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
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?
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.