I am looking for a solution to the following recurrence relation:
$$ D(x,1) = x $$
$$ D(x,n) = \min_{k=1,\ldots,n-1}{D\left({{x(k-a)} \over n},k\right)} \ \ \ \ [n>1] $$
Where $a$ is a constant, $0 \leq a \leq 1$. Also, assume $x \geq 0$.
This formula can be interpreted as describing a process of dividing a value of x to n people: if there is a single person ($n=1$), then he gets all of the value x, and if there are more people, they divide x in a proportional way, with some loss ($a$) in the division process.
For $a=0$ (no loss), it is easy to prove by induction that:
$$ D(x,n) = {x \over n}$$
For $a=1$, it is also easy to see that:
$$ D(x,n) = 0$$
But I have no idea how to solve for general $a$.
Additionally, it is interesting that small modifications to the formula lead to entirely different results. For example, starting the iteration from $k=2$ instead of $k=1$:
$$ D(x,2) = x/2 $$
$$ D(x,n) = \min_{k=2,\ldots,n-1}{D\left({{x(k-a)} \over n},k\right)} \ \ \ \ [n>2] $$
For $a=0$ we get the same solution, but for $a=1$ the solution is:
$$ D(x,n) = {x \over n(n-1)}$$
Again I have no idea how to solve for general $a$.
I created a spreadsheet with some examples, but could not find the pattern.
Is there a systematic way (without guessing) to arrive at a solution?
For $x>0$ the answer is $$D(x,n)=\frac{x(1-a)(2-a)\ldots (n-1-a)}{n!}$$
Proof: induction on $n$. Indeed,
$$D(x,n)=\min_{k=1,\ldots,n-1}\frac{x(k-a)(1-a)\ldots (k-1-a)}{k!n}$$ $$=\frac{x}{n}\min_{k=1,\ldots,n-1}\frac{(1-a)\ldots (k-a)}{k!}$$ andf it remains to prove
$$\frac{(1-a)\ldots (k-a)}{k!}>\frac{(1-a)\ldots (k+1-a)}{(k+1)!}$$ But this is evident.
For $x>0$ similarly.