Placing a fixed number of points on a curve to make the finite second derivative constant
In a nutshell

The problem can be solved algebraically by considering the equations as
defining $y_i=f(x_i)$. This gives $n-1$ equations with $n$ unknowns ($y_i$ for $i\in[2,n-1]$, and the values of $c$ and $u$ defined below),
resulting in the abitrary choice of one value. The answers are given
by the formula
$$x_i=f^{-1}(f(a)-(i-1)u+c\frac{(i-2)(i-1)}{2})$$

and the value of $u=f(a)-f(x_2)$ is determined from that of the
constant finite second derivative $c$, or the reverse, with the linear
relation
$$f(b)=f(a)-(n-1)u+c\frac{(n-2)(n-1)}{2}$$

However I wonder if the formula used is really that of a finite second
derivative.

Full proof

With the formula used, your question seems rather tractable, and solvable algebraically.

Like the previous answer, I note $c$ the value of the constant finite second derivative, as defined in the question (though this is further discussed below).

You should see it as a set of equations in $y$, with $y=f(x)$.
If you note $y_i=f(x_i)$, you get a set of $n-2$ equations
 $$y_{i-1}-2y_{i}+y_{i+1}=c\;\; \text{ for }i\in [2,n-1]$$
with $n-1$ unknown : $y_{i} \text{ for }i\in [2,n-1]$ and $c$, given
that you have $y_1=f(a)$ and $y_n=f(b)

This probably means an infinity of solutions, which may means that you
can choose arbitrarily one of the values, for example that of $c$.
I have not worked out the equations.

However, given the strictly monotonically decreasing function $f$ of your example, you have the
additional constaints that:
$$ \forall i\in[2,n-1], \;\;y_i\in(f(a), f(b)) \;\wedge\; y_i&gt;y_{i+1} $$

One way to start resolving it is by defining $u_i=y_i-y_{i+1}$ and $v_i=y_i-y_{i+2}$ for $i\in [1,n-2]$. Then the equations become
$y_i-2(y_i-u_1)+(y_i-v_i)=c$ which simplifies into
$2u_i-v_i=c$, from which we derive $v_i-u_i=u_i-c$.

From the definition of $u_i$, $u_{i+1}$, and $v_i$, we derive $u_{i+1}=v_i-u_i$. Hence $u_{i+1}=u_i-c$. From this recurrence relation we get $u_i=u-(i-1)c$, where $u$ is defined as $u=u_1=y_1-y_2$.

So we now have the simpler recurrence relation: $y_{i+1}=y_i-u_i=y_i-(u-(i-1)c)$, i.e., $y_{i+1}=y_i-u+(i-1)c$, from which we get
\begin{align}
 y_i&amp;=y_1+\sum_{k=1}^{i-1}{u_k} \\
 &amp;=y_1+\sum_{k=1}^{i-1}{-u+(k-1)c} \\
 &amp;=y_1-(i-1)u+c\sum_{k=0}^{i-2}{k} \\
y_i&amp;=y_1-(i-1)u+c\frac{(i-2)(i-1)}{2}
\end{align}
This last formula is valid for $i\in[1,n-1]$, since the recurrence formula is defined for $y_{i+1}$ and the equations used were defined for all values of $i$ up to $n-2$.
We must now consider $y_n$. Since $v_{n-2}-u_{n-2}=u_{n-2}-c$, we get $y_n=y_{n-1}-(u_{n-2}-c)$.
Since $u_{n-2}=u-(n-3)c$, we finally obtain
$\begin{align}
 y_n&amp;=y_1-(n-2)u+c\frac{(n-3)(n-2)}{2}-u+(n-3)c+c \\
 &amp;=y_1-(n-1)u+c\frac{(n-2)(n-1)}{2}
\end{align}
$

Hence our formula for $y_i$ is correct for all $i\in[1,n]$.

Note: a much simpler technique for $y_n$ would be to consider an extra point at $y_{n+1}$ on the curve, so as to have more equations. But all this was already written when I thought of it.

Considering that we have the last constraint $y_n=f(b)$, we can use it
with the equation for $y_n$ to determine $u=y_1-y_2$ if we wish to
choose $c$, or to determine $c$ if we rather choose $u$.

This is easily rephrased into the result given at the beginning of this answer. Then it should be easy to apply it to the example in the
question.
However, I do wonder whether all this is useful, because I am not sure this is the question you intended to
ask.

The reason is that the formula given in the question is not the formula of the finite
second derivative. The proper formula is something like:
$$4\times\frac{f(x_{i-1})-2f(x_i)+f(x_{i+1})}{(x_{i+1}-x_{i-1})^2}$$

See for example higher-order finite differences in wikipedia. Though I am not sure that even this corrected formula is meaningful as, the formula normally given assumes that $x_i= (x_{i-1}+x_{i+1})/2$. But the expression you use is somewhere between the formulae for forward and backward finite differences, and should thus be usable with some meaning.

Thus, if I am correct, you should edit your question to make it consistent, either by
removing the reference to the finite
second derivative, or by using the proper formula for it.