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A245688 Numbers n such that n^2 + 1 is the product of three distinct Fibonacci numbers > 1. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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

LINKS

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

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

EXAMPLE

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.

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):

CROSSREFS

Cf. A000032, A000045, A005248, A245236.

Sequence in context: A180550 A176134 A155844 * A211331 A141961 A142661

Adjacent sequences:  A245685 A245686 A245687 * A245689 A245690 A245691

KEYWORD

nonn,easy

AUTHOR

Michel Lagneau, Jul 29 2014

EXTENSIONS

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

STATUS

approved

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Last modified October 17 18:51 EDT 2019. Contains 328127 sequences. (Running on oeis4.)