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Returns \$(a_0,a_1,...,a_k)\$ with some possible trailing zeros.

v=>v.map((_,i)=>(g=(i,m=v.map((n,y)=>v.map((_,x)=>x==i?n:y**x)))=>+m||m.reduce((s,[v],i)=>v*g(0,m.map(([,...r])=>r).filter(_=>i--))-s,0))(i)/g())

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(removed the penultimate test case, which requires more precision than IEEE-754 provides)

How?

We use Cramer's rule to solve a system of linear equations based on a square Vandermonde matrix:

1. Given an input vector of length \$n\$, we build a Vandermonde matrix \$V_n\$ of size \$n\times n\$ with coefficients \$\alpha_i=i,0\le i <n\$:

   $$Vn=\begin{pmatrix} 1&0&0&...&0\\ 1&1&1&...&1\\ 1&2&4&...&2^{n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&n-1&(n-1)^2&...&(n-1)^{n-1} \end{pmatrix}$$

2. Using Cramer's rule, the coefficient \$a_i\$ of the polynomial is computed by taking the determinant of the matrix obtained by replacing the \$i\$-th column of \$V_n\$ with the input vector, and dividing by the determinant of \$V_n\$.

Example for \$(4,2,12,52,140)\$

The constant coefficient \$a_0\$ is given by:

$$a_0=\begin{vmatrix} \color{blue}4&0&0&0&0\\ \color{blue}2&1&1&1&1\\ \color{blue}{12}&2&4&8&16\\ \color{blue}{52}&3&9&27&81\\ \color{blue}{140}&4&16&64&256 \end{vmatrix}/|V_5|=\frac{1152}{288}=4$$

The coefficient \$a_1\$ is given by:

$$a_1=\begin{vmatrix} 1&\color{blue}4&0&0&0\\ 1&\color{blue}2&1&1&1\\ 1&\color{blue}{12}&4&8&16\\ 1&\color{blue}{52}&9&27&81\\ 1&\color{blue}{140}&16&64&256 \end{vmatrix}/|V_5|=\frac{-576}{288}=-2$$

And so on.