The local maxima

$\begin{aligned} f(x) & = \left(2+x+ax^2\right) \ln (x+1) - 2x \\ & = \left(2+x+ax^2\right) \left(x-\frac {x^2}2 + \frac {x^3}3-\frac {x^4}4 + \cdots \right)- 2x \\ & = \left(a-\frac 12 + \frac 23\right) x^3 - \left(\frac a2-\frac 13 + \frac 12\right) x^4 + O(x^5) \\ & = \left(a+\frac 16 \right) x^3 - \left(\frac a2 + \frac 16\right) x^4 + O(x^5) \end{aligned}$

For $f(0) = 0$ to be a local maximum, $f(\pm \Delta x) < 0$ , where $\Delta x$ is a small increment of $x$ . That is $f(\pm \Delta x) = \pm \left(a+\dfrac 16 \right) \Delta^3 x - \left(\dfrac a2 + \dfrac 16\right) \Delta^4 x + O(\Delta^5 x) < 0$ , which is always true when $a = - \dfrac 16 = A$ . Therefore $\lfloor 10000A \rfloor = \boxed {-1667}$ .