Fibonacci Code Golf

Answer 1

Microsoft Batch - 15 characters

Old challenge, but the world must know it is possible:

%1
%0 %1%2 %1 #

Output is to stderr in unary, counting only the # characters. Depending on the host system's space restrictions, it may produce only the first 14 numbers or so.

Answer 2

Language: dc, Char count: 20

Shorter dc solution.

dc -e'1df[dsa+plarlbx]dsbx'

Answer 3

13 chars of Golfscript:

2,~{..p@+.}do

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Update to explain the operation of the script:

1. 2, makes an array of [0 1]
2. ~ puts that array on the stack
3. So, at the time we run the do, we start the stack off with 0 1 (1 at top of stack)

The do loop:

1. Each . duplicates the top item of the stack; here, we do this twice (leaving us with 0 1 1 1 on initial run)
2. p prints the topmost value (leaving us with 0 1 1)
3. @ rotates the top 3 items in the stack, so that the third-topmost is at the top (1 1 0)
4. + adds the top 2 items in the stack (leaving 1 1)
5. . duplicates the top value, so that the do loop can check its truthiness (to determine whether to continue)

Tracing this mentally a couple of loops will be enough to tell you that this does the required addition to generate the Fibonacci sequence values.

Since GolfScript has bignums, there will never be an integer overflow, and so the top-of-stack value at the end of the do loop will never be 0. Thus, the script will run forever.

Answer 4

BrainF**k:

>+++++>+>+<[[>]<<[>>+>+<<<-]>>>[<<<+>>>-]<<[>+>+<<-]>>[<<+>>-]<[<]>-]

That'll generate the first 5. To generate more, replace the 5 + at the beginning with more: eg:

>++++++++++++++++++++++>+>+<[[>]<<[>>+>+<<<-]>>>[<<<+>>>-]<<[>+>+<<-]>>[<<+>>-]<[<]>-]

Answer 5

Not the shortest, but the fastest at the time of posting. :-)

float f(float n) {
    return (pow(1+sqrt(5.0))/2.0),n) - pow(1+sqrt(5.0))/2.0),n)/sqrt(n));
}

Answer 6

Here's my best using scheme, in 45 characters:

(let f((a 0)(b 1))(printf"~a,"b)(f b(+ a b)))