$$\left\lfloor \frac 12 + \sqrt{\frac 14 + \frac{2n}{s-2}}\right\rfloor + n$$
where $\lfloor x\rfloor$ represents the least integer less than or equal to $x$.
Answer : It is well known that
$$(s-2)\frac{n(n-1)}{2}+n$$
is the $n$-th $s$-gonal number.
Now, we will use the Lambek-Moser Theorem to get
$$f(n)=(s-2)\frac{n(n-1)}{2}$$
and hence
$$f^\ast (n) = \left\lfloor \frac 12 + \sqrt{\frac 14 + \frac{2n}{s-2}}\right\rfloor$$
and hence
$$F^\ast (n) = \left\lfloor \frac 12 + \sqrt{\frac 14 + \frac{2n}{s-2}}\right\rfloor + n$$
hence completing the proof.
Remark : A careful induction on $n$ (keeping $s$ fixed) would probably have also done the job.
Acknowledgement : I learnt about the Lambek-Moser Theorem from the book The Irrationals: a Story of the Numbers You Can't Count On by Julian Havil (section Theoretical Matters of chapter Does Irrationality Matter?) which I also reviewed for zbMATH.
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