Find the nth Fibonnaci Prime, in the shortest code

Question

The challenge is rather simple:

1. Take a positive whole number n as input.
2. Output the nth Fibonacci prime number.

Input can be as an parameter to a function (and the output will be the return value), or can be taken from the command line (and outputted there).

Note: Using built in prime checking functions or Fibonacci series generators is not allowed.

Good luck!

Answer 1

C, 66

f(n,a,b){int i=2;while(a%i&&i++<a);return(n-=i==a)?f(n,b,a+b):a;}

Answer 2

C, 85, 81, 76

f(n){int i=1,j=0,k;for(;n;n-=k==i)for(j=i-j,i+=j,k=2;i%k&&k++<i;);return i;}

• borrowed code style of simplified prime number check from @Gautam

• self contained C function (no globals)

Testing:

main(int n,char**v){printf("%d\n",f(atoi(v[1])));}

./a.out 10
433494437

./a.out 4
13

Answer 3

Mathematica 147 143 141 chars

f@0 = 0; f@1 = 1; f@n_ := f[n - 1] + f[n - 2]
q@1 = False; q@n_ := FactorInteger@n~MatchQ~{{_, 1}}
p = {}; k = 1; While[Length@p < n, If[q@f@k, p~AppendTo~f[k]]; k++];p[[-1]]

f is the recursive definition of Fibonacci number.

q detects primes.

k is a Fibonacci prime iff q@f@k is True.

For n=10, output is 433494437.

Answer 4

Ruby, 94 68 67

n=->j{a=b=1
while j>0
a,b=b,a+b
(2...b).all?{|m|b%m>0}&&j-=1
end
b}

Clojure, 112

Ungolfed:

(defn nk [n]
  (nth 
    (filter
      (fn[x] (every? #(> (rem x %) 0) (range 2 x)))    ; checks if number is prime
      ((fn z[a b] (lazy-seq (cons a (z b (+ a b))))) 1 2)) ; Fib seq starting from [1, 2]
    n)) ; get nth number

Golf: (defn q[n](nth(filter(fn[x](every? #(>(rem x %)0)(range 2 x)))((fn z[a b](lazy-seq(cons a(z b(+ a b)))))2 3))n))

Answer 5

Haskell 108

p=2 : s [3,5..]  where
    s (p:xs) = p : s [x|x<-xs,rem x p /= 0]
f=0:1:(zipWith (+) f$tail f)
fp=intersect p f

To get nth number call it fp !! n.

EDIT: Sic. Wrong answer, I fix it.

Answer 6

Groovy: 105 (134 with whitespaces)

b is the fibonacci function.

the closure inside the if is the prime check function. Update: a small fix on it

r is the prime fibonacci number.

r={ n->
  c=k=0
  while(1) {
    b={a->a<2?a:b(a-1)+b(a-2)}
    f=b k++
    if({z->z<3?:(2..<z).every{z%it}}(f)&&c++==n)return f
  }
}

Test cases:

assert r(0) == 0
assert r(1) == 1
assert r(2) == 1
assert r(3) == 2
assert r(4) == 3
assert r(5) == 5
assert r(6) == 13
assert r(7) == 89
assert r(8) == 233
assert r(9) == 1597

A readable version:

def fib(n) { 
  n < 2 ? n : fib(n-1) + fib(n-2)
}
def prime(n) {
  n < 2 ?: (2..<n).every { n % it }
}
def primeFib(n) { 
  primes = inc = 0
  while( 1 ) {
    f = fib inc++
    if (prime( f ) && primes++ == n) return f
  }
}

Answer 7

Here is a solution in APL:

{⊃({(+/⍵),⊃⍵}⍣(⍵-1))1}

It is iterative and uses the standard accumulative algorithm.

[Whoops! This is for Fibonnacci Primes...the above is just regular primes.]