Wanted: Polynomial $P(x)$ with $P(-l(l+1))=1/(2l+1)$, for $l\in \mathbb{N}$

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I'm looking for a polynomial

$P(x)=a_1+a_3 x+ a_5 x^2+\dots$

(numbering of $i$ in $a_i$ is due to the application of this) with sampling points

$P(-l(l+1))=\frac 1{2l+1}$, for $l=1,2,3,\dots$

When cuting off $i$ in $P(x)$ and the number of sampling points $l$ at some $N$, the $a_i$ can be found via the corresponding $NxN$ Vandermonde matrix. For instance, for $N=2$ one finds $a_1=\frac 2 5$ and $a_3=\frac 1{30}$. For $N\rightarrow \infty$, the $a_i$ seem to converg (see plot).

coefficients of $P(x)$ over $N$

For $N=1,2,3,4..$ one finds $a_1=\frac 1 3, \frac 2 5, \frac 3 7, \frac 4 9, ...$ which obviously converges to $\frac 1 2$. However, for the higher $a_i$ I cannot see any pattern. Is there a way to calculate the $a_i$?

[BTW, the sampling points are all on the curve $f(x)=\frac 1 {\sqrt{1-4x}}$.]

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There is no finite polynomial $P$ that gives you all the points you want, since the function $f(x)$ the points are sampled from has horizontal asymptote as $x \rightarrow - \infty$. (Polynomials do not have asymptotes).

It is also not possible to find a Taylor series for $f$ that converges on the entire real line, since $f$ is not defined at $x = 1/4$, and hence the Taylor series about any point $a < 1/4$ will have radius of convergence at most $\lvert a - 1/4 \rvert$. So the best one can do is indeed to approximate to only a finite (but arbitrary) number of sampling points for any given choice of coefficients $a_i$.