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A242103 Numbers m such that m^2 - 1 is the product of three distinct Fibonacci numbers > 1. 0

%I #15 Sep 06 2014 00:40:39

%S 7,9,11,14,29,76,121,199,329,521,659,1364,3571,4523,7307,9349,24476,

%T 64079,167761,212533,439204,1149851,3010349,7881196,20633239,54018521,

%U 141422324,370248451,969323029,2537720636,6643838879,17393796001,45537549124,119218851371

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

%C Conjecture : except the numbers 9, 14, 121, 329, 659, 4523, 7307 and 212533, a(n) is a Lucas number (A000204).

%e The non-Lucas number 9 is in the sequence because 9^2-1 = 80 = 2*5*8 is the product of three Fibonacci numbers.

%e The Lucas number 11 is in the sequence because 11^2-1 = 120 = 3*5*8 is the product of three Fibonacci numbers.

%p with(combinat,fibonacci):with(numtheory):nn:=150: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):

%o (PARI)

%o v=[];for(i=3,100,for(j=i+1,100,for(k=j+1,100,s=fibonacci(i)*fibonacci(j)*fibonacci(k);if(issquare(s+1),v=concat(sqrtint(s+1),v)))));v=vecsort(v);v \\ _Derek Orr_, Aug 27 2014

%Y Cf. A245688, A242074, A000204.

%K nonn

%O 1,1

%A _Michel Lagneau_, Aug 15 2014

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Last modified May 4 10:30 EDT 2024. Contains 372240 sequences. (Running on oeis4.)