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The Fibonacci polynomials are defined recursively by $F_0(x)=0, F_1(x)=1$ and $F_n(x)=xF_{n-1}(x)+F_{n-2}(x)$, for $n\geq2$.

While computing certain integrals, I observe the following (numerically) which prompted me to ask:

Question. For $n, k\in\mathbb{N}$, are these always integers? $$\int_0^1F_n(k+nz)\,dz$$

To help clarify, here is a list of the first few polynomials: $$F_2(x)=x, \qquad F_3(x)=x^2+1, \qquad F_4(x)=x^3+2x.$$

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    $\begingroup$ Did you check the case n=3 and k=1? $\endgroup$ – Cherng-tiao Perng Jun 26 '17 at 0:05
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    $\begingroup$ Perhaps you were misled by a typo: $F_0(x)=0$. $\endgroup$ – T. Amdeberhan Jun 26 '17 at 0:08
  • $\begingroup$ $F_3(1+3z)=9z^2+6z+2$, so $\int_0^1F_3(1+3z)dz=8$. $\endgroup$ – T. Amdeberhan Jun 26 '17 at 2:10
  • $\begingroup$ The first few $F_n=\sum F(n,k)x^k$ at least have the stronger property that every term of $\sum \int \ldots$ is an integer separately. If this is true in general, one can perhaps be optimistic about a proof since there is an explicit formula for the coefficients $F(n,k)$. $\endgroup$ – Christian Remling Jun 26 '17 at 2:52
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The integral of each individual monomial will be integral. First we have the identity $$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$ so my claim is that $$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz=\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)} \in \mathbb Z.$$ By the binomial theorem we can write $(k+n)^{n-2i}=k^{n-2i}+nk^{n-2i-1}(n-2i)+n^2d$ for some integer $d$. So we can write $$\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}=k^{n-2i-1}\binom{n-i-1}{i}+d\cdot\frac{n}{n-2i}\binom{n-i-1}{i}$$ it suffices to show that $\frac{n}{n-2i}\binom{n-i-1}{i}$ is an integer. However we can check that $$\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$$ and the claim follows.

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    $\begingroup$ Great job, as always! $\endgroup$ – T. Amdeberhan Jun 26 '17 at 17:58

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