Is there an increasing arithmetic sequence the digit sum of its terms forms again an increasing arithmetic sequence?

Question

Is there an increasing arithmetic sequence with $10000$ terms such that the digit sum of its terms forms again an increasing arithmetic sequence?

This problem is from the (Problems from the book) exercise ,It says here that the problem is the Tournament of the Towns exam,but read all years don't found it,problem And I’ve been trying to think about it for a long time, and I’ve come up with nothing, maybe using some kind of magical structure?

Answer 1

For simplicity, just consider initial values that look like $$ [N_1][N_2]\ldots[N_m] $$ and increments of the form $$ [1][1]\ldots[1], $$ where all the $[N_i]$ and $[1]$ blocks are the same number of digits long (left-padded with zeroes where necessary). For instance, you could consider the sequence of values $$ 9008007006005004003002001000 + 1001001001001001001001001001n. $$ Each $N_i$ behaves as an independent counter, and the digit sum increases by $m$ with each increment, minus $9$ times the number of carries performed in the sum. The digit sums will form an arithmetic sequence only as long as there are the same number of carries at each increment. In the example given, there is a single carry at each increment for sufficiently small $n$: first $N_1$ goes from $9$ to $10$, then $N_2$ goes from $9$ to $10$, then $N_3$ goes from $9$ to $10$, etc. The digit sums form an arithmetic sequence (increasing by $1$ at each increment) for the first $90$ terms, ending only when $N_1$ goes from $99$ to $100$ (requiring two carries).

This example can be generalized in a fairly obvious way, by stringing together the numbers from $N_1=10^n-1$ down to $N_{10^n}=0$. More precisely, take $$ A=\sum_{i=0}^{10^n-1}i\cdot10^{(n+1)i} \qquad B=\sum_{i=0}^{10^n-1}10^{(n+1)i}. $$ Then the digit sums of the arithmetic sequence $A, A+B, A+2B, A+3B, \ldots$ form an arithmetic sequence themselves, which persists for the first $9\cdot 10^n$ terms, when $N_1$ goes from $10^{n+1}-1$ to $10^{n+1}$. So for any $n\ge 4$, this solves OP's problem in the affirmative. Moreover, it proves

Thm. There are arbitrarily long arithmetic sequences ($[a_n]\equiv (a_0,a_1,a_2,\ldots)$) where the digit sums of the terms ($[d(a_n)]$) also form arithmetic sequences.

This leaves open the generalization:

Are there arbitrarily long arithmetic sequences $[a_n]$ where the digit sums of the terms ($[d(a_n)]$) and the digit sums of the digit sums of the terms ($[d^2(a_n)]$) both form arithmetic sequences?