Print the first n numbers of the fibonacci sequence in one expression

Question

So I've been messing around with Python a bit lately and I'm trying to find a way to output the nth number of the fibonacci sequence in a single expression. This is the code that I've written so far:

(lambda f: f if f<2 else (f-1)+(f-2))(n)
# n == 1 -> 1
# n == 2 -> 1
# n == 3 -> 3
# n == 4 -> 5
# n == 5 -> 7
....

However, as I commented above this simply outputs a set of odd numbers. I'm confused as to why this is happening because if I am to re-write this as a named lambda function, it would look something like this:

f = lambda n: n if n<2 else f(f-1)+f(f-2)
# f(1) -> 1
# f(2) -> 1
# f(3) -> 2
# f(4) -> 3
...
# f(10) -> 55
...

Now the reason I've added the Lambda Calculus tag is because I'm not sure if this question falls under the domain of simply understanding how Python handles this. I've read a tiny bit about the Y combinator in lambda calculus, but that's a foreign language to me and couldn't derive anything from resources I found for this about lambda calculus.

Now, the reason I'm trying to do this in one line of code, as opposed to naming it, is because I want to try and put this lambda function into list comprehension. So do something like this:

[(lambda f: f if f<2 else (f-1)+(f-2))(n) for n in range(10)]

and create an array of the first x numbers in the fibonacci sequence.

What I'm looking for is a method of doing this whole thing in one expression, and should this fall under the domain of Lambda calculus, which I believe it does, for someone to explain how this would work.

Feel free to offer an answer in JavaScript, C#, or other C-like languages that support Lambda functions.

EDIT: I've found the solution to what I was attempting to do:

[(lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v))))(lambda f:(lambda n: n if n<2 else f(n-1)+f(n-2)))(y) for y in range(10)]

I know that this is not at all practical and this method should never be used, but I was concerned with CAN I do this as opposed to SHOULD I ever do this.

Answer 1

You'll need to assign your lambda to an actual variable, and then call the lambda inside the lambda:

>>> g = lambda f: f if f < 2 else g(f-1)+g(f-2)
>>> [g(n) for n in range(10)]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]