A259503 Numbers k such that k^2+1 is the product of a Fibonacci number and a Lucas number.

0, 1, 2, 3, 5, 12, 133

Offset

1,3

Comments

Conjecture: the sequence is finite.

No more terms below 25*10^4. - Robert G. Wilson v, Jul 06 2015

No more terms below 10^7. - Manfred Scheucher, Aug 03 2015

Links

Table of n, a(n) for n=1..7.

Manfred Scheucher, Sage Script

Example

0^2+1 = 1 = 1*1 = F(1)*L(1);
1^2+1 = 2 = 2*1 = F(3)*L(1);
2^2+1 = 5 = 5*1 = F(5)*L(1);
3^2+1 = 10 = 5*2 = F(5)*L(0);
5^2+1 = 26 = 13*2 = F(7)*L(0);
12^2+1 = 145 = 5*29 = F(5)*L(7);
133^2+1 = 17690 = 610*29 = F(15)*L(7).

Maple

with(combinat, fibonacci):nn:=200: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)-1):
  if x=floor(x) then lst:=lst union {x}:
  else fi:
od:
od:
print(lst):

Mathematica

fibQ[n_] := (Fibonacci@ Round@ Log[ GoldenRatio, n*Sqrt@ 5 == n); fQ[n_] := Block[{k = 0, l}, While[l = LucasL@ k; l < n^2 + 2 && ! fibQ[(n^2 + 1)/l], k++]; If[l < 2 + n^2, True, False]]; k = 0; lst = {}; While[k < 250001, If[ fQ@ k, AppendTo[lst, k]; Print[k]]; k++]; lst (* Robert G. Wilson v, Jul 06 2015 *)

Crossrefs

Cf. A000032, A000045, A002522.

Sequence in context: A301929 A107475 A108225 * A193064 A133832 A328997

Adjacent sequences: A259500 A259501 A259502 * A259504 A259505 A259506

Keyword

nonn,more

Author

Michel Lagneau, Jun 29 2015

Status

approved