length $(k-1)$. Each word consists of $(k-2)$ symbols $``0"$ and a single symbol $``1"$. Let $S$ satisfy the following condition:

[*If for some $0 < s \leqslant k-1$ there exist $p_{n, d}$ words such that $``1"$ occupies the $s$th position then between the first and the $p_{n, d}$-th of these words there exists a word such that $``1"$ occupies a position with number strictly less than $s$*]{}; $L(k-1)=\sup\limits_S |S|$.

Then $L(k-1)\leqslant p_{n, d}^{k-1}-1$.

We have $L(1) \leqslant p_{n, d}-1$. Let $L(k-1) \leqslant {p_{n, d}^{k-1}} -
1$. We will show that $L(k) \leqslant {p_{n, d}^{k}} - 1$. Consider the words such that $``1"$ occupies the first position. Their number does not exceed $p_{n, d}-1$. Between any two of them as well as before the first one and after the last one, the number of words does not exceed $L(k-1)
\leqslant {p_{n, d}^{k-1}} - 1$. Hence $$L(k) \leqslant p_{n, d} - 1 +
\left(p_{n, d}\right)\left({\left(p_{n, d}\right)^{k-1}}-1\right) =
{\left(p_{n, d}\right)^{k}}-1,$$ as required.

We need a quantity which estimates the speed of "evolution” of sets $B^p(i)$:

Set $$\psi(p):= \left\{\max \ k:
B^p(i)=B^p(i+k-1)\right\}.$$ In particular, by lemma \[c:lem1.2\] we have $\psi(p_{n,d})\leqslant p_{n, d} d$.

For a given $\alpha$ we divide the sequence of the first 