[seqfan] Next integer ? A new multiple of (d+1)

[Eric Angelini]

Hello Seqfans,
This was fun to imagine:

Start S with a(n)=1

Now a(n+1) is always the smallest integer not yet present in S such 
that a(n+1) is a multiple of (d+1), d being the last digit of a(n):

S=1,2,3,4,5,6,7,8,9,10,11,12,15,18,27,16,14,20,13,24,25,30,17,32,21,...

S is a permutation of N(aturals), I hope (it was imagined with this 
purpose in mind).

Example:
 2 is the smallest integer not yet in S that is a multiple of (1+1)
 3 is the smallest integer not yet in S that is a multiple of (2+1)
 4 is the smallest integer not yet in S that is a multiple of (3+1)
...
10 is the smallest integer not yet in S that is a multiple of (9+1)
11 is the smallest integer not yet in S that is a multiple of (0+1)
12 is the smallest integer not yet in S that is a multiple of (1+1)
15 is the smallest integer not yet in S that is a multiple of (2+1)
18 is the smallest integer not yet in S that is a multiple of (5+1)
27 is the smallest integer not yet in S that is a multiple of (8+1)
16 is the smallest integer not yet in S that is a multiple of (7+1)
...                                                            |
This is d, the last digit of a(n) -----------------------------+

I'd like to see the graph of S... Any taker?
Best,
É.