Imitating the Imitator

 

Hello math-Fun,

The chess fairy piece « Imitator » is a gem that has led to beautiful problems for more than 80 years now (have a look here).

Could it be possible to build an integer sequence based on a similar idea?

Our Imitator is here a small black spot placed under the first digit of the sequence S.

We've decided that the Imitator must always slide from under an odd digit to under another odd digit of S.

The Imitator will be driven right or left by the successive terms of S.

When a(n) is odd, the Imitator moves under a(n) digits to the right.

When a(n) is even, the Imitator moves under a(n) digits to the left.

That’s it (for the Imitator).

We want as usual S to be the lexicographically earliest sequence of distinct positive integers with this property.

If I’m not wrong, S starts like this:

S = 1, 3, 5, 2, 7, 4, 6, 9, 11, 10, 13, 15, 8, 12, 17, 19, 21,…

Let’s check the behavior of the Imitator (the small black i under a digit of S).

(if needed, click the picture to enlarge)

Remarks

We might test those variants:

a— the Imitator must always be under an even digit;

b— the Imitator must always be under a prime digit;

c— the Imitator must always be under an odd term;

d— the Imitator must always be under an even term;
e— the Imitator must always be under a prime term.

What about integers with an inside Imitator (that stays inside when we're finished)?

Say we always place the Imitator under the leftmost digit of a(n), then "spell" the digits one after the other from there – and we finish with the rightmost one.

2023 is not an imitating integer, as the digit 3 doesn't bring back the Imitator inside 2023:

I guess the sequence T of such imitating integers is already somewhere in the OEIS (I couln't find it though). T starts like this, if I am not wrong:

T = 10, 23, 32, 45, 54, 67, 76, 89, 98, 100, 101, 110, 112, 121, 123, 134, 143, 145, ...