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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 $1$.

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    $\begingroup$ You can more than halve $135$ and find the true bound without much calculation. $\newcommand{rdig}{{\rm digi}\triangle~}$ For $x\in [100, 135]$, the largest $\rdig(x)$ can get is $1+6+45 = 52$. From there, note that if the last digit of $x$ is not $0$, then $\rdig(x) - \rdig(x-1)\geq 1$, so the highest number for which $\rdig(x)\geq x$ must end with a $9$. Similarily, we can see that if $\rdig(x)\geq x$ for some two-digit number ending in $9$, then the same is true for any smaller two-digit number ending in $9$. We get that $\rdig(59) = 60\geq 59$, and this is the largest for which it happens. $\endgroup$ – Arthur Nov 22 '17 at 12:50
  • $\begingroup$ Interesting if this is true for any radix. $\endgroup$ – rus9384 Nov 22 '17 at 13:13
  • $\begingroup$ @rus9384 It probably is, simply consider a number of the form $b-1,b-1,b-1,\dots$ in base $b$ to be the largest result possible. But then we'd have to test smaller cases for every base... hm... $\endgroup$ – Simply Beautiful Art Nov 22 '17 at 13:19
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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.

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