4
$\begingroup$

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!

$\endgroup$
9
$\begingroup$

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 $$P(n)=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}a^{n-2k}b^k.$$

$\endgroup$
  • $\begingroup$ @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
2
$\begingroup$

Try:

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.