I looked through some of my Project Euler solutions and found this
version. It is essentially the same as Uncle Bob's, but to my eye it
is a bit easier to read.
(defn least-nontrivial-divisor [n] ;; integer n > 1
(loop [k 2]
(cond
(zero? (rem n k)) k ;; k divides n, return k
(> (* k k) n ) n ;; k > sqrt n, return n
:else (recur (inc k)))))
(defn prime-factors [n] ;; integer n > 1
(loop [n n factors []]
(if (= 1 n)
factors
(let [d (least-nontrivial-divisor n)]
(recur (quot n d)
(conj factors d))))))
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