Discussion on: Challenge: find 'Kaprekar numbers'

[Josha von Gizycki]

Taking inspiration from Thomas Much's answer, I refactored my code to be more functional-esque and idiomatic:

(ns kaprekar)

(defn is-kap? [n]
  (let [square (* n n)
        ss (str square)
        length (count ss)
        ss (if (odd? length) (str "0" ss) ss)
        length (count ss)
        half-length (/ length 2)
        first-half (bigint (subs ss 0 half-length))
        second-half (bigint (subs ss half-length))]
    (= n (+ first-half second-half))))

(defn find-kaps []
  (->>
    (iterate inc 1)
    (filter is-kap?)
    (take 8)
    (run! println)))