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How about a simple relation like this?

$$P_{n}(t) = \frac{P_{n-1}(t)}{1+t} + \frac{t}{1+t}$$

with $P_0(t) = 1$, $b_{n}(t) = t/(1+t)$ and $a(t) = 1/(1+t)$? The solution is $P_n(t) = 1$ which is a polynomial with positive coefficient.

Or another example:

$P_{n}(t) = (1+t)^n$, a(t) = 1/(1-t), $b_n(t) = t^2(1+t)^{n-1}/(t-1)$

In general, you can find examples quite easily, given that you have $P_n(t)$ in mind, and function a(t) which is rational, and you solve for $b_n(t)$.

$$b_n(t) = P_n(t)-a(t)P_{n-1}(t)$$

As long as $b_n(t)$ is rational as you required, you have a valid example.

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How about a simple relation like this?

$$P_{n}(t) = \frac{P_{n-1}(t)}{1+t} + \frac{t}{1+t}$$

with $P_0(t) = 1$, $b_{n}(t) = t/(1+t)$ and $a(t) = 1/(1+t)$? The solution is $P_n(t) = 1$ which is a polynomial with positive coefficient.