TITLE: A possibly surprising appearance of $\sqrt{2}.$
QUESTION [20 upvotes]: Define $A=(a_n)$ and $B=(b_n)$ as follows: $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$, and $$a_n=a_1b_{n-1}-a_0b_{n-2} + 2n$$ for $n \geq 2$, where $A$ and $B$ are increasing and every positive integer occurs exactly once in $A$ or $B.$  Can someone prove that $$2n < a(n) - \sqrt{2} n < 3+2n$$ for $n \geq 2$ ?   
Evidence: 
$$A = (1, 2, 9, 12, 15, 18, 21, 26, 28, 33, 35, 40, 42, 47, 49, 54, 56,\dots)$$
$$B = (3, 4, 5, 6, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 24, 25, \dots)$$
I've checked that  $0.207 < a_n - (2+\sqrt{2})n < 2.914$ for $2 <= n <= 18000.$
This question is similar in form to Limit associated with complementary sequences. There P. Majer proves that $a_n - 4n$ is not bounded, whereas here, the claim is that $a_n - (2+\sqrt{2})n$ is bounded.

REPLY [7 votes]: An illustration for the answer by Gjergji Zaimi: the mysterious sequence $\epsilon_1,\epsilon_2,...,\epsilon_{5000}$

And here, in response to the comment by André Henriques, is the plot of lengths of gaps between consecutive points in the set $\{\epsilon_k\mid10000<k<20000\}$ rearranged in the increasing order. Some Cantor-set-like structure seems to be present indeed.