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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?

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  • $\begingroup$ Those look like binomial coefficients. $\endgroup$ Oct 21, 2019 at 20:03
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    $\begingroup$ Why did you delete your answer here? $\endgroup$ Nov 8, 2019 at 17:16

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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 $\color{red}{\text{yes}}$.

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I reckon that $$p_{2n-1}(x)=\sum_{k=0}^{n-1}(-1)^k \binom{n+1-k}k x^{n-1-k}$$ and $$p_{2n}(x)=\sum_{k=0}^{n-1}(-1)^k \binom{n+2-k}k x^{n-1-k}$$ and that these can be proved by induction.

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