Inspired by Digitangular numbers.
The triangle of a natural number is given by
$$\triangle(n)=\frac{n(n+1)}2$$
The digitangular counterpart of a natural number is given by the sum of the triangles of the digits of the number, e.g. $\newcommand{rdig}{{\rm digi}\triangle~}$
$$\rdig(613)=\sum_{k=6,1,3}\triangle(k)=21+1+6=28$$
A digitangular sequence of a natural number is given by repeated applications of the digitangular counterpart of that number (and for the sake of consistency, we'll include the original number too) until we reach $1$. Beginning with $613$, the digitangular sequence is given by
$$613,\rdig(613),\rdig(\rdig(613)),\dots\\613,28,39,51,16,22,6,21,4,10,1$$
I then proved that this sequence always terminates i.e. we'll always reach $1$ eventually with computer assistance. First show that the sequence always reaches $1$ for any starting natural number in $[1,135]$. For $x>135$, let $n=1+\lfloor\log_{10}(x)\rfloor$ be the number of digits in $x$. The largest value $\rdig(x)$ could be is if all of $x$'s digits were $9$'s, providing us with the bound
$$\rdig(x)\le45n=45(1+\lfloor\log_{10}(x)\rfloor)$$
And for any $x>135$, we have
$$45(1+\lfloor\log_{10}(x)\rfloor)<x$$
Hence,
$$x>135\implies\rdig(x)<x$$
So the sequence eventually goes below $135$, and that's where we brute force the proof.
I'm searching for (possibly more elegant) alternative proofs, that preferably do not require computer assistance, that this sequence, for every natural $n$, always goes down to $1$.
So for numbers below $136$ the highest possible next number is given by $99$ which goes to $90$ and clearly down to $90$ gives a lower sum.
Below $99$ we have $89$ which gives $36+45=81$ as the maximum.
Below $81$ we take $79$ which gives $28+45=73$
Then $69$ takes us down to $66$ and $59$ gives $60$.
For every number above $59$ therefore we get a decrease by applying the operation.
But for some small numbers we have a two step process, which requires more calculation. It is easy to see that these include $59,49,39,29,19,9$.
Next consider numbers which don't contain a $9$. $58$ takes us down to $51$ and $48$ goes down to $46$ but $38$ increases to $42$ and we have the cases $38,28,18,8$ to consider.
By this means you can reduce the computation, but you can't eliminate it.