Will terms in this sequence always have digital root 1? A recreational math problem, dubbed "Insert and Add", asks:  What is the least integer m that requires no less than n insertions of plus signs so that, after performing the addition(s), we arrive at a single digit?  (See the last page here:  http://orion.math.iastate.edu/butler/papers/16_03_insert_and_add.pdf)
It is similar to finding the additive persistence of n, but instead of merely counting the number of digital sums required to arrive at a single digit it counts the minimum number of plus signs inserted during that process.
10 is the smallest number that requires one plus sign: 1+0=1.  19 is the smallest to require two: 1+9=10 -> 1+0=1.  118 is the smallest to require three:  1+1+8=10 -> 1+0=1; alternatively we can try 1+18=19 -> 1+9=10 -> 1+0=1; and finally we can try 11+8=19 -> 1+9=10 -> 1+0=1.
3187, and 3014173 are the next two numbers in the sequence.
Now observe that all of these numbers (10, 19, 118, 3187, 3014173) have a digital root of 1.
Is it obvious that all future terms in this sequence will have digital root 1?
The sequence is https://oeis.org/A293929.

I am terrible at proofs, so I won't be surprised when someone points out a glaring hole in this, but what about:
Assume the terms $a(1)$ to $a(n)$ all have digital root $1$, but $a(n + 1) = x$ doesn't.  Increment $a(n)$ by one until we reach $x$.
Insert one plus sign into $x$ in the optimal way that guarantees the result of the addition, $y$, requires exactly $n$ more insertions of a plus sign to arrive at a single digit.   
Because $y$ requires $n$ insertions it cannot be less than $a(n)$, otherwise we would have found $y$ before $a(n)$.  Because $x$ has digital root greater than $1$, $y$ cannot equal $a(n)$.  So now $y$ must be in the range $a(n) < y < x$, but we already checked these before arriving at $x$, so no such number $y$ can exist, therefore no such $x$ can exist.  Clearly, $a(n + 1)$ cannot have digital root $0$. 
We have shown that no $a(n + 1) = x$ with digital root $0$, or $2$ through $9$ can exist, therefore $a(n + 1)$ must have digital root $1$.
