Can someone identify this polynomial sequence? Is it known in mathematics? I'm interested in various properties of this sequence.
I'd like to find $P(n)$, $n\in \mathbb{Z}^+$

\begin{align} P(0)&= 1\\ P(1)&= a\\ P(2)&= a^2+b\\ P(3)&= a^3+2ab\\ P(4)&= a^4+3a^2b+b^2\\ P(5)&= a^5+4a^3b+3ab^2\\ P(6)&= a^6+5a^4b+6a^2b^2+b^3\\ P(7)&= a^7+6 a^5 b+10 a^3 b^2+4 a b^3\\ P(8)&= a^8 + 7 a^6 b + 15 a^4 b^2 + 10 a^2 b^3 + b^4\\ P(9)&= a^9 + 8 a^7 b + 21 a^5 b^2 + 20 a^3 b^3 + 5 a b^4\\ P(10)&= a^{10} + 9 a^8 b + 28 a^6 b^2 + 35 a^4 b^3 + 15 a^2 b^4 + b^5 \end{align}

More steps upon request.

I'll be grateful for any hints!


Hint. Note that the following recurrence holds: for $n\geq 2$, $$P(n)=aP(n-1)+bP(n-2).$$ They are related to the Fibonacci polynomials. The wiki page gives a list of properties. For example we have that $$P(n)=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}a^{n-2k}b^k.$$

  • $\begingroup$ @BarryCipra Yes it's better to stick to OP's notation. Thanks for pointing out. $\endgroup$ – Robert Z Dec 17 '18 at 14:10
  • $\begingroup$ Many thanks! :) I must study these properties to find if I find something useful $\endgroup$ – Ender Dec 17 '18 at 14:25


$$-\frac{2^{-n} \left(\left(a-\sqrt{a^2+4 b}\right)^n-\left(\sqrt{a^2+4 b}+a\right)^n\right)}{\sqrt{a^2+4 b}}$$


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