In a [previous challenge](/learn/javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion), you learned how to use recursion to replace a `for` loop. Now, let's look at a more complex function that returns an array of consecutive integers starting with `1` through the number passed to the function.
As mentioned in the previous challenge, there will be a <dfn>base case</dfn>. The base case tells the recursive function when it no longer needs to call itself. It is a simple case where the return value is already known. There will also be a <dfn>recursive call</dfn> which executes the original function with different arguments. If the function is written correctly, eventually the base case will be reached.
For example, say you want to write a recursive function that returns an array containing the numbers `1` through `n`. This function will need to accept an argument, `n`, representing the final number. Then it will need to call itself with progressively smaller values of `n` until it reaches `1`. You could write the function as follows:
At first, this seems counterintuitive since the value of `n`*decreases*, but the values in the final array are *increasing*. This happens because the push happens last, after the recursive call has returned. At the point where `n` is pushed into the array, `countup(n - 1)` has already been evaluated and returned `[1, 2, ..., n - 1]`.
We have defined a function called `countdown` with one parameter (`n`). The function should use recursion to return an array containing the integers `n` through `1` based on the `n` parameter. If the function is called with a number less than 1, the function should return an empty array. For example, calling this function with `n = 5` should return the array `[5, 4, 3, 2, 1]`. Your function must use recursion by calling itself and must not use loops of any kind.