Choose some natural number and take $k$-th power of its digits and add that. Repeat that. How many cycles will be there? I have read in a book that has very much of a recreational flavour that if we take any natural number and square its digits and add that then if we repeat that long enough that only two outcomes can happen:
a) Either we stay forever in a trivial cycle $1$
or
b) We stay forever in a cycle $145, 42, 20, 4, 16, 37, 58, 89$.
But I do not now how a general situation looks like.
Do we have a formula that tells us how many exactly cycles do we have if we take $k$-th power, that is, how to calculate number of possible cycles as a function of $k$?
If we denote that function by $C(k)$ is it known is $C$ a bounded function? (In other words, is a set $\{C(1),C(2),...C(k),...\}$ bounded?)
 A: This is not at all an answer. First, the OP asks some interesting questions:
1.Can we compute the number of cycles for a given $k?$ 
 2. Can we have bounds for $C(K)?$
The a priori upper bound blows up very badly, so it is quite possible that the first question is computationally very hard ($\#P$ complete?)
For the second question, I don't even see an obvious lower bound. The upper bound can be gotten in the obvious way by seeing when the sum of the $k$-th powers is smaller than the number itself (but this will be doubly exponential in $k$ - is that the truth)?
A: Let $f(n)$ be the sum of the $k$'th powers of the digits of $n$.
If $d \in \{0,\ldots, 9\}$, we have $d^k \le 9^{k-1} d$.  Let $m$ be the least positive integer such that $10^m > 9^{k-1}$.  Then if
$d_r \ldots d_0$ is the decimal representation of $n$, we have
$d_j^k \le \dfrac{9^{k-1}}{10^m} d_j 10^j$ for $j \ge m$, and
$$ \eqalign{f(n) &\le \sum_{j < m} d_j^k + \frac{9^{k-1}}{10^m}  \sum_{j \ge m} d_j 10^j \cr 
&= \sum_{j<m} \left(d_j^k - \frac{9^{k-1}}{10^m} d_j 10^j \right) + \frac{9^{k-1}}{10^m} n\cr
&\le \sum_{j<m} 9^k (1-10^{j-m}) + \frac{9^{k-1}}{10^m} n\cr
&\le 9^k m + \frac{9^{k-1}}{10^m} n }$$
and so in particular $f(n) \le n$ if $n \ge \dfrac{9^k 10^m m}{10^m - 9^{k-1}}$.  This gives an a priori bound for all cycles.
Thus for $k=3$, the bound is $7673$, and I get the following cycles:
$$[1], [153], [370], [371], [407], [136, 244], [919, 1459],  [55, 250, 133],  [160, 217, 352]$$
