下面是我最新的算法，能迅速列出f(n)=n的全部解
求函数值f(n)的思路和shhgs的那个一样
主程序递增n的算法既保证不漏一个又很快，得意之作呵呵

def f(n):
    """f(n) = Sum{g(k)}, k=1..n
    where g(k) = number of '1's in decimal presentation of k
    """
    m = 1; r = 0
    # In the i-th loop, we sum No. of '1' showed in the i-th position
    # (start from the RIGHT) of all k's decimal presentation, k=1..n.
    # m == 10**(i-1), i.e. the order of i-th digit.
    while m <= n:
        head = n / (m *10) * (m * 10)   # digits to the left of digit_i
        digit_i = n / m % 10            # the i-th digit
        tail = n % m                    # digits to the right of digit_i
        # Example: for n=23456 and m=100, head=23000,digit_i=4,tail=56

        # This many '1's showed on the i-th position of k, for k=1..head-1.
        r += head / 10
        if digit_i == 1:
            r += tail + 1   # No. of '1's showed on i-th position for
k=head..n
        elif digit_i > 1:
            r += m          # same as above, but diff. calculation
        # Select the (i+1)-th position.
        m *= 10

    return r

#import pdb
#pdb.set_trace()

#look for some n2 satisfying f(n2) >= n2
n = 1; loops = 0
while n < 10**100:
    loops += 1
    F = f(n)
    if F == n:
        print "f(" + str(n) + ") = " + str(F) + ", loop " + str(loops)
        n += 1
    elif F < n:
        # find the order of n
        m = 10; d = 1   # m = 10**d holds
        while m <= n:
            m *= 10; d += 1
        # now d == No. of digits in str(n)

        # this jump will never be "too long", since there is no more than
        # n-F '1's between n+1 and n+jump. (f(n+jump) equals n+jump only
when
        # each number between n+1 and n+jump has d '1's
        jump = (n - F) / d
        if jump > 0:
            n += jump
        else:
            n += 1
    else:   # F > n
        # this is equivalent to new_n = n+(F-n),
        # and f(new_n) == f(n+(F-n)) >= f(n) == F == n+(F-n) == new_n
        n = F
    #print "not satisfied: f(" + str(n) + ") = " + str(F)

-- 
   regards,
yzhh