Hello Math-Fun, (I'm sorry to propose this here – but SeqFan didn't publish my last three posts, dunno why) The idea 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, 2(1), 15, (1)7, ... The eraser rule says that the last term of E and the coming sum cannot share any digit. The duplicate digit (parenthesized) 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, 2(1), 15, (1)7, 22, (2)9, 31, 40, 71, (111) stop. What about leading zeros in the "erased" sum? Let's see: E = 598, 414, (1)0(1)2,... Well, let's erase any leading zero – we will thus proceed here with: E = 598, 414, 2, 416, (41)8, 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. As usual, the same text as above is here -- with colors and pictures: http://cinquantesignes.blogspot.com/2020/05/the-eraser-game.html