Let $2\leq e\leq r$. I am trying to compute or estimate (from above) $$ \lim_{k\to\infty}\limsup_{n\to\infty}\frac{1}{n}\log\left[(e+r+1)^{2(n+k)-1}-((e+r)\cdot 2r+e+r+(e+r+1)\cdot r^2)^{c_{n,k}}\cdot (e+r+1)^{2(n+k)-1-3c_{n,k}}\right], $$ where $c_{n,k}=\lfloor\frac{2(n+k)-1}{3}\rfloor$.
The background is the following:
For the intervall $[n-k+1,n+k-1]$ around $0$, there are $a_{n,k}=(e+r+1)^{2(n+k)-1}$ possibilities to put elements from the alphabet $\left\{0,1,2,\ldots,e,e+1,\ldots,e+r\right\}$. Now, there are special triples that cannot appear and I divide the interval $[n-k+1,n+k-1]$ into $c_{n,k}$ groups of three. For each group of three consecutive positions I have $(e+r)2r+e+r+(e+r+1)$ possibilities to choose such a forbidden triple. For the remaining $2(n+k)-1-3c_{n,k}$ positions, I have $e+r+1$ possibilities.
Of course, we have $$ \lim_{k\to\infty}\limsup_{n\to\infty}\frac{1}{n}\log(a_{n,k})=2\log(e+r+1) $$ and this is an upper bound in my computations.
But I would like to improve this by finding a smaller upper bound. That's why I am asking for the computation above.