Common Lisp: How to return a list without the nth element of a given list?

A simple recursive solution:

(defun remove-nth (n list)
  (declare
    (type (integer 0) n)
    (type list list))
  (if (or (zerop n) (null list))
    (cdr list)
    (cons (car list) (remove-nth (1- n) (cdr list)))))

This will share the common tail, except in the case where the list has n or more elements, in which case it returns a new list with the same elements as the provided one.

Using remove-if:

(defun foo (n list)
  (remove-if (constantly t) list :start (1- n) :count 1))

butlast/nthcdr solution (corrected):

(defun foo (n list)
  (append (butlast list (1+ (- (length list) n))) (nthcdr n list)))

Or, maybe more readable:

(defun foo (n list)
  (append (subseq list 0 (1- n)) (nthcdr n list)))

Using loop:

(defun foo (n list)
  (loop for elt in list
        for i from 1
        unless (= i n) collect elt))

Here's an interesting approach. It replaces the nth element of a list with a new symbol and then removes that symbol from the list. I haven't considered how (in)efficient it is though!

(defun remove-nth (n list)
    (remove (setf (nth n list) (gensym)) list))