Sequence of Dudeney Numbers

Theorem
The only positive integers which are equal to the sum of the digits of their cube are:
 * $0, 1, 8, 17, 18, 26, 27$

two of which are themselves cubes, and one of which is prime.

Proof
We have trivially that:

Then:

A quick empirical test shows that when $n = 46$, it is already too large to be the sum of the digits of its cube.

For $46 < n \le 54$, $n^3 \le 54^3 < 200 \, 000$.

Hence the sum of the digits of $n^3$ is less than:
 * $1 + 5 \times 9 = 46 < n$

For $54 < n < 100$, $n^3 < 10^6$.

Hence the sum of the digits of $n^3$ is less than:
 * $6 \times 9 = 54 < n$

For $n \ge 100$, let $n$ be a $d$-digit number, where $d \ge 3$.

Then $10^{d - 1} \le n < 10^d$ and $n^3 < 10^{3 d}$.

Hence the sum of the digits of $n^3$ is less than:

so no numbers greater than $46$ can have this property.

Also reported as
Some sources (either deliberately or by oversight) do not include $0$ in this list.

Also see

 * Definition:Armstrong Number, with which the numbers in this entry appear frequently to be conflated