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I am seeking for an expression for the transformation of the Fibonacci-polynomials $F_n(x)$ from $x\to x+1,\; x\in\mathbb{R}$, ideally in terms of $F_m(x), m\le n$.

I have tried $$ F_n(x+1):= \sum_{k=0}^{n} {\frac{n+k-1}{2}\choose k}(x+1)^k = \sum_{0\le l\le k}^{n} {\frac{n+k-1}{2}\choose k} {k \choose l} x^l.$$

Here I see

$$ F_n(x+1) = F_n(x) + \sum_{0\le l\le k}^{n} {\frac{n+k-1}{2}\choose k} {k \choose l-1} x^{l-1}, $$ but then it gets somehow complicated. To me it seems plausible that $F_n(x+1)$ can be (uniquely?) expressed as a linear combination $\sum_{m\le n}\lambda_m F_m(x)$. And playing around with Mathematica I have found for $F_{10}(x+1)$ that $\vec{\lambda}_{10}=\{29,48,51,-98,-9,78,77,36,9,1\}$ for instance does the trick (the above confirms at least $\lambda_{10,10}=1$). Which looks a bit odd a sequence and gives no hit in OEIS.

The question emerged from a futile attempt of mine to attack this problem by induction over $x$.

Edit

Of course last coefficients $\lambda_{n,n}=1$ are one as noted above, the last but one are $\lambda_{n,n-1}=n-1$, then obviously $\lambda_{n,n-2}=\frac{(n-2)(n-3)}{2}$ and $\lambda_{n,n-3}=\frac{(n-4)(n-3)(n-7)}{4}$, at the moment I still fail to generalise that exactly. But it is clear that the power of $n$ is increasing in that series. Maybe this is known to someone?

Raphael J.F. Berger
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1 Answers1

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Interesting question! I don't have an explicit answer, but here's a path you can explore: because the polynomials $F_n(x)$ all have leading term $x^{n+1}$ and integer coefficients, the set $\{F_i(x): i\leq n\}$ forms a basis for the set of all degree $n+1$ polynomials; this implies immediately that there's always an expression for $F_n(x+1)$ (which is also just a polynomial of degree $n+1$ in $x$, after all) in terms of $F_i(x)$ for $i\leq n$ and that in fact this expression is unique.

What's more, you can probably find an explicit expression for the coefficients of the inverse matrix analagous to the explicit expression for the coefficients of the polynomial; that would then let you compute an expression for the coefficients you're after, although I'm not sure how 'closed-form' you would be able to get it.

Steven Stadnicki
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