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Timothy Chow
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EDITED. First let us define $$g_n(x) := \sum_{k=0}^{\lfloor n/2\rfloor} {n-k\choose k} x^{n-k}.$$ Then empirically, $$\eqalign{f_{2n}(x) &= g_n(x)^2\cr f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_n(x)\cr}$$$$\eqalign{f_{2n}(x) &= g_n(x)^2\cr f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_{n+2}(x)\cr}$$ where $h_n(x)$ is a Lucas polynomial. I think it should be possible to prove these formulas using the recurrence $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in Ira Gessel's comment below, the desired properties of the roots of $f_{2n}(x)$ and $f_{4n+1}(x)$ follow from the properties of the Chebyshev polynomials. Presumably the roots of Lucas polynomials are also well understood, but I'm not so familiar with them.

EDITED. First let us define $$g_n(x) := \sum_{k=0}^{\lfloor n/2\rfloor} {n-k\choose k} x^{n-k}.$$ Then empirically, $$\eqalign{f_{2n}(x) &= g_n(x)^2\cr f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_n(x)\cr}$$ where $h_n(x)$ is a Lucas polynomial. I think it should be possible to prove these formulas using the recurrence $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in Ira Gessel's comment below, the desired properties of the roots of $f_{2n}(x)$ and $f_{4n+1}(x)$ follow from the properties of the Chebyshev polynomials. Presumably the roots of Lucas polynomials are also well understood, but I'm not so familiar with them.

EDITED. First let us define $$g_n(x) := \sum_{k=0}^{\lfloor n/2\rfloor} {n-k\choose k} x^{n-k}.$$ Then empirically, $$\eqalign{f_{2n}(x) &= g_n(x)^2\cr f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_{n+2}(x)\cr}$$ where $h_n(x)$ is a Lucas polynomial. I think it should be possible to prove these formulas using the recurrence $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in Ira Gessel's comment below, the desired properties of the roots of $f_{2n}(x)$ and $f_{4n+1}(x)$ follow from the properties of the Chebyshev polynomials. Presumably the roots of Lucas polynomials are also well understood, but I'm not so familiar with them.

Added empirically guessed formulas for the remaining polynomials.
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Timothy Chow
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This is another comment. Empirically,EDITED. First let us define $$f_{2n}(x) = \biggl(\sum_{k=0}^{n/2} {n-k\choose k} x^{n-k}\biggr)^2.$$$$g_n(x) := \sum_{k=0}^{\lfloor n/2\rfloor} {n-k\choose k} x^{n-k}.$$ From thisThen empirically, $$\eqalign{f_{2n}(x) &= g_n(x)^2\cr f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_n(x)\cr}$$ where $h_n(x)$ is a Lucas polynomial. I think it should be easierpossible to prove these formulas using the recurrence $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in Ira Gessel's comment below, the desired facts aboutproperties of the roots. I think one should be able to guess a similar expression for of $f_{2n+1}(x)$$f_{2n}(x)$ and prove it using$f_{4n+1}(x)$ follow from the recurrenceproperties of the Chebyshev polynomials. Presumably the roots of Lucas polynomials are also well understood, but I'm not so familiar with them.

This is another comment. Empirically, $$f_{2n}(x) = \biggl(\sum_{k=0}^{n/2} {n-k\choose k} x^{n-k}\biggr)^2.$$ From this, it should be easier to prove the desired facts about the roots. I think one should be able to guess a similar expression for $f_{2n+1}(x)$ and prove it using the recurrence.

EDITED. First let us define $$g_n(x) := \sum_{k=0}^{\lfloor n/2\rfloor} {n-k\choose k} x^{n-k}.$$ Then empirically, $$\eqalign{f_{2n}(x) &= g_n(x)^2\cr f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_n(x)\cr}$$ where $h_n(x)$ is a Lucas polynomial. I think it should be possible to prove these formulas using the recurrence $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in Ira Gessel's comment below, the desired properties of the roots of $f_{2n}(x)$ and $f_{4n+1}(x)$ follow from the properties of the Chebyshev polynomials. Presumably the roots of Lucas polynomials are also well understood, but I'm not so familiar with them.

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Timothy Chow
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This is another comment. Empirically, $$f_{2n}(x) = \biggl(\sum_{k=0}^{n/2} {n-k\choose k} x^{n-k}\biggr)^2.$$ From this, it should be easier to prove the desired facts about the roots. I think one should be able to guess a similar expression for $f_{2n+1}(x)$ and prove it using the recurrence.