AD

 
AD is the absolute difference between a digit and the next one.
If AD is odd, we add it to the last integer.
If AD is even, we subtract it from the last integer.
We start with 10.
Does 10 enter a loop? Grow infinitely? End on a negative term?

S = 10,
AD = 1, odd – we add 1 to 10 and get 11:
S = 10, 11,
The above (underlined) AD is 1 again, odd – we add 1 to 11 and get 12:
S = 10, 11, 12,
The above (underlined) AD is 0, even – we subtract 0 from 12 and get 12 (again):
S = 10, 11, 12, 12,  
The above AD is 0 (again), even – we subtract 0 from 12 and get 12 (again):
S = 10, 11, 12, 12, 12,  
The above AD is 1, odd – we add 1 to 12 and get 13:
S = 10, 11, 12, 12, 12, 13,
The above AD is 1, odd – we add 1 to 13 and get 14:
S = 10, 11, 12, 12, 12, 13, 14,
The above AD is 1, odd – we add 1 to 14 and get 15:
S = 10, 11, 12, 12, 12, 13, 14, 15,
The above AD is 1, odd – we add 1 to 15 and get 16:
S = 10, 11, 12, 12, 12, 13, 14, 15, 16,
The above AD is 1, odd – we add 1 to 16 and get 17:
S = 10, 11, 12, 12, 12, 13, 14, 15, 16, 17,
The above AD is 1, odd – we add 1 to 17 and get 18:
S = 10, 11, 12, 12, 12, 13, 14, 15, 16, 17, 18,
The above AD is 2, even – we subtract 2 from 18 and get 16:
S = 10, 11, 12, 12, 12, 13, 14, 15, 16, 17, 18, 16,
The above AD is 2, even – we subtract 2 from 16 and get 14:
S = 10, 11, 12, 12, 12, 13, 14, 15, 16, 17, 18, 16, 14, … etc. 

I guess the first few terms of S look like this:
(...)
I don't know why, but all this reminds me the Random Walk 1–dimensional here.
____________________
Update and corrections, the next day, by Giorgos Kalogeropoulos:

I'm sending you the first 300 terms of my approach
S = 10, 11, 12, 12, 12, 13, 14, 15, 16, 17, 18, 16, 14, 17, 20, 16, 12, 17, 22, 16, 10, 17, 24, 29, 34, 37, 40, 34, 39, 37, 38, 43, 48, 49, 50, 44, 49, 49, 50, 55, 60, 61, 62, 56, 61, 59, 57, 64, 58, 59, 60, 56, 59, 55, 58, 59, 60, 54, 48, 44, 40, 45, 41, 42, 43, 39, 35, 40, 36, 41, 37, 37, 37, 42, 47, 52, 48, 53, 58, 58, 59, 53, 47, 52, 57, 53, 56, 57, 57, 62, 58, 54, 50, 48, 49, 47, 48, 51, 54, 50, 53, 47, 52, 53, 54, 50, 46, 46, 46, 49, 52, 48, 51, 45, 50, 51, 51, 47, 43, 43, 43, 39, 35, 36, 37, 40, 43, 41, 39, 40, 40, 34, 28, 26, 27, 23, 26, 29, 27, 30, 28, 24, 20, 16, 12, 8, 11, 9, 7, 10, 8, 11, 9, 5, 8, 6, 4, 7, 10, 13, 16, 12, 8, 6, 7, 10, 8, 11, 14, 12, 10, 8, 6, 7, 8, 6, 4, 2, 3, -1, 2, 5, 8, 9, 10, 15, 11, 7, 3, 8, 13, 16, 19, 15, 18, 14, 10, 11, 12, 17, 22, 20, 21, 24, 22, 25, 28, 26, 24, 25, 26, 31, 27, 25, 23, 21, 19, 17, 15, 20, 16, 19, 17, 13, 16, 12, 15, 16, 16, 21, 26, 22, 18, 14, 17, 20, 23, 24, 25, 26, 27, 28, 28, 22, 16, 14, 12, 15, 18, 14, 17, 13, 9, 10, 11, 14, 12, 6, 11, 7, 3, -1, 2, 3, 1, -5, -11, -15, -19, -14, -9, -8, -7, -11, -15, -8, -1, 4, 0, 3, 1, -5, -11, -13, -15, -17, -16, -11, -6, -5, -11, -4, -4, -12, -14, -20, -19, -27, -20, -20

The first time we meet a negative number is a(180) = -1
Here are the results for the first 100.000 terms:
ÉA:
> Giorgos, what about a graph starting with, say, 2023? Does a(1) have an influence on the general aspect of the "random walk"?

GK:
> Yes, I believe that different a(1) produce different random walks; here is a(1) = 2023 (for the first 100.000 terms, it remains positive):
GK:
> and here is a(1) = 12:
GK:
> I wonder why it seems to have a preference for positive numbers...
(...)
> I could not resist computing 1 million terms of the original seq. Things seem to be positive but anything can happen!
GK:
> the highest point is a(999999) = 211081 as a(1000000) = 211079 > the lowest point is a(85358) = -419 > there are 3 zeros at the positions {279, 84948, 84954}
> in total there are 517337 odds (addition) vs 482662 even ADs (subtraction).

> If we take 10^6 random pairs and compute the ADs, we get something like:
> So we are expecting equal even and odd terms. The fact that the sequence's histogram looks like this...

... maybe has to do with the fact that the pairs of digits are not "that random": i.e when a(n) = 100120, in that area, for many thousands of pairs we will always have the pairs (1,0) or (0,0) very often. But this is a local phenomenon. So, to summarize, although the pairs are not completely random and repeat, in very large scales I believe that all the ADs will take the right distribution – which means that we may see the sign changing an infinite number of times. For example when the sequence reaches 900.000 then we will count many 9s that now are missing.

EA:
Many thanks, Giorgos!
____________________
November 10th (late) update:

Hans Havermann (Oct. 24, 2023 on Math-Fun)


On Eric's updated website, Giorgos Kalogeropoulos computed one million terms and opined: 
"Things seem to be positive but anything can happen!" 
Here's my graph of one billion terms:

http://chesswanks.com/seq/AD1b.png

> On Oct 20, 2023, at 7:37 AM, Eric ANGELINI <eric50angelini@gmail.com> wrote:
>
https://cinquantesignes.blogspot.com/2023/10/ad.html
>
> AD is the absolute difference between a digit and the next one.
> If AD is odd, we add it to the last integer.
> If AD is even, we subtract it from the last integer.
> We start with 10.
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EA
Many thanks, Hans – and forgive my late discovery of your Math-Fun post.







 

 

 




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