Help me make this Clojure code about as fast as Java, please!

t.clj

(defn f [x y z n]
  (+ (* 2 (+ (* x y) (+ (* y z) (* x z))))
     (* 4 (+ x y z n -2) (- n 1))))

(defmacro from [[var ini cnd] & body]
  `(loop [~var ~ini]
     (when ~cnd
       ~@body
       (recur (inc ~var)))))

(defn g [n]
  (let [c (long-array (inc n))]
    (from [x 1 (<= (f x x x 1) n)]
      (from [y x (<= (f x y y 1) n)]
        (from [z y (<= (f x y z 1) n)]
          (from [k 1 (<= (f x y z k) n)]
            (let [l (f x y z k)]
              (aset c l (inc (aget c l))))))))
    c))

(defn h [x]
  (loop [n 1000]
    (let [^longs c (g n)]
      (if-let [k (some #(when (= x (aget c %)) %)
                       (range 1 (inc n)))]
        k
        (recur (* 2 n))))))

(time (println (h 1000)))

T.java

public class T {
  static int f(int x, int y, int z, int n) {
    return 2*(x*y+y*z+x*z) + 4*(x+y+z+n-2)*(n-1);
  }

  static int[] g(int n) {
    int[] count = new int[n+1];
    for (int x=1; f(x,x,x,1)<=n; x++)
      for (int y=x; f(x,y,y,1)<=n; y++)
        for (int z=y; f(x,y,z,1)<=n; z++)
          for (int k=1; f(x,y,z,k)<=n; k++)
            count[f(x,y,z,k)]++;
    return count;
  }

  static int h(int x) {
    for (int n=1000; true; n*=2) {
      int[] v = g(n);
      for (int i=1; i<=n; i++)
        if (v[i]==x)
          return i;
    }
  }

  public static void main(String args[]) {
    long t0 = System.currentTimeMillis();
    System.out.println(h(1000));
    long t1 = System.currentTimeMillis();
    System.out.println(t1-t0);

  }
}