How to calculate first 100 Fibonacci numbers?

clear
n(1) = 1;
n(2) = 1;
k = 3;
while k <= 100
    n(k) = [n(k-2) + n(k-1)];
    k = k + 1;
end

How do you get an answer that isn't this?

n =

     1.0e+20 *

    Columns 1 through 9

      0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000

    Columns 10 through 18

      0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000

    Columns 19 through 27

      0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000

    Columns 28 through 36

      0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000

    Columns 37 through 45

      0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000

    Columns 46 through 54

      0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000

    Columns 55 through 63

      0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000

    Columns 64 through 72

      0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000

    Columns 73 through 81

      0.0000    0.0000    0.0000    0.0000    0.0001    0.0001    0.0001    0.0002    0.0004

    Columns 82 through 90

      0.0006    0.0010    0.0016    0.0026    0.0042    0.0068    0.0110    0.0178    0.0288

    Columns 91 through 99

      0.0466    0.0754    0.1220    0.1974    0.3194    0.5168    0.8362    1.3530    2.1892

    Column 100

      3.5422

Answer 1

[Image Analyst]

Try removing the semicolon from this line

n(k) = [n(k-2) + n(k-1)];

Or display n with fprintf():

fprintf('%g, ', n);

Or put this at the beginning of your code

format compact
format short g

Answer 2

[Walter Roberson]

Giving the command

format long g

would help. You would still run into format problems, starting at about #74, as the numbers start to exceed the available resolution of double precision numbers.

Changing the first line to

n(1) = uint64(1);

also helps. You will notice, though, that the last 7 entries are all the same, 18446744073709551615, which is the maximum number that can be expressed as uint64.

Changing the first line to

n(1) = sym(1);

solves the problem, if you have the Symbolic Toolbox

Answer 3

[Maruf Ahamed]

 % code
  you can try this:
clc
clear all
n=input('n=')
j=0;
c=1;
h=2;
d(1)=1;
for i=1:n
   k=c+j;
   d(h)=k;
   h=h+1;
   j=k;
   c=c+j;
   d(h)=c;
   h=h+1;
end
 d  

[Walter Roberson]

This does not appear to solve the issue of limited precision of double() variables that will start happening at about 74.