[seqfan] a(n) + (next digit) = a(n+1)

[Eric Angelini]

Hello SeqFans,
To build a sequence S including the term 
<1000000> (one million) that satisfies:
a(n) + (next digit of S) = a(n+1)
is easy. Start with 1000000 (one million)
and work backwards, subtracting at each
step the first digit of a(n) to get the next integer, and so on.
You will reach 9 at the end. So the seq S above
will look like:
S=9,10,11,12,13,14,15,16,17,18,... 
,999972,999981,999990,999999,1000000,...
The funny thing here is that the lexico-first 
infinite such sequence is very difficult
to compute "upwards"! What is the
next term after 18, in the example
above? Is it 19 ou 20?
Building S "upwards" starting with 9
leads to a lot of such crossroads!
The contrast between "backwards from any a(n), no matter how huge
a(n) is" and the "upwards" method 
to reach said a(n) is stunning!

Backtracking mon amour!
Best,
É.