There's no reason to recur at all:

(defn factorial [n] (reduce * (range 1 (inc n))))

On Thu, Jul 15, 2010 at 9:49 AM, Frederick Polgardy <f...@polgardy.com>wrote:

> You don't need to recur to another function, just recur to a loop:
>
> (defn factorial [n]
>  (loop [x n acc 1]
>    (if (zero? x) acc (recur (dec x) (* acc x)))))
>
> (println (factorial 5))
>
> --
> Science answers questions; philosophy questions answers.
>
> On Jul 15, 2010, at 6:38 AM, Brisance wrote:
>
> > Here's a factorial function as found in the WikiBook "Learning
> > Clojure" <http://en.wikibooks.org/wiki/Learning_Clojure>:
> >
> > (defn factorial [n]
> >  (defn fac [n acc]
> >    (if (zero? n)
> >       acc
> >      (recur (- n 1) (* acc n)))) ; recursive call to fac, but reuses
> > the stack; n will be (- n 1), and acc will be (* acc n)
> >  (fac n 1))
> >
> > Question: how would I go about writing idiomatic Clojure to return
> > factorials of n, for large values of n. e.g. 1e6 or more? Preferably
> > without having to create another function.
> >
> > Thanks in advance for any insight.
> >
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