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Let $$ I_n(m)=\frac{1}{2\pi}\int_0^{2\pi}e^{ixm}(1+e^{-ix}+e^{-2ix})^n\, dx. $$

I am trying to compute $$ \limsup_{n\to\infty}\frac{1}{n}\log\left(\frac{1}{3^n}\sum_{m=0}^{2n}I_n(m)\lambda^m\right). $$

Here, $\lambda$ is the largest root of $\lambda^3-\lambda^2-1$, which is approximately $1.466$.

I did not have an idea how to compute this. Looks very hard to me.

Maybe you do see some trick.

Comparing it with

Is it possible to compute $\limsup_{n\to\infty}\frac{1}{n}\log\left(\frac{c\cdot\lambda^{2k}}{3^n}\sum_{m=0}^{2n}C(n,m)\cdot\lambda^m\right)$?

which is an alternative way of computation for thr same problem, the result should be that $$ \log\left(\frac{1+\lambda+\lambda^2}{3}\right) $$

I have exported this question from the comments of this post: How many possibilities?

John_Doe
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1 Answers1

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Just and observation ...

Considering path integral and Cauchy's integral formula $$I_n(m)=\frac{1}{2\pi}\int_0^{2\pi}e^{ixm}(1+e^{-ix}+e^{-2ix})^ndx=\\ \frac{-i}{2\pi}\int_0^{2\pi}e^{ix(m-1)}(1+e^{-ix}+e^{-2ix})^nd\left(e^{ix}\right)= \frac{-i}{2\pi}\int\limits_{|z|=1}z^{m-1}\left(1+\frac{1}{z}+\frac{1}{z^2}\right)^ndz=\\ \frac{1}{(2n-m)!}\frac{(2n-m)!}{2\pi i}\int\limits_{|z|=1}\frac{\left(z^2+z+1\right)^n}{z^{2n-m+1}}dz= \frac{f^{(2n-m)}(0)}{(2n-m)!}$$ where $f(z)=\left(z^2+z+1\right)^n=\left(z-e^{\frac{2i\pi}{3}}\right)^n\left(z+e^{\frac{i\pi}{3}}\right)^n$. Considering general Leibniz rule $$m=2n \Rightarrow I_n(2n)=\frac{f(0)}{0!}=1$$ $$m=2n-1 \Rightarrow I_n(2n-1)=\frac{f'(0)}{1!}=n$$ $$m=2n-2 \Rightarrow I_n(2n-2)=\frac{f''(0)}{2!}=\frac{n(n+1)}{2!}$$ $$m=2n-3 \Rightarrow I_n(2n-3)=\frac{f^{(3)}(0)}{3!}=\frac{n(n-1)(n+4)}{3!}$$ $$m=2n-4 \Rightarrow I_n(2n-4)=\frac{f^{(4)}(0)}{4!}=\frac{(n-1)n(n^2+7n-6)}{4!}$$ It's getting pretty irregular, unless I am missing something ... work in progress.

rtybase
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