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.

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