<s>[INST] Do the coefficients of these polynomials alternate in sign? Define polynomials $p_i(x)$ by the recurrence
\begin{align}
p_0(x)&=0 \\
p_1(x)&=1 \\
p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\
p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\
\end{align}
The first few are given by
\begin{align}
p_0(x)&=0\\
p_1(x)&=1\\
p_2(x)&=1\\
p_3(x)&=x-1\\
p_4(x)&=x-2\\
p_5(x)&=x^2-3x+1\\
p_6(x)&=x^2-4x+3\\
p_7(x)&=x^3-5x^2+6x-1\\
p_8(x)&=x^3-6x^2+10x-4
\end{align}
It is reasonable to ask whether the coefficients of these polynomials alternate in sign. Any thoughts?
 [/INST] Lord Shark's answer is good (but might contain some subtle mistakes) ... I reckon ...
\begin{eqnarray*}
p_{2i}(x) = \sum_{j=0}^{i-1} \binom{2i-j-1}{j} (-1)^j x^{i-j-1} \\
p_{2i-1}(x) = \sum_{j=0}^{i-1} \binom{2i-j-2}{j} (-1)^j x^{i-j-1}. \\
\end{eqnarray*}
Proof of the even formula ... 
\begin{eqnarray*}
&p_{2i-1}(x) -p_{2i-2}(x) =\\ & x^{i-1}+ \sum_{j=0}^{i-2} \binom{2i-j-3}{j+1} (-1)^{j+1} x^{i-j-2} - \sum_{j=0}^{i-1} \binom{2i-j-3}{j} (-1)^j x^{i-j-2} \\
&= x^{i-1}+ \sum_{j=0}^{i-2} \binom{2i-j-2}{j} (-1)^{j+1} x^{i-j-2}
= p_{2i}(x).
\end{eqnarray*}
And the answer to the question in the title is $\color{red}{\text{yes}}$.
</s>