Htykuut

Let $S=(1,1,a)$. Then construct a new, infinite sequence $X$ using the following process:

• First, $X_0=S_0=1$
  


• Then add $S_0$ to each element of $S$, with letters being converted to their position in the alphabet and back after addition. The sequence $S$ is now $(2,2,b)$
  


• Repeat with $S_1$. $X_1=S_1=2$.


• Then add $2$ ($S_1$) to each element of $S$. $S$ is now $(4,4,d)$
  


• Repeat. $X_2=S_2=d$, after addition, $S=(7,7,g)$
  


• This time, $X_3=S_0=7$ and so on

All operations on numbers are modulo 10. All operations on letters are modulo 26 (e.g. $y+3=b$, $9+1=0$).

Is there a sequence which when this process is applied to, generates an $X$ which repeats itself? Until now, I've found $(1,1,a,s)$ which generates an $X$ of $(1,2,d,y,1,2,h,g,7,4,...)$

Is there a sequence which when this process is applied to, generates an X which repeats itself?

Any finite starting sequence $S$ will generate an $X$ that's eventually periodic.

Your $S=(1,1,a)$ generates $X$ composed of the following period of length $117$ repeated infinitely:$tt 12d74r50n36j12v50v12t36v36z74j74d12j36b36l36f74v98h36x50z50d36p74p36n50f50p50j98j50x36d50l12z98p12h12r12l50b12f98b74n$

Your $S=(1,1,a,s)$ generates $X$ composed of the following period of length $96$ repeated infinitely:$tt 12dy12hg74fc50ra74te12te74bu98ns74dy50jk98te98hg98lo12ns50xm50lo50pw74lo36te50vi98pw36jk12fc12pw$

After working on it for a bit (yes, this is what I do in my free time), I found that the starting sequence just needs to sum up to a multiple of both $10$ and $26$. So for example, a sequence consisting of $130$ nines would work and would generate $(9,8,6,2,4,8,6,2,4,...)$.

When only dealing with letters, you can have the sequence sum to a multiple of only $26$. So for example, the sequence $(s,m,u,i,y,d,f,f,u)$ produces $(s,e,q,u,e,n,c,e,x,s,e,q,u,e,n,c,e,x,...)$. Note the $x$, it's a padder to get the shift it back into position so that the next time the original sequence is restored.