Computing $\limsup_{n\to\infty}\frac{1}{n}\log\left(\frac{1}{3^n}\sum_{m=0}^{2n}I_n(m)\lambda^m\right)$ 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?
 A: 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.
