@AlessandroDellaCorte, thank you for comment! Done.

Hello! Thank you for answer! Can I ask about a similar case? Let $g(n) = m$ with $m>0$ if $n$ belongs to A073169 ($\operatorname{composite}(n)-n$) where $m$ is the number of repetitions of $n$ in A073169. Let $b(n)$ be the $n$-th prime number, $b(1)=2$. Then I conjecture that $b(n) = 1 + b(n-1) + g(n)$. I also conjecture that this rule works for any complement sequences. Is there a way to prove it?

Hello Fedor! Thank you for answer! Of course you mean $a(n)=a(\lfloor k(1+\sqrt{2}\rfloor)=k$ instead of $a(n)=a(\lfloor k(1+\sqrt{2}\rfloor)=2$?