A245688 Numbers n such that n^2 + 1 is the product of three distinct Fibonacci numbers > 1.

47, 99, 123, 322, 843, 2207, 5778, 15075, 15127, 39603, 103682, 271443, 710647, 1860498, 4870847, 12752043, 33385282, 87403803, 228826127, 599074578, 1568397607, 4106118243, 10749957122, 28143753123, 73681302247, 192900153618, 505019158607, 1322157322203

Offset: 1

Author

Michel Lagneau, Jul 29 2014

Keywords

• nonn
• easy

Comments

Conjecture: except the numbers 99 and 15075, all the terms belong to A005248 (bisection of Lucas numbers).

The above conjecture holds for the first 1000 terms. - Jens Kruse Andersen, Aug 10 2014

Examples

			99 is in the sequence because 99^2+1 = 9802 = 2*13*377 where 2, 13 and 377 are three Fibonacci numbers, but 99 is not a Lucas number.
15075 is in the sequence because 15075^2+1 = 13*89*196418 where 13, 89 and 196418 are three Fibonacci numbers, but 15075 is not a Lucas number.
		

Links

• Jens Kruse Andersen, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000032, A000045, A005248, A245236.

Programs

Maple

with(combinat,fibonacci):with(numtheory):nn:=200:lst:={}:T:=array(1..nn):
   for n from 1 to nn do:
    T[n]:=fibonacci(n):
   od:
     for p from 1 to nn-1 do:
       for q from p+1 to nn-1 do:
          for r from q+1 to nn-1 do:
           f:=T[p]*T[q]*T[r]-1:x:=sqrt(f):
           if x=floor(x)and T[p]<>1
           then
           lst:=lst union {x}:
           else
           fi:
          od:
       od:
     od:
     print(lst):

Formula

Empirical g.f.: x*(9297*x^9-24320*x^8-52*x^7+52*x^3-127*x^2-42*x+47) / (x^2-3*x+1). - Colin Barker, Aug 13 2014

Extensions

Doubtful link, formula and PARI code deleted by Colin Barker, Jul 31 2014