2015

Status

approved