A259561 - OEIS

%I #13 Jul 15 2024 16:36:10

%S 0,1,2,3,4,10,11,29,40,76,199,521,1364,3571,9349,24476,64079,167761,

%T 439204,1149851,3010349,7881196,20633239,54018521,141422324,370248451,

%U 969323029,2537720636,6643838879,17393796001,45537549124,119218851371,312119004989,817138163596

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

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

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

%H Colin Barker, <a href="/A259561/b259561.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1).

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

%F a(n) = 3*a(n-1)-a(n-2) for n>11. - _Colin Barker_, Jun 30 2015

%F 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

%e 0^2+2 = 2 = 1*2 = F(1)*L(0);

%e 1^2+2 = 3 = 1*3 = F(1)*L(2);

%e 2^2+2 = 6 = 2*3 = F(3)*L(2);

%e 3^2+2 = 11 = 1*11 = F(1)*L(5);

%e 4^2+2 = 18 = 1*18 = F(1)*L(6);

%e 10^2+2 = 102 = 34*3 = F(9)*L(2).

%p with(combinat,fibonacci):nn:=100:lst:={}:

%p a:=n->2*fibonacci(n-1)+fibonacci(n):

%p for i from 0 to nn do:

%p for j from 0 to nn do:

%p   x:=sqrt(a(i)*fibonacci(j)-2):

%p   if x=floor(x) then lst:=lst union {x}:

%p   else fi:

%p od:

%p od:

%p print(lst):

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

%o (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

%Y Cf. A002878, A000032, A000045.

%K nonn,easy

%O 1,3

%A _Michel Lagneau_, Jun 30 2015