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We will say that the measure of a number is equal to the maximum degree in which it is possible to represent a number in the form of a sum of digits copied (You can not rearrange the numbers). For example, for $55$ this will be $5$, because $$ 55^1 = 55, \quad 55 = 55$$ $$ 55^2 = 3025, \quad 30+25 = 55 $$ $$ 55^3 = 166375, \quad 1+6+6+37+5 = 55$$ $$ 55^4 = 9150625, \quad 9+15+0+6+25 = 55 $$$$55 ^ 5 = 503284375, \quad 5 + 0 + 3 + 28 + 4 + 3 + 7 + 5 = 55.$$ Let $a_n$ be a sequence of numbers such that all smaller ones have a measure less than. What is the asymptotics of this sequence? Is it possible to somehow build numbers with a given measure? If not, what measures can not be built?

The task was put in a Russian forum, I put it a little different question, I will be glad to any help in its solution :)

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  • $\begingroup$ All natural large ones are considered, the first term in the sequence is obviously $2$ with measure $1$, and the second term is 9, since $$ 9 ^ 2 = 81, \quad 8 + 1 = 9 $$ $\endgroup$ – Vladislav Kharlamov Jun 8 '18 at 1:30
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    $\begingroup$ What exactly are you asking? $\endgroup$ – PotatoLatte Jun 8 '18 at 1:30
  • $\begingroup$ I ask the answer to one of the questions. Do you think it's bad when there are so many of them? $\endgroup$ – Vladislav Kharlamov Jun 8 '18 at 1:31
  • $\begingroup$ The Russian forum also presents some results, it seems to me amazing such large numbers on such relatively small results. $ n \quad b(n) \\ 675 \quad 50 \\ 945 \quad 68 \\ 964 \quad 71 \\ 990 \quad 107 \\ 991 \quad 71 \\ 1296 \quad 84 \\ 1702 \quad 114 \\ 2728 \quad 173 \\ 4879 \quad 285 \\ 5050 \quad 403 \\ 5149 \quad 300 \\ 5292 \quad 309 \\ $ $\endgroup$ – Vladislav Kharlamov Jun 8 '18 at 9:40
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    $\begingroup$ We shouldn't have an immediate obstacle $\mod 9$, so interesting numbers are $0$ or $1$ modulo $9$. Assuming that, one would expect that the sum of digits of $n^k$ is about $\frac 9{2\log 10} k\log n$, so the power may be as large as $\frac {2n\log 10}{9\log n}$ and, if you are lucky, you can go up from here for numbers divisible by $10$ because the growth rate of the sum of the digits for them is slower. For the numbers in the table that do not end with $0$, this is an almost exact match. However, the third (and last) term in your sequence is formally $10$, whose measure is $+\infty$. $\endgroup$ – fedja Jun 14 '18 at 2:40

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