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Let $\lambda$ be the largest (in absolute value) root of $x^3-x^2-1$, which is $\lambda\approx 1,466$, and $k\in\mathbb{N}_0=\{0,1,2,\ldots\}$ fixed.

Consider vectors $x$ of length $n$ the form $$ x=(x_0,x_1,\ldots,x_{n-1}),\qquad x_i\in\{0,1,2\}\text{ for }i=0,1,\ldots,n-1.\quad (*) $$ By $j(x)$ denote the sum of the components, i.e. $$ j(x) = \sum_{i=0}^{n-1}x_i. $$ We then have $$ 0\leq j(x)\leq 2n. $$ By $C(n,m)$ denote the coefficient of $x^m$ in $(1+x+x^2)^n$ which is the multinomial coefficient $$ C(n,m)=\binom{n}{m_1,m_2,m_3}\text{ with }\sum_{i=1}^3m_i=n\text{ and }m_2+2m_3=m. $$

($C(n,m)$ is the number of possible vectors $x$ of length $n$ of type $(*)$ such that $j(x)=m$, where $m_1$ is the number of zeros, $m_1$ is the number of ones and $m_3$ is the number of twos.)

I would like to know whether it is possible to determine $$ \limsup_{n\to\infty}\frac{1}{n}\log\left(\frac{c\cdot\lambda^{2k}}{3^n}\sum_{m=0}^{2n}C(n,m)\cdot\lambda^m\right), $$ where $c$ is some constant.

Unfortunately, I have no real idea how to compute this.

For an upper estimation, my idea would be to use that $$\sum_{m=0}^{2n}C(n,m)\leq\sum_{m_1+m_2+m_3=n}\binom{n}{m_1,m_2,m_3}=3^n $$ and $$ \max_{0\leq m\leq 2n}\lambda^m=\lambda^{2n}. $$ Thus, $$ \frac{c\lambda^{2k}}{3^n}\sum_{m=0}^{2n}C(n,m)\lambda^m\leq\frac{c\lambda^{2k}}{3^n}\lambda^{2n}\sum_{m=0}^{2n}C(n,m)\leq c\lambda^{2k+2n}. $$

Taking logarithm, dividing by $n$ and taking the limit superior, I therefore get $$ \limsup_{n\to\infty}\frac{1}{n}\log\left(\frac{c\cdot\lambda^{2k}}{3^n}\sum_{m=0}^{2n}C(n,m)\cdot\lambda^m\right)\leq 2\log\lambda. $$

In case this might be correct: Maybe it is also possible to get $2\log\lambda$ as a lower bound?

By the way: I exported this question from the comments of this post.

John_Doe
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1 Answers1

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Since $C(n, m)$ is the coefficient of $x^m$ in $(1+x+x^2)^n$, $$\sum_{m=0}^{2n}C(n,m)\lambda^m=(1+\lambda+\lambda^2)^n.$$ It is therefore not difficult to see that your sequence actually converges, with limit equal to $\log\left(\frac{1+\lambda+\lambda^2}3\right)$.

Jason
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  • Concretely, I have problems to see that $(1+\lambda+\lambda^2)^n=\sum_{m_1+m_2+m_3=n}\binom{n}{m_1,m_2,m_3}\lambda^{m_2}\cdot\lambda^{2m_3}=\sum_{m=0}^{2n}C(n,m)\lambda^m$. The first equality is the multinomial theorem. But the second equality is hard for me to understand. – John_Doe Nov 24 '17 at 21:15
  • Isn't this by definition...? – Jason Nov 24 '17 at 21:23
  • Also, I've added a small edit to my answer as originally I had left out the logarithm from the limit. Apologies. – Jason Nov 24 '17 at 21:24
  • I think not. Because $C(n,m)$ is the multinomial coefficient $\binom{n}{m_1,m_2,m_3}$ with $m_1+m_2+m_3=n$ and $m_2+2m_3=m$. So the sum on the left hand side has much more summands than the sum on the right hand side because of the extra restriction that $m_2+2m_3=m$. – John_Doe Nov 24 '17 at 21:26
  • "By $C(n, m)$ denote the coefficient of $x^m$ in $(1+x+x^2)^n$". This is what I'm taking as your definition. Everything else is unnecessary for your question. – Jason Nov 24 '17 at 22:12