Numbers that are the sums of powers of their digits (eg, $1^3+5^3+3^3=153$). What is known about them? I'm no mathematician, but there is interest. Today I coincidentally found out something about the number $153{:}$
$$1^3 + 5^3+3^3 = 153$$
So if we exponentiate each digit of the number 153 (1, 5, 3) with 3, its sum is the same origin number: 153.
THAT’S INTERESTING, I thought. That’s why I wrote a little script to determine other numbers like 153. Surprise: There are some, but not many. Here are the results:
Exponent: 3 — Number: 153
 Exponent: 3 — Number: 370
 Exponent: 3 — Number: 371
 Exponent: 3 — Number: 407
 Exponent: 4 — Number: 1634
 Exponent: 4 — Number: 8208
 Exponent: 4 — Number: 9474
 Exponent: 5 — Number: 4150
 Exponent: 5 — Number: 4151
 Exponent: 5 — Number: 54748
 Exponent: 5 — Number: 92727
 Exponent: 5 — Number: 93084
Here you’ll find the javascript script (JSFiddle link).

So … what can we learn out of this?
Is it something already known?
Does it mean something?

Any ideas?
 A: You verified that 153 is narcissistic_number like $370=3^3+7^3+0^3$ and $1634=1^4+6^4+3^4+4^4$. More examples are given at A005188.
A: This is called a narcissistic number, sometimes referred to as an Armstrong number. By definition, a narcissistic number is an n-digit number that is the sum of the nth powers of its digits. 
Other examples are $407 = 4^3 + 0^3 + 7^3$ or $371 = 3^3 + 7^3 + 1^3$.
See http://mathworld.wolfram.com/NarcissisticNumber.html
A: Very interesting question. We could choose a power $l$ and ask which numbers are the sum of the $l^{th}$ powers of its digits (for simplicity lets keep things in base $b=10$).
Since each digit is at most $9$, the sum of these powers is at most $n9^l$, where $n$ is the number of digits. On the other hand, a number with $n$ digits is at least $10^{n-1}$. So, if there exists such a n-digit number, we must have 
$$10^{n-1}\leq 9^ln.$$
We conclude that, for a fixed $l$, there can only be a finite number of such numbers.
For $l=3$, we must have $10^{n-1}\leq 729n$, which is true only for $n\leq 4$. So a number with more than $4$ digits cannot be the sum of the third powers of its digits.
The narcisistic numbers correspond to those $n$-digit numbers where we can take $l=n$. 
