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

%I #22 Aug 15 2014 10:25:35

%S 47,99,123,322,843,2207,5778,15075,15127,39603,103682,271443,710647,

%T 1860498,4870847,12752043,33385282,87403803,228826127,599074578,

%U 1568397607,4106118243,10749957122,28143753123,73681302247,192900153618,505019158607,1322157322203

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

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

%C The above conjecture holds for the first 1000 terms. - _Jens Kruse Andersen_, Aug 10 2014

%H Jens Kruse Andersen, <a href="/A245688/b245688.txt">Table of n, a(n) for n = 1..1000</a>

%F 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

%e 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.

%e 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.

%p with(combinat,fibonacci):with(numtheory):nn:=200:lst:={}:T:=array(1..nn):

%p for n from 1 to nn do:

%p T[n]:=fibonacci(n):

%p od:

%p for p from 1 to nn-1 do:

%p for q from p+1 to nn-1 do:

%p for r from q+1 to nn-1 do:

%p f:=T[p]*T[q]*T[r]-1:x:=sqrt(f):

%p if x=floor(x)and T[p]<>1

%p then

%p lst:=lst union {x}:

%p else

%p fi:

%p od:

%p od:

%p od:

%p print(lst):

%Y Cf. A000032, A000045, A005248, A245236.

%K nonn,easy

%O 1,1

%A _Michel Lagneau_, Jul 29 2014

%E Doubtful link, formula and PARI code deleted by _Colin Barker_, Jul 31 2014

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Last modified April 18 08:27 EDT 2024. Contains 371769 sequences. (Running on oeis4.)