Arithmetic mean of prime Fibonacci Numbers up to the x Fibonacci Number

Question

You should have heard about the Fibonacci Numbers, often called the Fibonacci Sequence. In this sequence the first two terms are 0 and 1, and every number after the first two is the sum of the two preceding ones. In other words, F(n) = F(n-1) + F(n-2).

Here are the first 20 Fibonacci numbers:

0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181

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Task:

Given an integer x, compute the arithmetic mean (the average) of the prime Fibonacci Numbers up to the x number of the Fibonacci Sequence.

Rules:

• the Fibonacci sequence starts with 0 and 1 for this challenge
• 3 < x < 40, because higher values of x might cause some huge execution time or overflows and smaller values have no output
• 1 is NOT prime, since it only has 1 divisor
• the arithmetic mean should include decimals, if it's the case, or should be displayed as an exact fraction
• you are only allowed to take x as input and the code needed to take the input does not count (e.g: if you need something like x = input(), you should not take it into consideration when counting the bytes)

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Examples:

Ex. 1: For x=10, the output is 5.75, because the 10th Fibonacci number is 55 and the prime Fibonacci numbers up to 55 are 2, 3, 5, 13, their average being 5.75

Following the explanation from example 1, other examples are:

Ex. 2: For x=15, the output is 57.5

Ex. 3: For x=20, the output is 277.428571428571, or any other close approximation. In this case 277.4286, for instance, is an accepted value

Ex. 4: For x=11, the output is 22.4

Ex. 5: For x=30, the output is 60536.4444444444, or any other close approximation, such as 60536.444

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Leaderboard:

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To change the leader, submit a shorter valid solution. Your code should be as short as possible, since this is code-golf, so the shortest answer in bytes wins. Good luck!

Answer 1

Actually, 10 bytes

Code:

R♂F;Mr♂P∩æ

Explanation:

R            # On the input, create the range [1 .. input].
 ♂F          # Map the nth Fibonacci command over it.
   ;M        # Duplicate and get the maximum of the list.
     r       # Create the range [0 .. maximum - 1].
      ♂P     # Map the nth prime operator over each element (zero-indexed).
        ∩    # Intersection of both.
         æ   # Get the mean and implicitly display.

Uses the CP-437 encoding. Try it online!

Answer 2

Python 2, 71 bytes

s=2.;a=b=c=1
exec"p=2**a%a==2;c+=p;s+=a*p;a,b=b,a+b;"*input()
print s/c

Try it online!

Python doesn't have useful arithmetic built-ins for this, so we do things by hand. The code iterates through Fibonacci numbers via a,b=b,a+b starting from a=b=1.

The Fermat primality test with base 2 is used to identify primes as a where 2^a == 2 (mod a). Though this only checks for probable primes, none of the false positives are within the first 40 Fibonacci numbers.

The current sum s and count c of primes are updated each time a prime is encountered, and their ratio (the mean) is printed at the end. Because the prime check misses a=2 and it's guaranteed to be in the input range, the sum starts at 2 and the count starts at 1 to compensate.

Answer 3

Jelly, 11 bytes

ÆḞ€ÆPÐfµS÷L

Try it online!

How it works

ÆḞ€ÆPÐfµS÷L  Main link. Argument: n (integer)

ÆḞ€          Map the Fibonacci atom over [1, ..., n].
   ÆPÐf      Filter by primality.
       µ     Begin a new chain. Argument: A (array of Fibonacci primes)
        S    Yield the sum of A.
          L  Yield the length of A.
         ÷   Compute the quotient.

Answer 4

Mathematica, 38 bytes

Mean@Select[Fibonacci@Range@#,PrimeQ]&

(* or *)

Mean@Select[Fibonacci~Array~#,PrimeQ]&

Explanation

Mean@Select[Fibonacci@Range@#,PrimeQ]&  
                                     &  (* For input # *)
                      Range@#           (* List {1, 2, 3, ... #} *)
            Fibonacci@                  (* Find the corresponding fibonacci numbers *)
     Select[                 ,PrimeQ]   (* Select the prime terms *)
Mean@                                   (* Take the Mean *)

Answer 5

Perl 6, 51 bytes

{sum($/=[grep *.is-prime,(0,1,*+*...*)[0..$_]])/$/}

Try it

Expanded:

{
  sum(
    $/ = [                # store as an Array in $/

      grep
        *.is-prime,       # find the primes
        (
          0, 1, *+* ... * # Fibonacci sequence
        )[ 0 .. $_ ]      # grab the indicated values in the sequence

    ]
  )

  /

  $/   # use the Array as a number (elems)
}

Answer 6

MATL, 16 bytes

lOi:"yy+]vtZp)Ym

Try it online!

Explanation

lO     % Push 1, then 0. This way the generated numbers with indices 1, 2, 3, ...
       % will be 1, 1, 2, ... as required
i      % Input n
:"     % Do the following n times
  yy   %   Duplicate top two numbers
  +    %   Add
]      % End
v      % Concatenate all numbers into a column vector
t      % Duplicate
Zp     % Vector of logical values: true for prime numbers, false otherwise
)      % Apply that vector as an index. This keeps only prime numbers
Ym     % Mean. Implicitly display

Answer 7

Octave, 75 71 bytes

@(n)mean((t=fix(((1+(s=5^.5)).^(x=1:n)-(1-s).^x)/s./2.^x))(isprime(t)))

Anonymous function that uses Binet's formula. Input and output are in the form of function arguments.

Try it online!

Answer 8

Maxima, 49 bytes

f(n):=mean(sublist(makelist(fib(x),x,n),primep));

Answer 9

Prolog (SWI), 269 264 254 218 bytes

• Thanks to Fatalize for saving 37 bytes!
• Thanks to Emigna for saving 9 bytes!

I'm fairly certain that I could golf some more bytes down.

X/X:-X<3.
N/R:-A is N-1,B is N-2,A/C,B/D,R is C+D.
2^0^0.
C^S^E:-G is C-1,G/M,G^Q^R,(p(M)->S=M+Q,E is R+1;S=Q,E is R).
a(X,S):-X^R^Q,S is R/Q.
p(N):-N*(N-1).
p(2).
p(3).
N*2:-N mod 2=\=0.
N*C:-N mod C=\=0,D is C-1,N*D.

Run it as a(15, R). for x = 15, R is the output variable.

Try it online!

---

A more readable version:

fibonacci(1, 1) :- !.
fibonacci(2, 2) :- !.

fibonacci(N, R) :-
  N0 is N - 1,
  N1 is N - 2,
  fibonacci(N0, R1),
  fibonacci(N1, R2),
  R is R1 + R2.

listed(2, 0, 0) :- !.

listed(C, S, Successes) :-
  C0 is C - 1,
  fibonacci(C0, Result),
  listed(C0, S0, S1),
  (
    is_prime(Result)
    ->
      S = Result + S0, Successes is S1 + 1
    ; S = S0, Successes is S1
  ).

f(X, S) :-
  listed(X, R, Q),
  S is R / Q.

is_prime(Num) :- is_prime(Num, Num - 1).

is_prime(2).
is_prime(3).

is_prime(Num, 2) :- Num mod 2 =\= 0, !.

is_prime(Num, C) :-
  Num mod C =\= 0,
  C0 is C - 1,
  is_prime(Num, C0).

Answer 10

Röda, 98 94 93 bytes

f n{z=0;i=1;j=1;s=0;seq 2,n-1|{|_|c=i+j;i=j;j=c;seq 2,c-1|{z+=1;s+=c}if[c%p>0]()for p}_[s/z]}

It's a function that returns the result as a floating point number.

Ungolfed version:

function f(n) {
    i = 1
    j = 1
    sum = 0
    num = 0
    seq(2, n-1) | for _ do
        /* calculate next fibonacci number: */
        c = i+j
        i = j
        j = c
        /* if it's prime: */
        {
            num += 1
            sum += c
        /* primality test: */
        } if [c%p > 0]() for p in [seq(2, c-1)]
    done
    return sum/num
}

Answer 11

05AB1E, 13 bytes

!ÅF¹£DpÏDOsg/

Try it online! or as a Test suite

Explanation

!ÅF             # get a list of fibonacci numbers up to fac(input)
   ¹£           # take the first input elements of that list
     D          # duplicate
      p         # isprime on the copy
       Ï        # keep the fibonacci numbers which are true in the isprime list
        D       # duplicate the list of fibonacci primes
         O      # sum the list
          s     # swap the other copy to the top of the stack
           g    # get its length
            /   # divide sum by length

Answer 12

Pyke, 11 bytes

S.b#_P)'sl/

Try it online!

Answer 13

dc, 129 bytes

?sa0sb1sd0ss0sz[[lF1+sF]sqlelG!=q]si[ls+sslz1+sz]sw[lddlb+dsdrsbdsG2se0sF[lGdle%0=ile1+dse<p]dspxlF0=wla1-dsa1<c]dscxEkls1-lz1-/p

A Fibonacci number generator and primality checker all in one. Nice.

Try it online!

Answer 14

Japt, 14 bytes

ò@MgXÃfj
x /Ul

Test it

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Explanation

Implicit input of integer U.
30

ò

Generate an array of integers from 0 to U inclusive.
[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30]

@   Ã

Pass each integer through a function.

MgX

Get the Xth Fibonacci number, where X is the current element.

[0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040]

fj

Filter (f) the array to those elements that return truthy when checked for primality (j). Implicitly assign the resulting array to variable U.
[2,3,5,13,89,233,1597,28657,514229]

x

Reduce the array by summing.
544828

/Ul

Divide the result by the array's length (l) and implicitly output the result.
60536.444444444445

Answer 15

perl, 91 bytes

$b=1;map{($a,$b)=($b,$a+$b),$p=("x"x$b)!~/^(.|(..+)\2+)$/,$s+=$b*$p,$c+=$p}2..$x;print$s/$c

Cuold have been 8 bytes shorter if the modulo pseudoprime test worked as well in perl as in the python answer:

$b=1;$s=2;map{($a,$b)=($b,$a+$b),$p=2**$b%$b==2,$s+=$b*$p,$c+=$p}2..$x;print$s/++$c

...but this gives wrong answer for input > 16 in perl.

Answer 16

Axiom, 104 bytes

g(n)==(y:=x:=r:=0;repeat(x>n=>break;j:=fibonacci(x);x:=x+1;if prime?(j)then(r:=r+j;y:=y+1));y=0=>-1;r/y)

ungolfed, test code and results

f(n)==
  y:=x:=r:=0
  repeat
     x>n=>break
     j:=fibonacci(x)
     x:=x+1
     if prime?(j) then(r:=r+j;y:=y+1)
  y=0=>-1
  r/y

(3) -> [[i,g(i)::Float] for i in [1,2,3,10,15,20,11,30,50]]
   Compiling function g with type PositiveInteger -> Fraction Integer
   (3)
   [[1.0,- 1.0], [2.0,- 1.0], [3.0,2.0], [10.0,5.75], [15.0,57.5],
    [20.0,277.4285714285 7142857], [11.0,22.4], [30.0,60536.4444444444 44444],
    [50.0,309568576.1818181818 2]]

i try to duplicate the matematica, octave etc languages entry, if one not count the mean() function, to implement it would be here 62 bytes so good too

mean(a:List Float):Any== 
    i:=1; s:=0.0
    repeat  
       if~index?(i,a)then break
       s:=s+a.i
       i:=i+1
    i=1=>[]
    s/(i-1)

--62 bytes
f(x:NNI):Any==mean(select(prime?,[fibonacci(i)for i in 1..x]))

Answer 17

JavaScript ES6, 137 136 118 113 bytes

m=

n=>(a=[],(f=x=>a[x]=x<2?x:f(--x)+f(--x))(n),eval((a=a.filter(x=>eval("for(y=x;x%--y;);y==1"))).join`+`)/a.length)

console.log(m(10))
console.log(m(15))
console.log(m(20))
console.log(m(11))
console.log(m(30))

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History

118 bytes

n=>(a=[0,1,1],(f=x=>a[--x]=a[x]||f(x)+f(--x))(n),eval((a=a.filter(x=>eval("for(y=x;x%--y;);y==1"))).join`+`)/a.length)

136 bytes

n=>(a=[0,1,1],(f=x=>a[--x]=a[x]||f(x)+f(--x))(n),eval((a=a.filter(x=>x>1&&!Array(x).fill().some((i,j)=>j>1&&!(x%j)))).join`+`)/a.length)

137 bytes

n=>(a=[0,1,1],(f=x=>a[--x]=a[x]||f(x)+f(--x))(n),eval((a=a.filter(x=>{d=x;while(--d>1)if(!(x%d))return 0;return x>1})).join`+`)/a.length)