A funny question: Taylor polynomials and series associated with the Lost numbers $4, 8, 15, 16, 23, 42$ The interpolation polynomial for the "Lost" numbers $4, 8, 15, 16, 23, 42$ is
$$
P(x)=60-\frac{612}{5}x+\frac{367}{4}x^{2}-\frac{235}{8}x^{3}+\frac{17}{4}x^{4}-\frac{9}{40}x^{5}.
$$
That is, $P(1)=4$, $P(2)=8$, $P(3)=15$, $P(4)=16$, $P(5)=23$, $P(6)=42$.
I would like to find the most "simple" analytic function $f(x)$ (that is not polynomial and does not contain "large" coefficients, "large" integer numbers or fractions with "large" numerators or denominators) such that $P(x)$ is the Taylor polynomial for $f(x)$ at zero (the Maclaurin polynomial). In other words, the Taylor series for $f(x)$ is
$$
60-\frac{612}{5}x+\frac{367}{4}x^{2}-\frac{235}{8}x^{3}+\frac{17}{4}x^{4}-\frac{9}{40}x^{5}+\ldots.
$$
As I see, the series of coefficients is alternating, and
$$
\frac{612}{5}>\frac{367}{4}>\frac{235}{8}>\frac{17}{4}>\frac{9}{40}.
$$
How to search such a function $f(x)$?

i think that the most "simple" function f(n) (that is not polynomial and does not contain "large" coefficients, "large" integer numbers or fractions with "large" numerators or denominators) is this:
$ f(n) = 4n+1-(15[\frac{n}{6}]+3)[\frac{1}{v(4n)}] $
$v(n) \ $ gives the numer of distinct primes in $  n $
$[n] $ is the $floor$ function.
but the series expansion at $ n=0 $ is:
$ \ \ \begin{cases}
12[\frac{1}{v(4n)}] +1  \ \ \ \  if \ \ n<0\\
1-3[\frac{1}{v(4n)}]  \ \ \ \ otherwise\\
\end{cases}$
