2
$\begingroup$

Here :

Digital root with squared digits

the possible cycles of repeated summation of the squares of the digits of a number are mentioned. What about cubes ?

$1$ , $153$ , $370$ , $371$ and $407$ are fixpoints

$(136/244)$ , $(919/1459)$ , $(160/217/352)$ and $(55/250/133)$ are cycles.

What is the complete set of the possible cycles ?

$\endgroup$
7
  • 3
    $\begingroup$ Since the sum of the cubes of the digits of $n$ is smaller than $n$ for $n \geqslant N_0$, where $N_0 < 3000$, a computer search should be quick. $\endgroup$
    – Daniel Fischer
    Commented Mar 8, 2018 at 21:40
  • $\begingroup$ @DanielFischer So, if every number $n\le 10^4$ arrives at one of the numbers above , does this prove that there are no more cycles ? $\endgroup$
    – Peter
    Commented Mar 8, 2018 at 22:04
  • $\begingroup$ @DanielFischer And upto which number do we have to check in the case of $k$-th powers ? $\endgroup$
    – Peter
    Commented Mar 8, 2018 at 22:07
  • 1
    $\begingroup$ Yes, and you can stop some time before $10^4$. $\endgroup$
    – Daniel Fischer
    Commented Mar 8, 2018 at 22:07
  • 1
    $\begingroup$ So e.g. for $k = 4$ one notes that $d\cdot 9^4 < 10^{d-1}$ for $d \geqslant 6$, thus one deals at most with $5$-digit numbers. That makes the maximal achievable sum $5 \cdot 6561 < 33000$, hence you have at most four nines and one three to deal with, which gives a bound of twenty-something thousand. $\endgroup$
    – Daniel Fischer
    Commented Mar 8, 2018 at 22:15

0

Reset to default

You must log in to answer this question.