The Eraser game

Hello Math-Fun,
The idea here is to always extend the sequence E with the sum of the last two terms – and to apply the "eraser rule" when needed:

E = 1, 2, 3, 5, 8, 13, 21, 15, 17, ...

The eraser rule says that the last term of E and the coming sum cannot share any digit. The duplicate digit (in yellow) are thus erased from the sum. (When needed, the remaining digits are concatenated to form a new integer).

Example:
We see that after 8 + 13 = 21 we must erase the 1 from 21 and proceed with 2;
this 2 will be added to 13 to produce 15;
but as 2 + 15 = 17 we will erase 1 again from 17 and extend E with 7; etc.

The above start stops quickly:
E = 1, 2, 3, 5, 8, 13, 21, 15, 17, 22, 29, 31, 40, 71, 111 stop.

What about leading zeros in the "erased" sum? Let's  see:
E = 598, 414, 1012,...
Well, let's erase any leading zero – we will thus proceed here with:
E = 598, 414,  2,  416, 418, 424, etc.

Questions:
What is the lexicographically earliest start pushing the sequence into a loop?
Is there a start leading to an infinite sequence?
Best,
É.
________________
Update, March 28th:
Hans Havermann was quick to answer the first question:
1, 4, 5, 9, 14, 23, 7, 30, 7, 3, 10, 3, 1, 4, ...
It seems that a lot of starting pairs either end in this loop... or in a dead end.
Bravo and thanks, Hans!