Clojure：列表中的复杂迭代？

user=> (func 20 '(1 2 3 4 5 6 7 8 9 10))
'(:0 1,:0 2,:2 3,... :20 0)

;; My idea on the best way to find a square root is simple.
;; If I want to find the square root of n,divide n in half
;; Then divide our initial number (n) by all numbers in the range 0...n/2 
;; Separate out a list of results that only only return a remainder of 0.
;; Then test the results in a comparison to see if the elements of our returned 
;; list when squared are equal with the number we want to find a square root of.
;; First I'll develop a function that works with evens and then odds

(defn sqroot-range-high-end [input] (/ input 2))
(sqroot-range-high-end 36) ; 18

(defn make-sqrt-range [input] (range (sqroot-range-high-end (+ 1 input))))
(make-sqrt-range 36) ; '(0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)

(defn zero-culler [input] (lazy-seq (remove zero? (make-sqrt-range input))))
(zero-culler 100) ; '(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)

(defn odd-culler [input] (lazy-seq (remove odd? (zero-culler input))))
(odd-culler 100) ; '(2 4 6 8 10 12 14 16 18)

;;the following is where I got stuck
;;I'm new to clojure and programming,;;and am just trying to learn in a way that I understand

(defn remainder-culler [input]
  (if
    (/ input (first odd-culler (input)))
  input)
  (recur (lazy-seq (input)))
)

(remainder-culler 100)