[seqfan] Digit-counters updating themselves

[Eric Angelini]

D=0,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,1,10,2,2,3,2,4,2,5,2,6,2,7,2,8,2,9,2,10,3,3,4,3,5,3,6,3,7,3,8,3,9,3,10,4,4,5,4,6,4,7,4,8,4,9,4,10,5,5,6,5,7,5,8,5,9,5,10,6,6,7,6,8,6,9,6,10,7,7,8,7,9,7,10,8,8,9,8,10,9,9,10,10,11,20,12,...

Hello SeqFans,
pick any comma in D.
Immediately to the left of the comma
there is a digit 'd'.
Immediately to the right of the comma
there is an integer d(n).
D is such that there are d(n) digit 'd'
so far in D [from the start of D up to the comma].

In other words, the rightmost digit of d(n) is present d(n+1) times in D, counting from d(1) to d(n).

I'm wondering: do all integers appear
at least once in D?

P.-S.
It is possible to compute similar sequences
for every base. I guess the binary-one is:

B = 0,1,1,10,10,11,110,100,110,111,1110,...

Best,
É.

Catapulté de mon aPhone