i was researching the numbers that are equal to the sum of their digits raised to the third power . 


like $153=1^3 + 5^3+3^3$ and i have 3 questions.


1) from any starting number $n$ ,do we always arrive to the sequence 0,1,153,370,371,407 or some finite cycles like the cycle $55 -> 250-> 133 ->55$
,or there is infinitely many cycles?


2) is there a method to construct such numbers for any given power $k$  

for example when k=3 => we get 153 or any other number that fulfill the condition and when k=4 => we get 1634 or any other number that fulfill the condition and so on ... ?


3) is there any power $k$ such that no number fulfill the condition that the sum of it's digits raised to the power $k$ equal to the initial number, or it does not have any simple cycles ???!!!! (i think this is very hard to answer)



the math behind the sequence 0,1,153,370,371,407
For given $k$, once you establish that $f(x) < x$ for $x > N$, you compute the fate of each $x \in [0,1,\ldots,N]$ as follows:


Start with some $x\_0$ whose fate is not yet known, and compute the values 
$x\_0, x\_1 = f(x\_0), x\_2 = f(x\_1), \ldots $ until either:


1. the fate of $x\_m$ is known, in which case $x\_0, x\_1, \ldots, x\_{m-1}$ all get the same fate, or
2. $x\_m$ has already appeared in the list $x\_0, x\_1, \ldots, x\_{m-1}$, say as $x\_k$, in which case the fate each of $x\_0, \ldots, x\_{m-1}$ is the cycle $[x\_k, x\_{k+1}, \ldots, x\_{m-1}]$.


When the fates of all $x\_i$ are known, you have enumerated all the cycles.


