Previous term + quantity of digits

[Eric Angelini]

Hello again SeqFans,

I was asking in my previous post:

> to compute the next term, add to n the term t -- t being the
> quantity of [last digit of n] used so far in the sequence.
>
> Examples:
> 0,1,2,3,4,5,6,7,8,9,10,12
> What follows 12?
> ake 12 + (2 = quantity of "2" used so far in the seq.) = 14
> What follows 31?
> Make 31 + (7 = quantity of "1" used so far in the seq.) = 38

Stefan Steinerberger kindly computed a few terms:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 23, 25, 27, 29,
31, 38, 41, 49, 52, 60, 64, 69, 73, 78, 82, 91, 100, 106, 112, 122,
134, 140, 148, 154, 163, 170, 179, 185, 190, 200, 212, 227, 234, 244,
256, 264, 277, 286, 296, 307, 317, 328, 337, 349, 358, 368, 379, 389,
400, 415, 423, 442, 469, 481, 508, 522, 551, 579, 592, 622, 654, 676,
693, 713, 734, 757, 776, 795, 812, 845, 863, 886, 907, 930, 948, 968,
989, 1011, 1044, 1071, 1107, 1132, 1166, 1190, 1213

I was wondering about the growth of this seq:
If a(n) denotes the nth term of the sequence,
is it true that a(n)/n^2 is unbounded while
a(n)/n^(2+e) is bounded for every e>0?

Best,
É.
(BTW forget about this remark in my first post -- the matter is
 more complicated:

> - If you want to know, say, how many "7" are used in the range
> 0--52, go down from 52 to the first "7" (in "27"), substract
> 27 from the next term (29) and you have the answer : two "7"
> are used between 0 and 52.)

(But this is ok -- it is the definition):
> At any moment you can check how many digits of a certain kind
> are used so far in the seq. with a simple substraction:
>
> - Take two consecutive integers, say 49 and 52; substract them:
> 52-49=3; there are three "9" used so far in the seq.