Consider the sequences of numbers $\left\{0, 1, 2\right\}$ with length $n$. There are $3^n$ such sequences. I define each sequence like a function. If a function consists of {0,1,2} elements of the length $«n»$, let's consider this function $\phi (n).$
Because, we can deduce all sequences for only finite number $n$, If we accept any infinity sequence equal to a specific function, problematic points occur. Because, we have infinitely number of function, which is we can not deduce all functions.
Definition: In mathematics, a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations. It may contain constants, variables, certain "well-known" operations (e.g., + − × ÷), and functions (e.g., nth root, exponent, logarithm, trigonometric functions, and inverse hyperbolic functions), but usually no limit.
For infinity sequence , which consist of elements $\left\{0,1,2\right\}$, if we have a specific $n-$th term any formula, we have a "specific mathematical function" (algebraic closed form expression or non-elementary function), which is we are looking for.
I can choose an infinitely number of functions that are Closed-Form Expression in term of a $"n".$
Examples: let $n\in\mathbb{Z^{+}} \bigcup \left\{0\right\}$
$$\phi(n)=n+2-3\lfloor \frac{n+2}{3}\rfloor=\left\{0,1,2,0,1,2\cdots \right\}$$
$$\phi(n)=n+1-3\lfloor \frac{n+1}{3}\rfloor=\left\{ 2,0,1,2,0,1\cdots \right\}$$
$$\phi(n)=n-3\lfloor \frac{n}{3}\rfloor=\left\{ 1,2,0,1,2,0\cdots\right\}$$
$$\phi(n)=n^n-3\lfloor \frac{n^n}{3}\rfloor=\left\{1,1,0,1,2,0,1,1,0,1,2,0\cdots \right\}$$
$$\phi(n)=n^n+n-3\lfloor \frac{n^n+n}{3}\rfloor=\left\{2,0,0,2,1,0,2,0,0,2,1,0\cdots \right\}$$
$$\phi(n)=\lfloor 10^n \pi \rfloor - 3 {\lfloor \frac{ \lfloor 10^n \pi \rfloor }{3}}\rfloor$$
$$\phi(n)=\lfloor 10^n e \rfloor - 3 {\lfloor \frac{ \lfloor 10^n e \rfloor }{3}}\rfloor$$
$$\phi(n)=\lfloor 10^n \sqrt2 \rfloor - 3 {\lfloor \frac{ \lfloor 10^n \sqrt2 \rfloor }{3}}\rfloor$$
$$\phi(n)=\lfloor 10^n \log \pi \rfloor - 3 {\lfloor \frac{ \lfloor 10^n \log \pi \rfloor }{3}}\rfloor$$
$$\cdots \cdots \cdots \cdots \cdots$$
For these periodic and non-periodic sequences there are exist $n-$th term "closed-form expression."
Then, we can define an infinitely number of "specific mathematical functions", (non-elementary, non-algebraic) which is non-periodic.
Example:
$$ \phi(n)=\lfloor 10^n \displaystyle\int_0^\infty e^{-x^n}dx \rfloor - 3 {\lfloor \frac{ \lfloor 10^n \displaystyle\int_0^\infty e^{-x^n}dx \rfloor }{3}}\rfloor$$
$$\cdots \cdots \cdots \cdots \cdots$$
Claims: (only for infinitely number sequences)
A) There exist infinitely number of sequences that, consist of elements $\left\{0,1,2\right\}$, which is can not express by the any "closed-form expression" or any "specific mathematical function".
B) There exist infinitely number of sequences that, consist of elements $\left\{0,1,2\right\}$, which is can express by the any closed-form expression.
C) There exist infinitely number of sequences that, consist of elements $\left\{0,1,2\right\}$, which is can express by the any "specific mathematical function" (non-elementary, non-algebraic).
Which of these claims are true? I'm looking for a proof that confirms or denies the claims.
Thank you!
Others have already addressed the claims. Here I discuss a bit about finite vs infinite sequences, as you seem to be a bit confused.
For any fixed $n$, there are $3^n$ sequences of length $n$ with symbols from $S = \{0,1,2\}.$
For any fixed $n$, there are $3^n$ functions from $\{1,2,...,n\}$ to $S$. Note that the function domain (possible input values) is only the set $\{1,2,...,n\}$, not all $\mathbb{N}$ (all positive integers). E.g. if $n=4$ then $f(7)$ is undefined, since the input value is invalid for this $f$. These $3^n$ functions are in bijective mapping with the $3^n$ sequences: the sequence $(f(1), f(2), ..., f(n))$ fully specifies $f$.
If you union together all the sets of all finite sequences (of any length), the resulting set is of course infinite. It is countably infinite.
Similarly, if you union together all the sets of all functions from a finite domain $\{1,2,...,n\}$ to $S$, the resulting set is again countably infinite, and again in bijective mapping with the set of finite sequences.
Nothing above involve infinite sequences. Here they come now:
Any infinite sequence is $(s_1, s_2, s_3, ...)$ where each $s_i \in S$. The set of all infinite sequences is of course infinite, but it is uncountably infinite.
The set of functions $f: \mathbb{N} \rightarrow S$ is in bijective mapping with the set of infinite sequences, because again $(f(1), f(2), ...)$ fully specifies the function. This set of function is again uncountably infinite.
So, the status of your claims depends on whether you mean finite or infinite sequences, or equivalently, functions with finite domains (of the form $\{1,2,...,n\}$ for some $n$) or functions with domain $\mathbb{N}$.
If you include infinite sequences, then the answers are (A) Yes, (B) Yes, (C) Yes -- see the answer by Ross Milikan, for example.
If you mean only finite sequences (remember: there are still countably infinite such finite sequences), then, IMHO every such finite sequence has a closed form expression because IMHO the usual written form of the finite sequence itself is a closed form expression -- it is made from these five symbols: $\color{red}{0,1,2,(}$ and $\color{red}{)}$. Thus the answers would be (A) No, (B) Yes, (C) Yes.