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programmingpraxis - Scheme, pasted on Sep 18:
; kaprekar numbers

(define-syntax fold-of
  (syntax-rules (range in is)
    ((_ "z" f b e) (set! b (f b e)))
    ((_ "z" f b e (v range fst pst stp) c ...)
      (let* ((x fst) (p pst) (s stp)
             (le? (if (positive? s) <= >=)))
        (do ((v x (+ v s))) ((le? p v) b)
          (fold-of "z" f b e c ...))))
    ((_ "z" f b e (v range fst pst) c ...)
      (let* ((x fst) (p pst) (s (if (< x p) 1 -1)))
        (fold-of "z" f b e (v range x p s) c ...)))
    ((_ "z" f b e (v range pst) c ...)
      (fold-of "z" f b e (v range 0 pst) c ...))
    ((_ "z" f b e (x in xs) c ...)
      (do ((t xs (cdr t))) ((null? t) b)
        (let ((x (car t)))
          (fold-of "z" f b e c ...))))
    ((_ "z" f b e (x is y) c ...)
      (let ((x y)) (fold-of "z" f b e c ...)))
    ((_ "z" f b e p? c ...)
      (if p? (fold-of "z" f b e c ...)))
    ((_ f i e c ...)
      (let ((b i)) (fold-of "z" f b e c ...)))))

(define-syntax list-of (syntax-rules ()
  ((_ arg ...) (reverse (fold-of
    (lambda (d a) (cons a d)) '() arg ...)))))

(define (ilog b n)
  (let loop1 ((lo 0) (b^lo 1) (hi 1) (b^hi b))
    (if (< b^hi n) (loop1 hi b^hi (* hi 2) (* b^hi b^hi))
      (let loop2 ((lo lo) (b^lo b^lo) (hi hi) (b^hi b^hi))
        (if (<= (- hi lo) 1) (if (= b^hi n) hi lo)
          (let* ((mid (quotient (+ lo hi) 2))
                 (b^mid (* b^lo (expt b (- mid lo)))))
            (cond ((< n b^mid) (loop2 lo b^lo mid b^mid))
                  ((< b^mid n) (loop2 mid b^mid hi b^hi))
                  (else mid))))))))

(define (kaprekar? k)
  (let* ((k2 (* k k))
         (n (+ (quotient (ilog 10 k2) 2) 1))
         (p (quotient k2 (expt 10 n)))
         (q (modulo k2 (expt 10 n))))
    (= (+ p q) k)))

(display (list-of k (k range 1 1000) (kaprekar? k)))


Output:
1
(1 9 45 55 99 297 703 999)


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