I have discovered a closed form function for the Look and Say sequence, which relies only on the operators $+,\ -,\ \div,\ \times,$ infinite sums, and finite products. Call this function $L$. It takes in a term as input and outputs the next term. That is, $L(x_n) = x_{n+1}$, where $x_n$ is the $n^{\text{th}}$ term in the Look and Say sequence. Is this useful to any field or any specific problem?
Here is a link to the entire function.
$$ \begin{array} {|c|c|c|c|} \hline \text{Equality} & eq(x,y): & \mathbb{R}\times\mathbb{R}\to\mathbb{B} & 1-|\operatorname{sgn}(x-y)| \\ \hline \text{Digit At} & dat(x,b,i): & \mathbb{R}\times\mathbb{N}\times\mathbb{Z}\to\mathbb{N} & \left\lfloor\frac{|x|}{b^i}\right\rfloor\mod b \\ \hline \begin{aligned} \text{Number} \\ \text{of Digits}\end{aligned} & nd(x,b): & \mathbb{N}\times\mathbb{N}\to\mathbb{N} & \lceil\log_b(x+1)\rceil \\ \hline \text{Reverse} & rev(x,b): & \mathbb{N}\times\mathbb{N}\to\mathbb{N} & \sum_{i=0}^{nd(x,b)-1}dat(x,b,I)\cdot10^{nd(x,b)-i-1} \\ \hline \text{Sum Digits} & sd(x,b): & \mathbb{N}\times\mathbb{N}\to\mathbb{N} & \sum_{i=0}^{nd(x,b)-1}dat(x,b,i)\\ \hline \begin{aligned} \text{Look and Say} \\ \text{Counter} \;\;\;\; \end{aligned} & C_\lambda(x,i): & \mathbb{N}\times\mathbb{N}\to\mathbb{N} & sd(x\bmod10^i,10) \\ \hline \begin{aligned} \text{Unpadded} \\ \text{Difference} \end{aligned} & \delta(x): & \mathbb{N}\to\mathbb{N} & \begin{aligned} \sum_{i=0}^{nd(x,10)-2}10^i\cdot \left[ 1-eq(dat(x,10,i), \\ dat(x,10,i+1))\right] \end{aligned} \\ \hline \begin{aligned} \text{Padded}\;\; \\ \text{Difference} \end{aligned} & \Delta(x): & \mathbb{N}\to\mathbb{N} & 10^{nd(x,10)}+10\cdot\delta(x)+1 \\ \hline \begin{aligned} \text{Leftmost}\\ \text{Index} \;\;\; \end{aligned} & il(x,i): & \mathbb{N}\times\mathbb{Z}\to\mathbb{Z} & nd(x\bmod10^{i+1},10)-1 \\ \hline\begin{aligned} \text{Rightmost}\\ \text{Index} \;\;\;\; \end{aligned} & ir(x,i): & \mathbb{N}\times\mathbb{Z}\to\mathbb{Z} & nd(x,10)-nd\left(rev\left(\left\lfloor\frac{x}{10^{i+1}}\right\rfloor,10\right),10\right) \\ \hline \text{Look and Say} & L(x): & \mathbb{N}\to\mathbb{N} & \begin{aligned}\sum_{i=0}^{nd(\Delta(x),10)-2}\left[dat(\Delta(x),10,i)\cdot \\ \left( \left(ir(\Delta(x),i)-i\right)\cdot10^{1+2\cdot C_\lambda(\Delta(x),i)} \\ + dat(x,10,i)\cdot10^{2\cdot C_\lambda(\Delta(x),i)}\right)\right] \end{aligned} \\ \hline \end{array} $$
Here is a link to desmos, an online calculator, showing the correctness of the function for a few terms. Because of the nature of base $10$, $L(1111111111)=101$ ("ten ones"). Also, due to the limit of computability depth of desmos, large terms are unlikely to work, as desmos rounds intermediate results.