3
$\begingroup$

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)

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ You are asking about the recursion $x_{n+1}=f(x_n)$, where $x_{n+1}$ is the sum of the digits of $x_n$ raised to the power $k$. It is easy to see that $x_{n+1}<x_n$ for large $x_n$. From this it follows that the iteration must always terminate in a periodic cycle, and there can only be a finite number of such cycles. $\endgroup$
    – Michael Renardy
    Nov 3, 2016 at 1:02
  • 1
    $\begingroup$ OEIS sequence A046197. $\endgroup$
    – Robert Israel
    Nov 3, 2016 at 1:02
  • 4
    $\begingroup$ Hardy wrote, "There are just four numbers (after 1) which are the sums of the cubes of their digits, viz. 153 .... These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals much to a mathematician....it is plain that one reason is the extreme speciality of both the enunciations and the proofs, which are not capable of any significant generalization." Page 25 of math.ualberta.ca/mss/misc/A%20Mathematician's%20Apology.pdf $\endgroup$
    – Gerry Myerson
    Nov 3, 2016 at 1:14
  • 1
    $\begingroup$ Nitpickery, but important pedantry: the values you give are more properly considered a set rather than a sequence; there's not really any sequence generation or ordering to them other than just the usual linear ordering on numbers. This is especially important here since you do consider sequences gotten by repeating the sum-of-digit-$k$th-powers operation in the body of your question. $\endgroup$
    – Steven Stadnicki
    Nov 3, 2016 at 1:26
  • $\begingroup$ That's basically true, though you could say the same thing about the sequence of primes in increasing order, and yet there is a lot to be said about the asymptotics of the $n$-th prime. (Also, though such sets are listed as increasing-order sequences in OEIS $-$ thus the S in the acronym.) $\endgroup$
    – Noam D. Elkies
    Nov 3, 2016 at 3:22

2 Answers 2

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
4
$\begingroup$

If a number $n$ has $5$ or more digits, $n$ is greater than the sum of the cubes of the digits. Clearly, $5\cdot 9^3<10000$, so repeating the process eventually leads to a number with fewer than five digits.

As someone wrote in the comment, this implies that there are a finite set of cycles, and every number ends in such a cycle.

Improve this answer
$\endgroup$
2
  • $\begingroup$ what about the second and the third questions $\endgroup$
    – Ahmad
    Nov 3, 2016 at 1:05
  • 1
    $\begingroup$ For the third question, for any value of $k$ and any starting number $n$, you will always end up at a repeating sequence by this same argument. Taking $f_k(n)$ to be the function you describe, the question "$\forall k \in \mathbb{N}, \exists n \in \mathbb{N} \setminus \{1\}$ such that $f_k(n) = n$?" (i.e. is there a always a number which maps onto itself?) is probably a bit trickier. $\endgroup$
    – Philip C
    Nov 3, 2016 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.