3
$\begingroup$

Is there a formula for the sequence of integers n such that the Catalan number $C_n=\frac{1}{n+1} \binom{2n}{n}$ and the Fibonacci number $F_n$ have Gcd equal to one?

Or more general for a number z>=1, is there a formula for the sequence of integers n such that the Catalan number $C_n$ and the Fibonacci number $F_n$ have Gcd equal to z? Another question: Define $A_z:=$ cardinality of $ \{ n \leq z | Gcd(C_n,F_n,n)>1 \}$. Is there a formula for $A_z$ and does $A_z /z$ converge ? Computer suggests it converge against 1/3.

Here my GAP output (which also makes clear how the sequences start) . Here [a,b,c] means that a is such that the gcd is one and b is the value of the catalan sequence $C_a$ and c is the value of the fibonacci sequence $F_a$. [ 1, 1, 1 ], [ 2, 2, 1 ], [ 3, 5, 2 ], [ 4, 14, 3 ], [ 5, 42, 5 ], [ 8, 1430, 21 ], [ 10, 16796, 55 ], [ 11, 58786, 89 ], [ 13, 742900, 233 ], [ 14, 2674440, 377 ], [ 17, 129644790, 1597 ], [ 22, 91482563640, 17711 ], [ 23, 343059613650, 28657 ], [ 25, 4861946401452, 75025 ], [ 26, 18367353072152, 121393 ], [ 28, 263747951750360, 317811 ], [ 29, 1002242216651368, 514229 ], [ 31, 14544636039226909, 1346269 ], [ 34, 812944042149730764, 5702887 ], [ 38, 176733862787006701400, 39088169 ] , [ 41, 10113918591637898134020, 165580141 ], [ 43, 150853479205085351660700, 433494437 ], [ 46, 8740328711533173390046320, 1836311903 ], [ 47, 33868773757191046886429490, 2971215073 ] ]

  • 2
    Plan of attack: calculate the first few terms, then look it up in the Online Encyclopedia of Integer Sequences. – 2017-02-06
  • 0
    Done, nothing found. – 2017-02-06
  • 0
    It seems that those are exactly the indices of Fibonacci primes (i.e. $n$ such that $n$ is prime). I've checked it up to $n=18$. – 2017-02-06
  • 0
    Since you've evidently done some work on this, Mare, perhaps you could share it with us, instead of hiding it. Give us the first few (dozen) terms of the sequence, please, so we don't all have to do it by ourselves. – 2017-02-06
  • 0
    I get 1,2,3,4,5,8,10,11,13,14,17 for gcd=1 and thats not found in the database – 2017-02-06
  • 0
    Actually $\gcd(F_{10}, C_{10})=11$, $\gcd ( F_8 , C_8 )=3$ and $\gcd( F_2, C_2)= \gcd (2,2) = 2$. You made some mistakes. – 2017-02-06
  • 0
    @Crostul: Seems to depend to the numbering, I have $F_{10}=55, \quad C_{10}=16796$ and $gcd=1$ – 2017-02-06
  • 0
    Thanks for checking, I included my GAP output. Maybe we dont beginn to count the same for F_1 or C_1 ? – 2017-02-06
  • 0
    I found that result by starting to count with $1$. Fibonacci numbers are $1,1,2,...$ and so on starting with $F_1=F_2=1$. Changing the indices, everything changes. – 2017-02-06
  • 0
    @Crostul I start with the same for fibonacci so you might start different for catalan numbers? See my edit with GAP output. – 2017-02-06
  • 0
    Very minor point: type `\gcd(a, b)` to get $\gcd(a, b)$. – 2017-02-06
  • 0
    @Crostul, in one comment you write $\gcd(F_2,C_2)=\gcd(2,2)$, in another, you write $F_1=F_2=1$, so you're not being consistent in your Fibonaccis. – 2017-02-07
  • 0
    Now posted to MO, http://mathoverflow.net/questions/261620/gcd-of-fibonacci-and-catalan – 2017-02-07

0 Answers 0