The Problems of Ukrainian Authors

Answer: b) such sequence exists; c) such sequence doesn't exist.

a. Show the example of such a "consecutive" sequence which is different from $a_n=n$. Let $n=2^{{\alpha }_1}{3}^{{\alpha }_2}{5}^{{\alpha }_3}\dots p_r^{{\alpha }_r}$ be the unique representation of $n$ as a product of primes, then the sequence $\{a_n\}$, $a_n=2^{{\alpha }_2}{3}^{{\alpha }_1}{5}^{{\alpha }_3}\dots p_r^{{\alpha }_r}$ satisfies the statement of the problem.

Really, if $n=2^{{\alpha }_1}{3}^{{\alpha }_2}{5}^{{\alpha }_3}\cdots p_r^{{\alpha }_r}{p}_{r+1}^0{p}_{r+2}^0\cdots$, $m=2^{{\beta }_1}{3}^{{\beta }_2}{5}^{{\beta }_3}\cdots p_s^{{\beta }_s}{p}_{s+1}^0{p}_{s+2}^0\cdots$, $k=2^{{\gamma }_1}{3}^{{\gamma }_2}{5}^{{\gamma }_3}\cdots p_t^{{\gamma }_t}{p}_{t+1}^0{p}_{t+2}^0\cdots$ and $a_n<a_m$, then $a_{kn}<a_{km}$, as $$a_{kn}=2^{{\alpha }_2+{\gamma }_2}{3}^{{\alpha }_1+{\gamma }_1}{5}^{{\alpha }_3+{\gamma }_3}\cdots p_{\max (r,t)}^{{\alpha }_{\max (r,t)}+{\gamma }_{\max (r,t)}}=a_k{a}_n,$$ $$a_{km}=2^{{\beta }_2+{\gamma }_2}{3}^{{\beta }_1+{\gamma }_1}{5}^{{\beta }_3+{\gamma }_3}\cdots p_{\max (s,t)}^{{\beta }_{\max (s,t)}+{\gamma }_{\max (s,t)}}=a_k{a}_m$$ and $a_{kn}=a_k{a}_n<a_k{a}_m=a_{km}$. The constructed sequence is non-trivial, because $a_2=3$, and is constructed using a permutation of the exponents of $n$ in its prime factorization by means of shifting the exponents of two and three.

b. Next, let $$N(k)=\{\begin{array}{ll}k-2,& \text{if }k=4,6,8,\dots ,\\ 1,& \text{if }k=2,\\ k+2,& \text{if }k=1,3,5,\dots \end{array}$$ Then $N(k_1)\ne N(k_2)$ if $k_1\ne k_2$. If now we construct a sequence in which for the number $n$ represented as a product of primes, we shift the exponents of $p_k$ to $p_{N(k)}$, then the constructed sequence satisfies the statement of the problem. Really, let for $n=2^{{\alpha }_1}{3}^{{\alpha }_2}{5}^{{\alpha }_3}{7}^{{\alpha }_4}\dots p_{r-1}^{{\alpha }_{r-1}}{p}_r^{{\alpha }_r}$ $$a_n=5^{{\alpha }_1}{2}^{{\alpha }_2}11^{{\alpha }_3}{3}^{{\alpha }_4}\cdots p_{N(r-1)}^{{\alpha }_{r-1}}{p}_{N(r)}^{{\alpha }_r}.$$ Then the sequence $\{a_n\}$ is "consecutive", because $a_{kn}=a_k{a}_n$ and $a_{kn}=a_k{a}_n<a_k{a}_m=a_{km}$.

Moreover, if $n\ge 2$ then $a_n\ne n$, because if $a_n=n$, then the following equalities hold: ${\alpha }_{N(k)}={\alpha }_k$, ${\alpha }_{N(N(k))}={\alpha }_{N(k)}={\alpha }_k$, $\dots$, ${\alpha }_{N(...N(N(k)))}={\alpha }_k$. But $N(...N(N(k)))$ is unrestrictedly large beginning with some iteration, so if there exists nonzero ${\alpha }_k$ in the prime factorization of $n$, then in its representation must exist nonzero powers of arbitrarily large prime numbers, which is a contradiction.

c. If for some sequence holds $a_1\ne 1$, then there exists $l$ such that $a_l=1$. Then $a_l<a_1$ and $a_{l2}<a_l=1$ — contradiction, because $a_{l2}\in \mathbb{N}$.