[seqfan] The last digit "d" of a(n) is present in a(n+1+d)

[Eric Angelini]

Hello SeqFans,

... as said above, the last digit "d" of a(n)
must be present in a(n+1+d).

Two more easy rules:

a) When there is no writing constraint on a(n),
   take a(n) = a(n) + 1
b) When there is a writing constraint on a(n),
   take the smallest a(n) > a(n-1) not leading
   to a contradiction

---

We then have (if I don't miss smtg):

     n = 1  2   3   4   5    6    7    8    9    10   11   12   13   14   
     S = 1, 2, 10, 20, 102, 103, 104, 112, 113, 123, 124, ...
ObDigit:    -   1   0   02   -    -     2   -     3    2    4    3    3    

Could please someone compute S (as this is the 
third time I make a mistake in the first terms :-(

Best,
É.