Seq and first diff show the same "digit pattern"

[Eric Angelini]

Hello SeqFan,
could someone compute a few more terms of this seq:

1  12  14  155  160  211  271  292  419  548  572  691 ...

The principle is:

- Seq and first diff show the same "digit pattern".

S = 1  12  14   155  160  211  271  292   419   548  572  691 ...
d =  11   2  141    5   51   60   21   127   129   24   19 ...

Rules:
- start S with "1"
- add to the last term of S the smallest integer d no yet
  added and not present in S such that the concatenation
  of S's terms and the concatenation of all ds are the
  same string of digits

So, never twice the same integer in sequence or first
differences. 

I'm quite sure that all N's will be split between S and d.

Best,
É.

http://www.research.att.com/~njas/sequences/A110621  
has a close Mathematica pgm by Robert G. Wilson.

(thanks again to him!)




Something is wrong with my email, and I haven't been receiving all emails 
have.

I will post the solution now, even though only a short while has passed 

Original email as spoiler-space.

>Since people are posting sequence puzzles on seq.fan lately, I thought I 
>would post this sequence puzzle of a different varity.
>
>I suspect this 'puzzle' is easy, and I'll probably regret I posted this.
>
>---
>
>Let {c(k)} be as defined at sequence A022940. ({c(k)} itself is not in 
>the EIS.)
>
>Define sequence {a(k)} as follows:
>
>Let b(n) = c(n) - n + 1.
>
>a(1) = the number of 1's in {b(k)}. a(2) = the number of 2's in {b(k)}.
>In general, a(n) = the number of n's in {b(k)}.
>
>So, {a(k)} begins: 0,1,1,3,5,6,7,9,...
>
>Define {a(k)}. (Define it in a simpler way than by the steps given above.)
>
>Thanks,
>Leroy Quet


I get that a(1)=0, a(2)=1. a(n) = c(n-2) -1, for n >= 3.

I don't know if there is a "closed form" (nonrecursive representation) 
for {c(k)}.

Thanks,
leroy Quet