Recursive Fibonacci printing

Question

I’m having problems with the recursive function Fibonacci, in the exercise you ask to print inside the function or even using an auxiliary recursive function, but you can’t print in the main function, I tried everything to print, but they keep repeating, the recursive function Fibonacci I know how to do, just missing the impression, follows the enunciation and ask someone to give me a light.

"The Fibonacci sequence is defined by:

• F0 = 0
• F1 = 1
• Fn = F_{n - 1} + F_{n - 2}, for n > 1

Develop a recursive function that calculates and prints the first n Fibonacci sequence numbers. The function must have the following signature:

int fibonacci (int n);

The function shall print the first n numbers separated by spaces, by example, for n = 10:

0 1 1 2 3 5 8 13 21 34"

Answer 1

Remember what it is about the function it is the sum of the predecessor by the previous predecessor , it begins with two terms the 0 and the 1.

• fórmula : 0, 1,(1+0),((1+0)+1),(((1+0)+1)+(1+0))....

transforming this expression into a function we can arrive at Fibro(n - 1) + Fibro(n - 2) where n is the number of terms considering , of course , that it is being done in a decreasing way (so that need to pass only one parameter value).

recursively made each sum of your terms will be stored in a stack which after arriving at the base condition will be terminated and will have your terms unsealed.

Obs: I leave an example in C# the reasoning is the same

  static void Main(string[] args)
    {
        int n=0;
        Console.WriteLine("Digite o tamanho da sequencia");
        n = int.Parse(Console.ReadLine());

        Console.WriteLine("resultado");
        for (int c=0; c<=n; c++) {
          Console.Write(" "+Fibro(c + 1));//imprimi aqui mas você pode imprimir dentro da recursão mas vai perder o objetivo da recursão de mostrar uma pilha 
        }
        Console.ReadKey();

    }

    public static int Fibro(int n){
        if (n == 1 || n == 0)
        {
            return 1;
        }
        else {

            return Fibro(n - 1) + Fibro(n - 2);//recursão 

        }

     }

Answer 2

If only the printing is missing, each time you calculate the value of the number is enough

printf("Fibonacci é: %d", fibonacci);

Where %d is just a "Holder" for your number, which is passed in the second parameter.

Answer 3

This is a version of the Fibonacci algorithm using an auxiliary vector, which enables the calculation of values up to 50, instantaneously. For larger values, it will be necessary to use long int, long int or others.

#include <stdio.h> 
#include <stdlib.h> 

void calculaFibonnaci();
int fibonnaci();

int vetor[1000];

void calculaFibonnaci(int i) { 

    if(i==1 || i==2) {
        vetor[i] = 1;
    }
    else { 
        vetor[i] = fibonnaci(i-1) + fibonnaci(i-2);
    }
}

int fibonnaci(int i)
{
    if(vetor[i]==0) {
        calculaFibonnaci(i);
        printf("%d ", vetor[i]);
    }
    else {
        return vetor[i];
    }   
} 

int main(int argc, char* argv[]) 
{
    int i, N;
    int *vetor;
    printf("Digite quantos termos da série de fibonacci você deseja: ");
    scanf("%d", &N);

    for(i=1; i<=N; i++)
        fibonnaci(i);

    return 0;
} 

Answer 4

#include<stdio. h> #include<stdlib. h> #include<string. h>

int Fib(int n){

if(n == 0){
    
    return 0;
    
}
if(n < 3){
    
    return 1;   
    
}else{
    
    return fib(n-1) + fib(n-2);

}

}

int main(void){

int n;
int r;
int i;

scanf("%d", &n);

for(i = 0; i <= n; i++){
    
    r = fib(i);
    
    printf("%d ", r);       
}



return 0;

}

Answer 5

Fibonacci series in C:

#include <stdio.h>
#include <stdlib.h>

unsigned long fibonacci();
unsigned long v[sizeof(int)] ;

unsigned long fibonacci(int n) {
   if (n <= 0)
      return -1;
   if((n <= 2)  && v[n] == 0) {
      v[n] = 1;
      printf("%lu ", v[n]);
   }
   else if (v[n] == 0) {
      v[n] = fibonacci(n-2) + fibonacci(n-1);
      printf("%lu ", v[n]);
   }

   return v[n];
}

int main(int argc, char* argv[]) {
   int in;
   printf("Entre com o tamanho da série: ");
   scanf("%d", &in);
   fibonacci(in);
   return 0;
}

Fibonacci series in Python:

d = {}
def fib(n):
  if (n < 2):
    d[n] = n
    return d[n]
  if (d.keys().__contains__(n)):
    return d[n]    
  d[n] = fib(n - 2) + fib(n - 1)    
  return d[n]  

for i in range(0,101):
  print('fib(',i,'): ', fib(i))