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Suppose we have a vector $x=(x_0,x_1,x_2,\ldots,x_{n-1})$ with $x_i\in\{0,1,2\}$ for all $i=0,1,...,n-1$.

Define $$ x'=\sum_{i=0}^{n-1}x_i, $$ then $$ 0\leq x'\leq n\cdot 2.. $$

I would like to know how many possibilities of vectors there are, for fixed $0\leq m\leq n\cdot 2$, to have $$ x'=m. $$

I think, for example, for $x'=0$ and $x'=n\cdot 2$, we only have one possibility, namely $(0,...,0)$ and $(2,...,2)$. But maybe there is some general formula.

I don’t know whether there is some Trick.

John_Doe
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2 Answers2

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The expression providing the result you need is $$ I_n(m)=\sum_{x_0=0}^2\cdots \sum_{x_{n-1}=0}^2 \delta_{\sum_i x_i,m}\ , $$ where $\delta_{a,b}=1$ if $a=b$ and $=0$ otherwise. Basically this is a counting device: you get a '$+1$' for every configuration $\mathbf{x}$ such that the sum of the components is equal to $m$. Using the integral representation $$ \delta_{a,b}=\int_0^{2\pi}\frac{d\xi}{2\pi}e^{\mathrm{i}\xi (a-b)} $$ you get $$ I_n(m)=\int_0^{2\pi}\frac{d\xi}{2\pi}e^{\mathrm{i}\xi m}\left(\sum_{x=0}^2 e^{-\mathrm{i}\xi x}\right)^n=\int_0^{2\pi}\frac{d\xi}{2\pi}e^{\mathrm{i}\xi m}(1+e^{-\mathrm{i}\xi}+e^{-2\mathrm{i}\xi})^n\ . $$ Expanding with the multinomial theorem, you can get a more explicit expression. Mathematica can handle this integral for specific values of $m$ and $n$. For example, for $m=38$ and $n=29$ you get $I_n(m)=774369291150$.

Pierpaolo Vivo
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    Nice solution, though I personally feel that this is a bit of an overkill; there are far simpler methods that would give a solution that could be calculated in WolframAlpha, for example. $\$E.g. we could observe that, the 'answer' the OP seeks is the coefficient of $x^m$ in $(1 + x + x^2)^n$. Plugging into WolframAlpha then gives the solution as it does in your answer. (I haven't posted this as an answer, as I don't feel it is particularly complete or satsfactory enough to be a full answer.) – John Don Nov 23 '17 at 21:37
  • @Pierpaolo Is it possible to compute $\limsup_{n\to\infty}\frac{1}{n}\log (\frac{1}{3^n}\sum_{m=0}^{2n}I_n(m)\cdot c_m)$, where $c_m$ is some constant depending on $m$? – John_Doe Nov 24 '17 at 07:53
  • @JohnDon I have the same question to your answer. By $a(n,m)$ denote the coefficient of $x^m$ in $(1+x+x^2)^n$. Is it then possible to determine $\limsup_{n\to\infty}\frac{1}{n}\log(\frac{1}{3^n}\sum_{m=0}^{2n}a(n,m)\cdot c_m)$? – John_Doe Nov 24 '17 at 08:24
  • @JohnDon By the way: If I see it right, the coefficient of $x^m$ in $(1+x+x^2)^n$ is the multinomial coefficient $\binom{n}{n_1,n_2,n_3}$ with $n_1+n_2+n_3=n$ and $n_2+2n_3=m$. Here, $n_1$ is the number of zeros, $n_2$ the number of ones and $n_3$ the number of twos. – John_Doe Nov 24 '17 at 08:40
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    @John_Doe That observation is correct; with regards to your follow-up question, it is somewhat hard to say with no knowledge of what $c_m$ is. You say it is a constant (depending on $m$), but as it is inside a sum in which $m$ varies, this is like saying that $a(n,m)$ is just a constant that depends on $n$ and $m$. – John Don Nov 24 '17 at 10:15
  • @John_Doe Sorry, I was incorrect. It has to be $c_{2k+m}$ for some fixed $k$ and I have $c\lambda^{2k+m}\leq c_{2k+m}\leq d\lambda^{2k+m}$. Here $c$ and $d$ are constants and $\lambda$ is the largest (in absolute value) eigenvalue of the matrix $\begin{pmatrix}1 & 1 & 0\0 & 0 & 1\1 & 0 & 0\end{pmatrix}$. I think its not that important where this comes from. My question is if we can compute $\limsup_{n\to\infty}\frac{1}{n}\log\left(\frac{c\lambda^{2k}}{3^n}\sum_{m=0}^{2n}a(n,m)\lambda^m\right)$, c constant. – John_Doe Nov 24 '17 at 13:25
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I don't know why John Don didn't post this as an answer. This is a combinatorial problem, covered by generating functions (for example this one or this one and a really good reading here). So, we want $$\sum\limits_{i=0}^{n-1} x_i =m$$ $$0\leq x_i\leq 2, \forall x_i=0..n-1$$ The generating function is $$(1+x+x^2)(1+x+x^2)...(1+x+x^2)=(1+x+x^2)^n$$ and the coefficient near $x^{m}$ term is the answer. Considering multinomial theorem: $$(1+x+x^2)^n=\sum\limits_{k_1+k_2+k_3=n}\binom{n}{k_1,k_2,k_3}1^{k_1}\cdot x^{k_2}\cdot x^{2k_3}=\\ \sum\limits_{k_1+k_2+k_3=n}\frac{n!}{k_1!k_2!k_3!} x^{k_2+2k_3}$$ In other words, the answer is $$\sum\limits_{k_1+k_2+k_3=n\\k_2+2k_3=m}\frac{n!}{k_1!k_2!k_3!}$$

rtybase
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    Yes ... true, thank you for pointing out! Fixing ... – rtybase Nov 26 '17 at 16:26
  • Denoting this sum by $C(n,m)$, i.e. $$C(n,m):=\sum_{k_1+k_2+k_3=n\k_2+2k_3=m}\frac{n!}{k_1!k_2!k_3!}$$, I think that $$\frac{1}{3^n}\sum_{m=0}^{2n}C(n,m)x^{2n}=x^{2n}$$ since $\sum_{m=0}^{2n}C(n,m)=3^n$. Am I right? – John_Doe Nov 26 '17 at 18:48
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    @John_Doe I looks so, from $(1+x+x^2)^n=\sum\limits_{k_1+k_2+k_3=n}\binom{n}{k_1,k_2,k_3}1^{k_1}\cdot x^{k_2}\cdot x^{2k_3}$ just by $x=1$. – rtybase Nov 26 '17 at 19:06